group06.ipynb

Kernel: SageMath (stable)

[r

Abstract Algebra: An Interactive Approach, 2e

©2015 This notebook is provided with the textbook, "Abstract Algebra: An Interactive Approach, 2nd Ed." by William Paulsen. Users of this notebook are encouraged to buy the textbook.

Chapter 6
Building Larger Groups from Smaller Groups

Initialization: This cell MUST be evaluated first:

%display latex
try:
    load('absalgtext.sage')
except IOError:
    load('/media/sf_sage/absalgtext.sage')

The Direct Product
The Fundemental Theorem of Finite Abelian Groups
Automorphisms
Semi-Direct Products
Sage Interactive Problems

The Direct Product

In this chapter, we will focus on forming larger groups using the smaller groups previously studied as building blocks. In this way, many important groups can be constructed. In fact, we will be able to construct all finite abelian groups using only the groups that we have already studied.

The first step in building larger groups is to understand how two groups can be combined. Let H and K be two groups, and let G be the set of all ordered pairs (h, k) such that h ∈ H and k ∈ K.

We need to describe how to multiply two ordered pairs together. If (h₁, k₁) and (h₂, k₂) are two ordered pairs in G, the product could be defined in the natural way:

(h₁, k₁)·(h₂, k₂) = (h₁·h₂, k₁·k₂).

It is clear that the associate law would hold. If e₁ is the identity element for H and e₂ the identity element for K, then (e₁, e₂) would be the identity element of G.

Finally we can define the inverse of an ordered pair (h, k) to be (h^-1, k^-1). This is so

(h, k)·(h^-1, k^-1) = (h^-1, k^-1)·(h, k) = (e₁, e₂).

Since the four properties of groups hold for the set G of ordered pairs (h, k), G forms a group.

DEFINITION 6.1
Given two groups H and K, the direct product of H and K, denoted H × K, is the group of ordered pairs (h, k) such that h ∈ H and k ∈ K, with multiplication defined by

(h₁, k₁)·(h₂, k₂) = (h₁·h₂, k₁·k₂).

EXAMPLE:
Let H = Z₄, and K = Z₂. Consider the direct product G = Z₄ × Z₂. Since Z₄ consists of the elements {0,1,2,3} and Z₂ consists of {0, 1}, the set of all ordered pairs (h, k) with h in Z₄ and k in Z₂ is

{(0,0), (0,1), (1,0), (1,1), (2,0), (2,1), (3,0), (3,1)}.

Thus, we will have a group of order 8. Multiplication is performed component-wise in the two groups. In order to define this group in Sage, we first define the groups Z₄ and Z₂.

Z4 = ZGroup(4)
Z2 = ZGroup(2)
G = DirectProduct(Z4, Z2); G

We can now look at the Cayley table.

MultTable(G)

Recall that in Chapter 4, we found that there were only 5 non-isomorphic groups of order 8: Z₈, Z₁₅^*, Z₂₄^*, D₄, and Q. Hence, this group must be isomorphic to one of these groups. Notice that this group is commutative, so it is neither Q nor D₄. There is an element of order 4, namely (1,0), but there are no elements of order 8. Hence this group cannot be either Z₈ nor Z₂₄^*. Since Z₁₅^* is the only remaining possibility, this group must be isomorphic to Z₁₅^*.

EXPERIMENT:
Compare the group Z₄ × Z₂ given by the table

MultTable(G)

with the table for Z₁₅^*, generated by the commands

M = ZStar(15); M

MultTable([M[0], M[1], M[2], M[3], M[4], M[5], M[6], M[7]])

Can you, by rearranging the elements in the last MultTable command, get the color pattern of this table to resemble the pattern for Z₄ × Z₂?

Hint: Which element of Z₁₅^* corresponds to (2,0)? What are the other elements of order 2?

In this example, we saw that the direct product of two commutative groups turned out to be commutative. This is, in fact, what generally happens.

PROPOSITION 6.1
Let H and K be two groups. Then H × K is commutative if, and only if, both H and K are commutative.

Click here for the proof.

It is easy to find the number of elements in a direct product. If H has order n, and K has order m, then the number of ordered pairs (h, k) would be n·m.

We can generalize the direct product to a set of more than two groups. Let

G₁, G₂, G₃, …, G_n

be a collection of n groups. Then we define

G₁ × G₂ × G₃ × ··· × G_n

to be the set of ordered n-tuples (g₁, g₂, g₃, …, g_n) with multiplication defined by

(g₁, g₂, …, g_n)·(h₁, h₂, …, h_n) = (g₁·h₁, g₂·h₂, …, g_n·h_n).

The direct product of more than two groups can also be defined by taking the direct product of direct products. That is, given three groups G, H, and K, we can define

(G × H) × K and G × (H × K)

using only the definition for the direct product of two groups. But it is trivial to see that the mappings

ƒ: (G × H) × K → G × H × K
ϕ : G × (H × K) → G × H × K

given by

ƒ(((g, h), k)) = (g, h, k) and  ϕ((g, (h, k))) = (g, h, k)

are both surjective isomorphisms. Thus,

(G × H) × K ≈ G × H × K ≈ G × (H × K).

It also should be noted that we have the natural mapping

ϕ : H × K → K × H

given by

ϕ((h, k)) = (k, h).

This is clearly one-to one and onto, and is a homomorphism. Thus,

H × K ≈ K × H.

This shows that the direct product between groups is a commutative operation, as well as associative. This suggests that some groups may be able to be expressed as a direct product of two or more smaller groups. If this is the case, then the order in which the smaller groups are combined would be irrelevant.

DEFINITION 6.2
Let G be a group. We say that G has a decomposition if

G ≈ H × K,

where neither H nor K is the trivial group.

For example, we saw above that Z₁₅^* had a decomposition, namely Z₄ × Z₂.

EXPERIMENT:
Does the group S₃ have a decomposition into a direct product? Why or why not?

We would like to find a way of testing whether a general group can be decomposed into smaller groups. In the case of S₃, we could use the fact that all smaller groups are abelian, along with Proposition 6.1 to show that S₃ cannot have a decomposition. But for other groups, the problem is more difficult. The following theorem gives us a way to determine whether a given group has a decomposition.

THEOREM 6.1:The Direct Product Theorem
Let G be a group with identity e, and let H and K be two subgroups of G. Suppose the following two statements are true:

1)  H ∩ K = {e}.

2)  h·k = k·h for all h ∈ H and k ∈ K.

Then H · K ≈ H × K.

Click here for the proof.

We can use this theorem to find the direct product of two groups in Sage.

EXAMPLE:
Generate the direct product S₃ × Z₈^* in Sage.

We first must define the two groups in Sage using the same identity and different letters for the generators. This way, the identity will be the only element in common between the two groups.

Let us define S₃ in Sage:

InitGroup("e")
AddGroupVar("a","b")
Define(a^3, e)
Define(b^2, e)
Define(b*a, a^2*b)
H = Group(a, b); H

Now let us define Z₈^*, using c and d for the two generators.

AddGroupVar("c","d")
Define(c^2, e)
Define(d^2, e)
Define(d*c, c*d)
H = Group(a, b)
K = Group(c, d); K

Of course we did not use the InitGroup command before defining the second group because we want to define a single group involving all four generators. By using e as the identity for both groups, we have that the intersection of H and K is

Intersection(H, K)

so the first condition of the direct product theorem is satisfied.

In order for the second condition of the direct product theorem to be satisfied, every element of H must commute with every element of K. This will be true as long as all of the generators of H commute with all of the generators of K. Since there are 2 generators of H and 2 of K, we can tell Sage that the generators commute using 2·2 = 4 definitions:

Define(c*a, a*c)
Define(c*b, b*c)
Define(d*a, a*d)
Define(d*b, b*d)
H = Group(a, b); H

K = Group(c, d); K

According to the direct product theorem H·K is now the same as H × K. Here, then, is the direct product:

H * K

Alternatively, we could find the smallest group that contains all of the generators:

G = Group(a, b, c, d); G

We can check that the order of this direct product is

len(G)

as expected, since S₃ is of order 6, and Z₈^* is of order 4. The multiplication table of this new group is given as

MultTable(G)

We saw two other groups of order 24, namely S₄ and the octahedronal group. Yet these turned out to be isomorphic. Is this new group of order 24 possibly the same group?

Recall that the main tool we used for finding the isomorphism between the octahedronal group and S₄ was determining the orders of the elements in the groups. We defined S₄ to be:

S4 = Group(P(2,1), P(2,3,1), P(2,3,4,1)); S4

To determine how many elements are of order 2, we can use the RootCount function

RootCount(S4, 2)

and see that there are 10 solutions to x² = e, with one being the identity, so there are 9 elements of order 2. Now let us try this with G = S₃ × Z₈^*:

RootCount(G, 2)

So there are 15 elements of G which are of order 2, as opposed to 9 elements in S₄. Hence, these two groups are not isomorphic.

This example shows it would be helpful to know the number of elements of order n of the direct product H×K. Then we could determine whether or not two groups expressed as direct products were isomorphic.

Recall in §2.3 we denoted the number of solutions to xⁿ = e by R_n(G). In particular, if G is cyclic, R_n(G) = gcd(|G|, n). It is also rather easy to calculate R_n(G) for direct products.

PROPOSITION 6.2
Let H and K be finite groups, and let n be a positive integer. Then

R_n(H × K) = R_n(H)·R_n(K).

Click here for the proof.

Let us use this proposition on the example S₃ × Z₈^*. The number of elements of this group solve x² = e is the product of the number of elements solving the equation in S₃ and the number of such elements in Z₈^*. Note that four of the six elements of S₃ satisfy x² = e, while all four elements of Z₈^* satisfy the equation. Thus, there are 4·4 = 16 elements of the product S₃ × Z₈^* whose square is the identity. One of these is the identity element, but the other 15 elements are of order 2. Thus, we can find the number of elements of a direct product of given order without having to use Sage.

EXPERIMENT:
Use Proposition 6.2 to predict the number of elements of S₃ × Z₈^* which are of order 3 and of order 6. Then use Sage to verify your predictions.

We will end this section by showing another way to test whether a group is a direct product of two of its subgroups.

COROLLARY 6.1
Let G be a group with identity e, and let H and K be two normal subgroups of G. Then

if H ∩ K = {e}, then H·K ≈ H × K.

Click here for the proof.

This corollary is sometimes more useful than the direct product theorem, even though for abelian groups the two are equivalent, since all subgroups of abelian groups are normal. In the next section we will continue to study the decomposition of abelian groups, and find that all finite abelian groups can be decomposed uniquely into a certain form.

Fundamental Theorem of Finite Abelian Groups

In the last section we promised that we will be able to construct any finite abelian group using as building blocks the groups that we have already learned. In this section, we will show how that can be done by considering the direct products of the cyclic groups Z_n. We will even be able to find the number of abelian groups of a given order. Let us begin with a simple example.

EXAMPLE:
Consider the group Z₆ = {0, 1, 2, 3, 4, 5}. Can we express this as the direct product of two smaller groups?

Note that for abelian groups, the second condition of the direct product theorem is automatically satisfied, so we merely have to find two subgroups of Z₆ whose intersection is just the identity {0}, and whose product gives us the whole group. With a group this small, it is easy to find all of the subgroups:

{0},  {0, 3},  {0, 2, 4},  and  {0, 1, 2, 3, 4, 5}.

Thus, the only two candidates are the subgroups

H = {0, 3} ≈ Z₂ and K = {0, 2, 4} ≈ Z₃.

Since the intersection of these two subgroups is the identity, we have by the direct product theorem that H·K ≈ H × K. Since H × K is of order 6, we have that H · K is of order 6, and hence is Z₆. Therefore,

Z₆ ≈ Z₂ × Z₃.

Let us check this using Sage: Z₂ and Z₃ are easy to define.

Z2 = ZGroup(2); Z2

Z3 = ZGroup(3); Z3

We can form the direct product

G = DirectProduct(Z2, Z3); G

Here is the multiplication table:

MultTable(G)

We want to see if this group is isomorphic to Z₆. Thus, we want to find an element of order 6. By computing

RootCount(G, 2)

RootCount(G, 3)

we see that there is only one element of order 2, and 2 elements of order 3, so there must be two elements of order 6. In fact, it is not hard to see that (1, 1) and (1, 2) are these elements. From this information we can create an isomorphism. We let F be the homomorphism between Z₆ and Z₂ × Z₃.

Z6 = ZGroup(6); Z6

F = Homomorph(Z6, G)

HomoDef(F, Z6[1], G[5])

FinishHomo(F)

So this in a homomorphism. The image is

Image(F, Z6)

so F is surjective. The kernel is

Kernel(F)

So F is also injective. Therefore, we have found the isomorphism from Z₆ to Z₂ × Z₃.

GraphHomo(F)

EXPERIMENT:
Using the same technique as above, see if you can find a decomposition of Z₁₀ into a direct product of smaller groups. Define the homomorphism between Z₁₀ and the direct product, and show that the homomorphism is surjective and injective.

By observing the patterns in the decomposition of Z₆ and Z₁₀, let us see if we can find a decomposition of any commutative group.

Observe the groups H = {0, 3} and K = {0, 2, 4} in the example of the decomposition of Z₆. Notice that H consists of all of the elements of Z₆ such that [removed]h² = 0. Also, K consists of all the elements such that k³ = 0. The fact that these two subgroups have only the identity in common is not a coincidence, as the next lemma shows:

LEMMA 6.1
Let G be an abelian group of order m n, where m and n are coprime. Then

H = {h ∈ G | h^m = e}

and

K = {k ∈ G | kⁿ = e}

are both subgroups of G, and G ≈ H × K.

Click here for the proof.

This lemma tells us that if an abelian group has an order that is a product of two coprime numbers, this group can be written as a direct product of two groups. Unfortunately, the lemma does not tell us that neither H nor K are trivial groups. As a result, we still have to ask whether this is a decomposition.

Here is an example to illustrate the possible problem that could occur. Suppose we know G is an abelian group of order 24. Since 24 = 8·3, we could let m = 8, and n = 3 in Lemma 6.1. Then H would consist of all elements of order 1, 2, 4, or 8, while K would consist of the elements of order 1 or 3. Then we would have

G ≈ H × K.

But what if G had no elements of order 3? Then K would be just the identity element, and H would have to be all of G. Lemma 6.1 would hold, but since H and K are not proper subgroups, this would not give a decomposition of G. The next lemma will show that, in fact, G must have an element of order 3.

LEMMA 6.2
If G is a finite abelian group and p is a prime that divides the order of G, then G has an element of order p.

Click here for the proof.

This lemma is known as Cauchy's theorem for abelian groups. In fact we will see that Cauchy's theorem is true for all groups, not just abelian groups. However, the result for abelian groups is good enough for us to proceed.

EXPERIMENT:
Since the Euler totient function of 35 is

EulerPhi(35)

the group

G = ZStar(35); G

is an abelian group with 24 elements. Lemma 6.2 then tells us that there are elements of order 2 and of order 3. We can find the set of elements H such that h³ = 1, with the command

[x^3 for x in G]

and the set of elements K such that k⁸ = 1 with

[x^8 for x in G]

and comparing this list with G. Show that H·K = Z₃₅^*, and hence Z₃₅^* has the decomposition H × K.

This experiment demonstrates how Lemma 6.2 guarantees that the subgroups H and K generated by Lemma 6.1 must be proper subgroups. Notice that in this experiment, H contained 3 elements while K contained 8 elements. There seems to be a pattern: the number of elements of G satisfying

x^(pⁿ) = e,

is pⁿ whenever p divides the order of G exactly n times. This apparent pattern can be shown to exist for all abelian groups. We thus have the following lemma:

LEMMA 6.3
Let G be an abelian group of order pⁿ·k where p is prime, k is not divisible by p, and n > 0. Then there are subgroups P and K of G such that

G ≈ P × K,

where |P| = pⁿ and |K| = k.

Click here for the proof.

Lemma 6.3 is a tremendous help in finding the structure of abelian groups. For example, suppose we have an abelian group G of order 24. Since 24 = 2³·3, Lemma 6.3 states that G is isomorphic to a direct product of a group of order 8 and a group of order 3. Since Z₃ is the only group of order 3, and there are 3 abelian groups of order 8, namely

Z₈, Z₁₅^*, and Z₂₄^*,

we have that G must be isomorphic to one of the three groups

Z₈ × Z₃, Z₁₅^* × Z₃, and Z₂₄^* × Z₃,

EXPERIMENT:
Since Z₃₅ is an abelian group with 24 elements, this must be isomorphic to one of the three groups mentioned above. Which of the three is it?
(Hint: first use Sage to find Z₃₅, and to determine the number of elements of order 2. Then use Proposition 6.2 to find the number of elements of order 2 for the three groups listed above.)

We have seen that if we can find all abelian groups of order pⁿ for p a prime number, then we will be able to find all finite abelian groups. The next lemma is the key to finding all abelian groups whose order is a power of a prime.

LEMMA 6.4
Suppose P is an abelian group of order pⁿ, where p is a prime. Let x be an element in P that has the maximal order of all of the elements of P. Then

P ≈ X × T,

where X is the cyclic group generated by x, and T is a subgroup of P.

Click here for the proof.

To give an illustration of how Lemma 6.4 is applied, consider the groups of order 8. Lemma 6.4 states that these can be decomposed into cyclic groups. We have already seen that Z₁₅^* was isomorphic to Z₄ × Z₂, and Z₈ is already a cyclic group. Let's use Lemma 6.4 on the group Z₂₄^*.

All non-identity elements of Z₂₄ are of order 2, so this is the maximal order. Thus, Lemma 6.4 states that Z₂₄ can be decomposed into Z₂ and a group of order 4. Since Z₂ × Z₄ is isomorphic to Z₁₅, the only other choice is Z₂₄ ≈ Z₂ × Z₈^*.

Now we can apply Lemma 6.4 again to Z₈. This is of order 4, and all elements besides the identity are of order 2, so Z₈ can be decomposed into Z₂ and a group of order 2, which must be Z₂. Thus,

Z₈^* ≈ Z₂ × Z₂,

and so

Z₂₄^* ≈ Z₂ × Z₂ × Z₂.

It is clear how repeated use of Lemma 6.4 can decompose any abelian group into cyclic groups. But now we want to address the issue as to whether a decomposition is unique. Is it possible for a group to be represented as a direct product of cyclic groups in two different ways?

In one sence, the answer is yes, for

Z₁₂ × Z₂ ≈ Z₆ × Z₄.

Yet this doesn't give the whole story, since both of these can futher decomposed:

Z₁₂ × Z₂ ≈ Z₆ × Z₄ ≈ Z₂ × Z₃ × Z₄.

So the question that we should ask is whether an abelian group can be decomposed in two different ways into groups which can not be decomposed further, that is, to the point where each factor is a cyclic group whose order is a power of a prime. A way to reword this question is to ask whether two different decompositions are isomorphic.

The main tool for testing whether two groups are isomorphic is to count elements of a given order. This can be accomplished by computing R_n(G) for various values of n. It is natural to compute R_n(G) for a decomposition of cyclic groups.

LEMMA 6.5
Let p be a prime number, and G be the direct product of cyclic groups

G = Z_(pₕ_¹) × Z_(pₕ_²) × Z_(pₕ_³) × ⋯ × Z_(pₕ_ⁿ ) × Z ₖ_₁ × Z ₖ_₂ × ⋯ × Z ₖ_ₘ

where h₁, h₂, …, h_n are positive integers, and k₁, k₂, …, k_m are coprime to p. Then if q = p^x,

	⎛	∑	n	Min(hi, x)	⎞
	⎝		i=1		⎠
Rq(G) = p					,

where Min(h_i , x) denotes the smaller of the two numbers h_i and x.

Click here for the proof.

An example here would help to illustrate the power of Lemma 6.5. Suppose we are given the group

G = Z₃ × Z₉ × Z₉ × Z₂₇.

How many elements are there of order 9? Note that we can rewrite G as

G = Z_3¹ × Z_3² × Z_3² × Z_3³.

Thus, h₁ = 1, h₂ = 2, h₃ = 2, and h₄ = 3. Since we want the elements of order 9 = 3², we let x = 2. Then

∑_i  Min(h_i , x) = 1 + 2 + 2 + 2 = 7,

so R₉(G) = 3⁷, meaning there are 2187 solutions to g⁹ = e. However, not all of these elements will be of order 9, since this count also includes elements of order 3. In order for g to be of order 9, it must satisfy g⁹ = e and g³ ≠ e. Thus, we will apply the formula again with x = 1 to get

∑_i  Min(h_i , x) = 1 + 1 + 1 + 1 = 4.

So R₃(G) = 3⁴, and the total number of elements of G of order 9 is 3⁷ − 3⁴ = 2106.

EXPERIMENT:
How many elements of G have order 27? (Use x = 3 instead of x = 2.)

We are now ready to show that all finite abelian groups can be represented as the direct product of cyclic groups. However, we would like to show at the same time that such a representation is unique. To this end we will use the previous lemma in conjunction with the following.

LEMMA 6.6
Let h₁, h₂, h₃, … h_n be a set of positive integers, and define ƒ(x) as

	n
ƒ(x) =	∑	Min(hi, x)
	i = 1

where Min(h_i, x) denotes the minimum of h_i and x. Then the number of times that the integer x appears in the set of integers h₁, h₂, h₃, … h_n is given by

2 ƒ(x) − ƒ(x − 1) − ƒ(x + 1).

Click here for the proof.

We can now use Lemmas 6.3 through 6.6 to show that not only can a finite abelian group be decomposed into a direct product of cyclic groups, but that it can be decomposed into a standard form in a unique way, much as a positive integer can be uniquely decomposed into a product of prime numbers.

THEOREM 6.2:The Fundamental Theorem of Finite Abelian Groups
A nontrivial finite abelian group is isomorphic to

Z_(p₁ₕ_¹) × Z_(p₂ₕ_²) × Z_(p₃ₕ_³) × ⋯ × Z_(pₙₕ_ⁿ ),

where p₁, p₂, p₃, … p_n are prime numbers (not necessarily distinct). Furthermore, this decomposition is unique up to the rearrangement of the factors.

Click here for the proof.

From this theorem, we can easily find all non-isomorphic abelian groups of a given order. For example, to find all non-isomorphic abelian groups of order 16, we note that all such groups are direct products of the cyclic groups of orders 2, 4, 8, or 16. We want to find all possible combinations of 2, 4, 8, and 16 which will multiply to give 16. With a little experimenting, we find that there are five such combinations:

2·2·2·2,  2·2·4,  4·4,  2·8,  and 16.

Thus, there are 5 possible abelian groups of order 16:

Z₂ × Z₂ × Z₂ × Z₂,  Z₂ × Z₂ × Z₄,  Z₄ × Z₄,  Z₂ × Z₈,   and Z₁₆.

Since the fundamental theorem (6.2) also states that the representation is unique, these five groups must be non-isomorphic to each other. Notice that there is a correlation between these five groups, and the five ways we can express the number 4 as a sum of positive integers:

1 + 1 + 1 + 1 = 4
1 + 1 + 2 = 4
2 + 2 = 4
1 + 3 = 4
4 = 4

This leads us to a way of finding the number of non-isomorphic groups of order p^m for any m.

COROLLARY 6.2
Let P(m) denote the number of ways in which m can be expressed as a sum of positive integers, without regard to order. Then if p is a prime number, there are exactly P(m) non-isomorphic abelian groups of order p^m.

Click here for the proof.

We call P(m) the number of partitions of m. We can have Sage count the number of partitions for us, using the command PartitionsP(m). Thus, to find the number of partitions of the number 4, we can enter

PartitionsP(4)

We can even have Sage list all of the paritions for us

[i for i in Partitions(4)]

Note that in the last command, we used Partitions instead of PartitionsP.

According to Corollary 6.2, the number of non-isomorphic abelian groups of order 3¹⁰⁰ is

PartitionsP(100)

We won't ask for the list of all of them here! As we can see, the number of partitions of m grows very rapidly as m gets large. How rapidly? We can make a logarithmic plot of the first 100 values of P(m) with the following command:

list_plot([ln(PartitionsP(m)) for m in range(100)])

This looks like a sideways parabola, indicating that P(m) grows in about the same rate as e^√m.

We can now find the number of non-isomorphic abelian groups of any order, as the following corollary shows.

COROLLARY 6.3
Let m > 1 be an integer with prime factorization

p₁^h¹·p₂^h²·p₃^h³ ⋯ p_n^hⁿ

where p₁, p₂, p₃, …, p_n are distinct primes. Then the number of non-isomorphic abelian groups of order m is given by

P(h₁)·P(h₂)·P(h₃) ⋯ P(h_n).

Click here for the proof.

EXAMPLE:
Suppose we want to find the number of non-isomorphic abelian groups of order 180 billion. Since 180,000,000,000 = 2¹¹·3²·5¹⁰, we have that the number of groups is

PartitionsP(11)*PartitionsP(2)*PartitionsP(10)

So there are 4704 abelian groups of order 180 billion.

From these two corollaries, we see that all finite abelian groups have been classified. One of the outstanding problems in group theory is to classify all finite groups. This is as yet an unsolved problem although much progress has been made through the use of computers. In the next two sections we will show some other ways of generating larger groups which have become a key to some of the recent work that has been done in group theory.

Automorphisms

We have already studied several examples of homomorphisms and isomorphisms between two groups, but suppose we considered a mapping from a group to itself.

EXAMPLE:
Find an isomorphsm from Z₈ onto itself.

We could consider the following mapping:

Z8 = ZGroup(8)
CircleGraph(Z8, Pow(3))

Notice that this mapping is one-to-one and onto, so in fact we could consider this as a permutation of the numbers 0 through 7, namely (1 3)(2 6)(5 7). However, to make this into a homomorphism in Sage, we have to define a mapping that sends Z8[1] to Z8[3].

F = Homomorph(Z8, Z8)

HomoDef(F, Z8[1], Z8[3])

FinishHomo(F)

This shows that we have a homomorphism. Now we can see that F is the same mapping as Pow(3)

CircleGraph(Z8, F)

We see that F is a homomorphism from Z₈ onto itself. We give such a homomorphism a special name.

DEFINITION 6.3
An automorphism of the group G is a homomorphism from G to G which is one-to-one and onto.

If we study the above automorphism ƒ on Z₈, we discover why this works. Recall that the operation of this group is addition modulo 8. Hence the mapping x → x³ in Z₈ will send each number x to (3 x) mod 8. Therefore,

ƒ(x·y) = ƒ((x + y) mod 8) = (3(x + y) mod 8 = (3x + 3y) mod 8 = ƒ(x)·ƒ(y).

By observing this pattern, we find another automorphism G of Z₈ by letting G send x into x⁵ = (5 x) mod 8. The graph of G would be given by

G = Homomorph(Z8, Z8)
HomoDef(G, Z8[1], Z8[5])
FinishHomo(G)

CircleGraph(Z8, G)

So we have found 2 automorphisms on Z₈, one which acts like the permutation P(3,6,1,4,7,2,5) and the other acting like P(5,2,7,4,1,6,3). But we can multiply these permutations together to form a third permutation.

EXPERIMENT:
Is the permutation

P(3,6,1,4,7,2,5)*P(5,2,7,4,1,6,3)

represent a homomorphism of Z₈? Define H to be the new function sending Z8[1] to Z8[7].

This experiment suggests that the product of two automorphisms is again an automorphism. We can formally define the product of two automorphisms ƒ(x) and ϕ(x) as (ƒ·ϕ)(x) = ƒ(ϕ(x)).

But there are some other patterns of the automorphisms that we can observe. Because an automorphism ƒ(x) on G is a one-to-one and onto mapping, we can consider the inverse mapping ƒ⁻¹(x). For example, the inverse to the automorphism F is denoted by the permutation

P(3,6,1,4,7,2,5)^(-1)

In this case, the inverse of F is again F, which is an automorphism.

EXPERIMENT:
Are the inverses of G and H again automorphisms? Which automorphisms are they?

Since we seem to be able to multiply two automorphisms together and take inverses, the natural question comes up, which is answered in the next proposition.

PROPOSITION 6.3
Given a group G, the set of all automorphisms on G forms a group, denoted Aut(G). In fact, Aut(G) is a subgroup of the group of permutations on the elements of G.

Click here for the proof.

EXAMPLE:
Find the automorphism group for Z₈.

We have two elements already, which we can express in terms of permutations. Let's find the group generated by these two permutations:

A = Group(P(3,6,1,4,7,2,5),P(5,2,7,4,1,6,3)); A

So Aut(Z₈) has at least four elements. Is this the maximum number of automorphisms? Notice that 1 is of order 8 in Z₈, so it must be mapped to an element of order 8, namely 1, 3, 5, or 7. But since 1 is a generator of the group, any automorphism is determined by where the generator is mapped to. Hence, there are at most four automorphisms of Z₈. Since we have already found four, we must have found all of them!

Since we have Aut(Z₈) is a group of order 4, the natural question is which group is this isomorphic to? Here is a multiplication table for A:

MultTable(A)

From this, we see that all elements of A besides the identity are of order 2. So

Aut(Z₈) ≈ Z₂ × Z₂ ≈ Z₈^*.

We can use a similar argument to find the automorphism group for any cyclic group.

PROPOSITION 6.4

Aut(Z_n) ≈ Z_n^*.

Click here for the proof.

So far, the automorphism group is smaller than the original group, but the goal of this chapter is to form larger groups. Let us consider a non-cyclic group.

EXAMPLE:
Find the set of automorphisms on the group Z₈^*.

First define this group in Sage, using two generators:

InitGroup("e")
AddGroupVar("a","b")
Define(a^2, e)
Define(b^2, e)
Define(b*a, a*b)
G = Group(a, b); G

A good strategy for finding all of the automorphisms is to first determine an upper bound for the number of automorphisms. Suppose ƒ is an automorphism. Then ƒ(e) = e, but all other elements are of order 2. Hence, any of the other elements might map to each other in any way. For example, ƒ(a) might be a, b, or a·b. Once we know where a is mapped, ƒ(b) might be either of the other two elements. However, once we know ƒ(a) and ƒ(b), then ƒ(a·b) must be ƒ(a)·ƒ(b). Thus, there are at most 3·2 = 6 elements of Aut(Z₈^*). If we find that there are indeed this many automorphisms, then Aut(Z₈^*) would be larger than Z₈^*.

Here would be a possible automorphism:

ƒ(e)	=	e
ƒ(a)	=	b
ƒ(b)	=	a
ƒ(a·b)	=	a·b

We can define this as a homomorphism is Sage.

F = Homomorph(G, G)
HomoDef(F, a, b)
HomoDef(F, b, a)
FinishHomo(F)

CircleGraph(G, F)

Let us represent this automorphism as a permutation on the elements of Z₈^*. There is not a natural numbering of the elements, so it would be difficult to use the P[ ] notation. However, we still can use the notation of cycles, and say that ƒ is the 2-cycle (a b). Note that we are using the cycle notation with elements instead of numbers. We are allowed to do this in Sage, and so we can enter this function as follows:

f = C(a, b); f

In fact, we can still use the CircleGraph command to graph this function:

CircleGraph(G, f)

Let us consider another automorphism having ƒ(a) = a, but having ƒ(b) = a·b.

H = Homomorph(G, G)
HomoDef(H, a, a)
HomoDef(H, b, a*b)
FinishHomo(H)

CircleGraph(G, H)

We see from the circle graph that this is indeed an automorphism, and can be represented by the cycle (b a·b).

h = C(b, a*b)
CircleGraph(G, h)

The advantage of using cycles instead of functions is that we can now form the group generated by these two automorphisms

A = Group(f, h); A

This gives us six elements, which is the maximum number of automorphisms we can have. Thus, A is Aut(Z₈^*). Therefore, we have found that the automorphism group is sometimes larger than the original group.

Which group is this? It is not hard to recognize A as the permutation group on the elements a, b, and a·b. Thus, Aut(Z₈^*) is isomorphic to S₃.

For non-commutative groups, there is a quick way to find many of the automorphisms. Let G be a non-commutative group, and let x be any element in G. The mapping f_x : G → G defined by

f_x(y) = x·y·x^-1

will always be an automorphism, for

f_x(y·z) = x·y·z·x^-1 = (x·y·x^-1)·(x·z·x^-1) = f_x(y)·f_x(z).

So f_x(y) is a homomorphism. Since the inverse homomorphism can easily be found,

y ∈ f_x^-1(v) ⟺ x·y·x^-1 = v ⟺ y = x^-1·v·x ⟺ y = f_x₋₁(v),

we have that f_x(y) is one-to-one and onto, therefore f_x(y) is an automorphism.

DEFINITION 6.4
An automorphism ϕ(y) of a group G is called an inner automorphism if there is an element x in G such that

ϕ(y) = x·y·x^-1 for all  y ∈ G.

The set of inner automorphisms of G is denoted Inn(G).

EXAMPLE:
Find the inner automorphisms of the quaternion group Q.

First, we load Q into Sage:

Q = InitQuaternions(); Q

Let us begin by determining an upper bound for the number of automorphisms. Suppose ƒ is an automorphism of Q. Then ƒ(1) = 1, as always, but ƒ(−1) must be −1, since −1 is the only element of Q of order 2. All of the other elements are of order 4, so ƒ(i) could be any one of the remaining six elements. Once ƒ(i) is determined, though, so this will be determined. Then ƒ(j) could be any one of the remaining four elements. Since i and j generate Q, once ƒ(i) and ƒ(j) is determined, ƒ will be determined for all of Q. Thus, there is a maximum of 6·4 = 24 possible automorphisms.

Let us try to find some inner automorphisms on Q. If we choose x = i, then ƒ_i(y) is the following mapping:

ƒi(1)	=	i·1·(−i)	=	1
ƒi(i)	=	i·i·(−i)	=	i
ƒi(j)	=	i·j·(−i)	=	−j
ƒi(−1)	=	i·(−1)·(−i)	=	−1
ƒi(−i)	=	i·(−i)·(−i)	=	−i
ƒi(k)	=	i·k·(−i)	=	−k
ƒi(−j)	=	i·(−j)·(−i)	=	j
ƒi(−k)	=	i·(−k)·(−i)	=	k

So ƒ_i exchanges j and −j, and exchanges k and −k, leaving all other elements alone. Thus, we can express this automorphism as

F = C(j, -j)*C(k, -k); F

Here is the graph of this automorphism:

CircleGraph(Q, F)

From the above descussion. we know that this is a homomorphism. If we use x = j instead of x = i, we get a different automorphism:

ƒj(1)	=	j·1·(−j)	=	1
ƒj(i)	=	j·i·(−j)	=	−i
ƒj(j)	=	j·j·(−j)	=	j
ƒj(−1)	=	j·(−1)·(−j)	=	−1
ƒj(−i)	=	j·(−i)·(−j)	=	i
ƒj(k)	=	j·k·(−j)	=	−k
ƒj(−j)	=	j·(−j)·(−j)	=	−j
ƒj(−k)	=	j·(−k)·(−j)	=	k

We can represent this automorphism in Sage by

G = C(i, -i)*C(k, -k); G

Here is the graph:

CircleGraph(Q, G)

Since F and G are both automorphisms, we can find the group generated by these two:

Group(F, G)

Therefore, we have found 4 automorphisms on Q. But is this all of the automorphisms?

EXPERIMENT:
Find the automorphism ƒ_x(y) with x = k. Find the corresponding permutation on Q, and have Sage draw a circle graph of this automorphism. Is this automorphism one that we have seen before?

If we try the above procedure with x equaling any of the eight elements of Q, we will always get one of the four automorphisms. Hence, we have found all of the inner automorphisms of Q, which we can express in terms of cycles as

{( ), (i, −i)(j, −j), (j, −j)(k, −k), (i, −i)(k, −k)}.

EXAMPLE: Find all of the automorphisms of Q.

We found 4 inner automorphisms, but there certainly could be more. Suppose [removed]ƒ(i) = i, and ƒ(j) = k? Since i, j, and k are all of order 4, it is conceivable that this would produce an automorphism. This time, we will use Sage's Homomorph function to define the mapping.

X = Homomorph(Q, Q)
HomoDef(X, i, i)
HomoDef(X, j, k)
FinishHomo(X)

So X is a homomorphism, but is X an automorphism? By finding

Image(X, Q)

we see that X is onto, and by finding

Kernel(X)

we see that X is one-to-one. Therefore, X is an automorphism. By looking at the circle graph:

CircleGraph(Q, X)

we see that the automorphism can be represented by the 4-cycle

H = C(j, k, -j, -k); H

Let us add this to our group of automorphisms:

Group(F, G, H)

len(_)

So far, we have 8 automorphisms. Could there be any others?

Notice that all 8 of the automorphisms that we have come up with so far send i either to i or −i. What if we sent i to, say, k? We could let j be sent to j for simplicity. Let's try it.

Y = Homomorph(Q, Q)
HomoDef(Y, i, k)
HomoDef(Y, j, j)
FinishHomo(Y)

We can check that this is one-to-one and onto by looking at the circle graph.

CircleGraph(Q, Y)

So Y is an automorphism of Q. It can be represented by the 4-cycle

J = C(i, k, -i, -k); J

So here is an updated list of automorphisms:

A = Group(F,G,H,J); A

This is a long list. Let's see how many automorphisms we have.

len(A)

So we have 24 automorphisms of Q. Since we began by observing that there could be at most 24 automorphisms, we finally must have them all. Is this group isomorphic to one of the groups of order 24 that we have studied before?

Rather than proving an isomorphism directly as we have done in the past, in this case there is a geometrical correlation between this automorphism group and the octahedronal group. Note that any automorphism of Q must fix 1 and −1, but the other six elements can go to any other of the six elements. Suppose we were to label the six vertices of the octahedron with the six remaining elements of Q, as follows:

ShowOctahedronWithQuaternions()

The vertex in the back is labeled −j. Now notice that each rotation of the octahedron gives an automorphism of Q, and vice versa. For example, rotating the red face 120° clockwise corresponds to the automorphism

(i j k)(−i −j −k).

So the automorphism group is isomorphic to the octahedronal group, which we saw was isomorphic to S₄. Thus, Aut(Q) is isomorphic to S₄.

Although the inner automorphisms did not produce the full automorphism group, this set of inner automorphisms turns out to be a very important subgroup of the automorphism group. Notice that we found that the product of the two inner automorphism was an inner automorphism. This suggests that Inn(G) is a subgroup of Aut(G). In fact we can even say more!

PROPOSITION 6.5
Let G be a group. Then Inn(G) is a normal subgroup of Aut(G).

Click here for the proof.

We found two inner-automorphisms of Q, namely F and G. We found one more inner automorphism by considering

Inn = Group(F, G); Inn

If we examine the automorphism ƒ_x(y) = x·y·x^-1 for different elements x, we will find that we will always get one of these four automorphisms. Thus, this group is the group of inner automorphisms on Q. Here is the multiplication table:

MultTable(Inn)

Since the square of every element is the identity, we recognize Inn(Q) as isomorphic to Z₈^*.

Because the inner automorphism group is always a normal subgroup, we could consider the quotient group.

DEFINITION 6.5
We define the outer automorphism group to be the group

Out(G)= Aut(G)/Inn(G).

For example, the outer automorphism group of Q is given by

Z = LftCoset(A, Inn); Z

This gives a group of order 6.

As complicated as the elements of Z are, we can form a multiplication table.

MultTable(Z)

Since this group is non-abelian, we see that the group is isomorphic to S₃.

If G is an abelian group, then the only inner automorphism is the identity automorphism. Thus, for abelian groups,

Inn(G) ≈ {e} and Out(G) ≈ Aut(G).

EXPERIMENT:
Load S₃ into Sage with the following commands:

InitGroup("e")
AddGroupVar("a", "b")
Define(a^2, e)
Define(b^3, e)
Define(b*a, a*b^2)
G = Group(a, b); G

Note that for any automorphism, the element a must map to one of the three elements of order 2, and b must map to one of the two elements of order 3. Since the elements a and b generate the group, there can be at most six automorphisms. By finding a few of the inner automorphisms, and having Sage create a multiplication table, identify Inn(S₃). What is Aut(S₃)?

Let us look at one last example where the automorphism group turns out to be much larger than the original group.

EXAMPLE:
Find the automorphism group of Z₂₄^*.

The following loads the group into Sage:

InitGroup("e")
AddGroupVar("a", "b", "c")
Define(a^2, e)
Define(b^2, e)
Define(c^2, e)
Define(b*a, a*b)
Define(c*a, a*c)
Define(c*b, b*c)
Y = ListGroup(); Y

Once again, we will begin by determining an upper bound for the number of automorphisms. Suppose ϕ(x) is an automorphism of Z₂₄^*. Naturally, ϕ(e) = e, but the other 7 elements are of order 2. Thus, ϕ(a) could be any of the 7 remaining elements. Now ϕ(a) and ϕ(b) are unrelated so ϕ(b) could be any of the 6 elements besides e and ϕ(a). We then would have ϕ(a·b) = ϕ(a)·ϕ(b), so four elements will be accouted for. But ϕ(c) could be any of the 4 elements left over. Once ϕ(a), ϕ(b), and ϕ(c) are known, the entire automorphism is determined. Therefore, there are at most 7·6·4 = 168 possible automorphisms.

Now let us try to find an automorphism. Since a, b, and c are interchangeable in the definition of the group, we could consider a mapping ƒ that maps a to b, b to c, and c back to a.

F = Homomorph(Y, Y)
HomoDef(F, a, b)
HomoDef(F, b, c)
HomoDef(F, c, a)
FinishHomo(F)

CircleGraph(Y, F)

We see that this produces an automorphism, and that a·b maps to b·c, which maps to c·a, which maps back to a·b. The element a·b·c is fixed by this mapping. Thus, this mapping would be the permutation

f = C(a, b, c)*C(a*b, b*c, a*c); f

Can we find other automorphisms? Our success should indicate that there are many of them! Suppose we consider ϕ(a) = a, ϕ(c) = c, but let ϕ(b) = a·b?

G = Homomorph(Y, Y)
HomoDef(G, a, a)
HomoDef(G, b, a*b)
HomoDef(G, c, c)
FinishHomo(G)

CircleGraph(Y, G)

This automorphism works, and would be represented by the permutation

g = C(b, a*b)* C(b*c, a*b*c); g

We can now use Sage to find other automorphisms generated by these two.

Because of the large number of possible elements in Aut(Z₂₄), we will be better off if we number the elements in the group Z₂₄. This will allow us to use the P notation. (Sage works much faster using the P notation than using cycles.) Since all automorphisms fix the element e this element need not be numbered. The other 7 elements can be numbered as follows, using the same order as the elements appear in ListGroup:

1)	a
2)	b
3)	a·b
4)	c
5)	a·c
6)	b·c
7)	a·b·c

Let us convert f and g to permutations using the numbering scheme we gave to the elements.

f = CycleToPerm(C(1,2,4)*C(3,6,5)); f

g = CycleToPerm(C(2,3)*C(6,7)); g

That is, f maps element 1 (a) to element 2 (b), which is mapped to element 4 (c), etc.

Likewise, g exchanges the 2nd and 3rd elements, and exchanges the 6th and 7th elements of Z₂₄^*. The circle graphs will still looks the same:

CircleGraph([0,1,2,3,4,5,6,7], f)

CircleGraph([0,1,2,3,4,5,6,7], g)

Since f and g both represent automorphisms of Z₂₄^*, we can find other automorphisms be considering the group generated by f and g.

A = Group(f, g); A

Sage came up with quite a few automorphisms. Let's see how many we have.

len(A)

Therefore, we have found 168 different automorphisms of Z₂₄^*. We saw that this was the maximum number of automorphisms we could have, so we must have all of them. However, this is a difficult way to see the elements of the group! It is a challenge just to see if a given permutation is somewhere in this list. Is there an easier way of viewing this group?

One way to tidy up this long list is to convert the permutations to numbers. Recall that in Chapter 5 we ordered all of the permutations, thereby converting every permutation to a number. Setting DisplayPermInt to true causes the permutations to be displayed as integers.

DisplayPermInt = true

A

This looks much more manageable. By displaying the permutations f and g,

f

g

we see that f and g represent the 187th and the 723rd permutations.

This is the largest group with which we will work using Sage. In fact we will discover in the next chapter that this group has some very special properties. We will explore this group in detail using Sage in the next chapter.

We have now seen several examples where the group of automorphisms is larger than the original group. But this group of automorphisms can also be used as a tool for connecting two groups to form an even larger group, in much the same way that two groups formed the direct product. The next section will explore this methodology.

Semi-Direct Products

So far in this chapter we have seen how we can combine two groups H and K together to form a larger group H × K using the direct product. We then saw that H × K had isomorphic copies of H and K as normal subgroups.

In this section we will see another way to combine two groups H and K. Once again the larger group will have isomorphic copies of H and K as subgroups, but only one of the two subgroups will be a normal subgroup. For this reason, the combination of groups is called a semi-direct product.

Suppose that H and K are any two groups, and suppose that we have a homomorphism

ϕ: H → Aut(K).

That is, for every h in H, ϕ(h) will be a mapping from K into K. Because the output of the function ϕ is again a function, we will write ϕ_h instead of ϕ(h). This way we can write ϕ_h(k) to express the function ϕ_h evaluated at k. That is, if h₁ and h₂ are two elements of H, then and ϕ_h₂(k) will be two automorphisms of K, and also

ϕ_h₁·h₂(k) = (ϕ_h₁·ϕ_h₂)(k) = ϕ_h₁(ϕ_h₂(k)).

(Recall that ϕ_h₁·ϕ_h₂ means we do ϕ_h₂ first, then do ϕ_h₁.)

There will always be at least one possible homomorphism from H to Aut(K) since we could let ϕ_h be the identity automorphism for all h. However, there will usually be several other possible homomorphisms from H to Aut(K). For each such homomorphism, we can define a product of K and H.

DEFINITION 6.6
Let K and H be two groups, and let G to be the set of all ordered pairs (k, h), where k is in K and h is in H. Let ϕ be a nontrivial homomorphism from H to Aut(K). Then the semi-direct product of K with H through ϕ, denoted

K ⋊_ϕH,

is the set G with multiplication defined by

(k₁, h₁)·(k₂, h₂) = (k₁·ϕ_h₁(k₂), h₁·h₂).

PROPOSITION 6.6
The semi-direct product of K with H through ϕ is a group.

Click here for the proof.

A semi-product of two groups acts in many ways like the direct product. One property that is in common is that there are copies of the two original groups within the product. In fact, we have the following:

LEMMA 6.7
Let G = K ⋊_ϕH be the semi-direct product of K with H through the homomorphism ϕ. Suppose that e₁ is the identity element of K, and e₂ is the identity element of H. Then

H = { (e₁, h) | h ∈ H }

is a subgroup of G, and

K = { (k, e₂) | k ∈ K }

is a normal subgroup of G. Furthermore,

H ≈ H,   K ≈ K,

and H ∩ K is the identity element of G.

Click here for the proof.

Since the semi-direct product contains copies of the two smaller groups within itself, the natural question is whether an arbitrary group G can be expressed as a semi-direct product of two of its subgroups. The conditions for when this will happen is set forth in the following theorem.

THEOREM 6.3: The Semi-Direct Product Theorem
Suppose that a group G has two subgroups N and H whose intersection is the identity element. If N is a normal subgroup of G and H is not a normal subgroup of N·H, then there exists a nontrivial homomorphism ϕ from H to Aut(N) such that

N·H ≈ N ⋊_ϕH.

Click here for the proof.

Enough of the technical proofs, let us learn how to define a semi-direct product with Sage. We will use this result to define the semi-direct product of two groups in Sage. We first must define the two groups in Sage using the same identity as we did for the direct product, using different letters for the generators. We must also know the homomorphism ϕ from H to Aut(N).

We will want to express every element into the form k·h, where k is in N, and h is in H.

From the definition, we see that

(k, e₂)·(e₁, h) = (k·ϕ_e₂(e₁), e₂·h) = (k, h).

Thus, we see that k·h can represent the ordered pair (k, h). We need to tell Sage how to handle expressions of the form h·k.

For each generator a of N, and each generator b of H, we can calculate how b·a should be defined by evaluating

(e₁, b)·(a, e₂) = (e₁·ϕ_b(a), b·e₂) = (ϕ_b(a), b).

So for each generator a of H, and each generator b of N, we make a definition of the form

Define(b * a, ϕ_b(a) * b )

where we replace the expression ϕ_b(a) with its element of N.

EXAMPLE:

Use Sage to find a semi-direct product of Z₅ with Z₂.

We first must define Z₅ and Z₂ into Sage using the same identity but different generators.

InitGroup("e")
AddGroupVar("a", "b")
Define(a^5, e)
Define(b^2, e)
Z5 = Group(a); Z5

Z2 = Group(b); Z2

Next, we must find a nontrivial homomorphism ϕ from Z₂ to Aut(Z₅). But

Aut(Z₅) ≈ Z₅^* ≈ Z₄.

Since the element b is of order 2, ϕ(b)² = ϕ(b²) = e, so ϕ(b) must be of order 2 to keep the homomorphism from being trivial. We need ϕ(b) to map to the only element of Aut(Z₅) which is of order 2.

Which element is this? With a moment's reflection, we can see that

ƒ: N → N, ƒ(k) = k^-1

will always be an automorphism of whenever N is an abelian group. As long as N has an element that is not its own inverse, this automorphism will be of order 2.

Therefore, to find a semi-direct product of Z₅ by Z₂, we must let ϕ_b be the automorphism ƒ(k) = k^-1. Then ϕ_b(a) = a^-1 = a⁴. So the definition

Define(b*a, a^4*b)

completes the definition of the semi-direct product. We now have the group along with the multiplication table:

G = ListGroup(); G

MultTable(G)

This is the semi-direct product Z₅ ⋊ Z₂, a non-abelian group of order 10. Note that when there is only one possible semi-direct product between two groups, we can leave out the ϕ in the notation.

We can ask Sage what this group is, using the command StructureDescription(). This command analyzes the last group defined using the InitGroup and Define commands.

StructureDescription()

Sage says that this group is D₅, which introduces a new class of groups, known as the dihedral groups.

EXPERIMENT:
Find a semi-direct product Z₆ ⋊ Z₂. Then ask Sage what the group is using the StructureDescription() command.

We can easily generalize these examples to form an important class of groups.

DEFINITION 6.7:
Let n > 2, and let ϕ be the homomorphism from Z₂ = {e, b} to Aut(Z_n) given by

ϕ_e(k) = k, ϕ_b(k) = k^-1.

Then the semi-direct product Z_n ⋊_ϕZ₂ is called the dihedral group of order 2n. It is denoted D_n, and is a non-abelian group of order 2n.

Following this example, we see that the dihedral group can be defined in Sage by the commands

InitGroup("e")
AddGroupVar("a", "b")
Define(a^n, e)
Define(b^2, e)
Define(ba, a^-1b)
Dn = Group(a, b)

The symbol n must be replaced with an integer before executing these commands. When n = 3, we get a non-abelian group of order 6, so we must have that D₃ is isomorphic to S₃. We also have already seen D₄, since this was one of the 5 groups of order 8 that we found in chapter 4.

It should be noted that the semi-direct product may greatly depend on the choice of the homomorphism ϕ.

EXAMPLE:
Consider finding the semi-direct products of Z₈ with Z₂. The dihedral group D₈ is certainly one such product, but there could be others. The homomorphism ϕ would have to map the element b of order 2 in Z₂ to an element of order 2 in Aut(Z₈). But since

Aut(Z₈) ≈ Z₈^* ≈ Z₂ × Z₂

has three elements of order 2, we have three choices of what ϕ_b could be. Do these produce different groups?

To explore this possibility, let us first load in the dihedral group D₈.

InitGroup("e")
AddGroupVar("a", "b")
Define(a^8, e)
Define(b^2, e)
Define(b*a, a^7*b)
D8 = ListGroup(); D8

MultTable(D8)

We can check that we have defined D₈:

StructureDescription()

Notice that the main diagonal contains 10 e's. This means that there are 9 elements of D₈ which are of order 2. This will be useful for later comparisons.

What are the other automorphisms on Z₈? Since Z₈^* = {1, 3, 5, 7}, we have the automorphisms

ϕ(a) = a,   ϕ(a) = a³,   ϕ(a) = a⁵,   and   ϕ(a) = a⁷ = a^-1.

The first automorphism is the trivial automorphism, and the last is the one we just considered. Suppose we let ϕ_b(a) = a³. Then b·a = a³·b, so we can enter this into Sage with the commands

InitGroup("e")
AddGroupVar("a", "b")
Define(a^8, e)
Define(b^2, e)
Define(b*a, a^3*b)
G = ListGroup(); G

MultTable(G)

Is this the same group as the dihedral group? Look along the main diagonal. We can count six occurances of the identity elenent, so only five elements of this group are of order 2. Therefore this cannot be isomorphic to D₈. What does Sage call this group?

StructureDescription()

Sage calls this group "QD16", since this is the quasidihedral group of order 16.

What if we let ϕ_b(a) = a⁵? We would have b·a = a⁵·b, and so we would use the following commands to enter this group into Sage:

InitGroup("e")
AddGroupVar("a", "b")
Define(a^8, e)
Define(b^2, e)
Define(b*a, a^5*b)
M = ListGroup(); M

MultTable(M)

How many elements of this group are of order 2? This time there are only four occurances of the identity along the main diagonal, so there are only three elements of order 2. Thus we have yet another non-isomorphic group. Sage's name for this group:

StructureDescription()

The colon denotes the semi-direct product symbol, ⋊. Sage is identifying this as a semi-direct product, but not telling us which one. In fact, two non-isomorphic groups can sometimes produce the same StructureDescription() result.

We see in these three examples that the semi-direct product N ⋊_ϕH depends on the choice of the homomorphism ϕ form H into Aut(N). In fact, even though the three non-identity elements of Z₈ are essentially equivalent (The automorphism group of Z₈ included all permutations of these three elements), we see that the three elements produced three different semi-direct products of Z₈ with Z₂.

This example is really more of an exception rather than a rule. Part of what makes this example unusual is that the automorphism group Z₈^* is abelian, and hence does not have any nontrivial inner automorphisms. We will see in the next proposition that if two homomorphisms ϕ and ƒ from H to Aut(N) are related through an inner automorphism of Aut(N), then the corresponding semi-direct products will be isomorphic.

PROPOSITION 6.7
Let ϕ be a homomorphism from a group H to the group Aut(N). Suppose that ƒ is another homomorphism such that

ƒ_h(k) = w(ϕ_h(w^-1(k))),

where w(k) is an automorphism of N. Then

N ⋊_ƒ H ≈ N ⋊_ϕH

Click here for the proof.

It is also clear that two homomorphisms ϕ and ƒ are related through an automorphism of H, the semi-direct products must be isomorphic since we are merely relabeling the elements of H. As a result there will be many instances in which there will be only one non-isomorphic semi-direct product of N by H. Whenever this happens, we can denote the semi-direct product as

N ⋊ H

without having to specify the homomorphism ϕ.

We will find that we can describe many groups in terms of semi-direct products that would be hard to describe in any other way. With Sage, the structure of these semi-direct products can easily be studied.

Proofs:

Proof of Proposition 6.1:

First, suppose that H and K are both abelian. Then for two elements (h₁, k₁) and (h₂, k₂) in H × K, we have

(h₁, k₁)·(h₂, k₂) = (h₁·h₂, k₁·k₂) = (h₂·h₁, k₂·k₁) = (h₂, k₂)·(h₁, k₁).

So the two elements in H × K commute. Hence, H × K is abelian.

Now suppose that H × K is commutative. We then have

(h₁·h₂, k₁·k₂) = (h₁, k₁)·(h₂, k₂) = (h₂, k₂)·(h₁, k₁) = (h₂·h₁, k₂·k₁).

Comparing components, we see that h₁·h₂ = h₂·h₁ and k₁·k₂ = k₂·k₁. Since this is true for all h₁ and h₂ in H, and all k₁ and k₂ in K, both H and K are abelian.

Return to text

Proof of Theorem 6.1:

First, let us show that every element in H·K can be uniquely written in the form h·k, where h ∈ H and k ∈ K. Suppose that

h₁·k₁ = h₂·k₂.

Then h₂^-1·h₁ = k₂·k₁^-1. Since this element must be in both H and K, and the intersection of H and K is the identity element, we have that

h₂^-1·h₁ = k₂·k₁^-1 = e.

Thus, h₁ = h₂ and k₁ = k₂. Therefore, every element of H·K can be written uniquely as h·k, where h is in H, and k is in K.

Next, we need to show that H·K is a group. Since h·k = k·h for all h ∈ H and k ∈ K, we have that H·K = K·H. Thus, by Lemma 4.2, H·K is a subgroup of G.

We can now define a mapping

ϕ : H·K → H × K

by ϕ(x) = (h, k), where h and k are the unique elements such that h ∈ H, k ∈ K, and x = h·k. It is clear that ϕ; is one-to-one, since the element (h, k) can only have come from h·k. Also, ϕ is onto, for the element h·k maps to (h, k). All that remains to show is that ϕ(x·y) = ϕ(x)·ϕ(y). Let x = h₁·k₁ and y = h₂·k₂. Then

ϕ(x·y) = ϕ(h₁·k₁·h₂·k₂) = ϕ(h₁·h₂·k₁·k₂) = (h₁·h₂, k₁·k₂) = (h₁, k₁)·(h₂, k₂) = ϕ(x)·ϕ(y).

Thus, ϕ is an isomorphism, and so H·K ≈ H × K.

Return to text

Proof of Proposition 6.2:

Let e₁ denote the identity element of H, and e₂ denote the identity element of K. An element x = (h, k) in H × K solves the equation xⁿ = (e₁, e₂) if and only if

hⁿ = e₁ and kⁿ = e₂.

Since there are R_n(H) solutions to the former, and R_n(K) solutions to the latter, there are R_n(H)·R_n(K) ordered pairs (h, k) that solve both of these equations. Thus, there are R_n(H)·R_n(K) elements of H × K for which xⁿ = (e₁, e₂).

Return to text

Proof of Corollary 6.1:

The first condition of the direct product theorem (6.1) is given, so we only need to show that the second condition holds. That is, we need to show that h·k = k·h for all h in H, and k in K. Let h ∈ H and k ∈ K.

Since K is a normal subgroup of G, h·k·h^-1 is in K. Thus, h·k·h^-1·k^-1 is also in K.

But H is also a normal subgroup of G, so k·h^-1·k^-1 is in H. Hence, h·k·h^-1·k^-1 is also in H.

We now use the fact that the only element in both H and K is e. Thus, h·k·h^-1·k^-1 = e, which implies h·k = k·h.

Therefore, the second condition of the direct product theorem (6.1) holds, and so by this theorem, H·K ≈ H × K.

Return to text

Proof of Lemma 6.1:

To check that H and K are indeed subgroups simply observe that since G is commutative the functions ϕ(x) = x^m and ƒ(x) = xⁿ are both homomorphisms of G. Then H and K are the kernels of the mappings ϕ and ƒ.

To show that H and K have only the identity element in common, we consider an element x in the intersection. By the Chinese remainder theorem (0.7), there exists a non-negative number k < m·n such that

k mod m = 1  and  k mod n = 0.

Then k = 1 + m b for some number b. Thus,

x^k = x^{(1+m b)} = x·(x^m)^b = x·e^b = x

since x is in H. Yet k = n c for some number c, so

x^k = x^{n c} = (xⁿ)^c = e^c = e

since x is in K. Thus, x = e, and so H ∩ K = {e}. Since G is abelian, the direct product theorem (6.1) proves that

H·K ≈ H × K.

All that is left to prove is that G = H·K. Let g be an element in G. Since m and n are coprime, by the greatest common divisor theorem (0.4) there exists a and b such that

a n + b m = gcd(m, n) = 1.

Then

g = g¹ = g^{(a n+b m)} = g^{a n}·g^{b m}.

Now, (g^{a n})^m = (g^a)^{n m} = e, so g^{a n} is in H. Likewise, g^{b m} is in K. Thus, every element of G is in H·K, and so

G ≈ H × K.

Return to text

Proof of Lemma 6.2:

We will proceed using induction on the order of G. If |G| is a prime number, then p must be |G|, and G must be isomorphic to Z_p. So there would be an element of order p in G.

Suppose that the assumption is true for all groups of order less than |G|. If G does not have any proper subgroups, then G would be a cyclic group of prime order (which we have already covered.) Thus, we may assume that G has a subgroup N that is neither G nor {e}.

Since G is abelian all subgroups are normal. Thus we could consider the quotient group G/N. Since |G| = |N| · |G/N|, p must divide either |N| or |G/N|. If p divides N, then because N is a smaller group than G, by induction N must have an element of order p, which would be in G.

If p does not divide |N| it must divide |G/N|. Since G/N is a smaller group than G, by induction G/N must have an element of order p. This element can be written a N for some a in G.

Since a N is of order p, a cannot be in N, yet a^p must be in N. If |N| = q, we would have by Corollary 3.2 that (a^p)^q = e.

If b = a^q is not the identity, then b^p = e, and so b would be the required element. But if b = e, then (a N)^q = N. But a N was of order p, and so p must divide q. But we assumed that p did not divide q = |N|. Hence, b is not the identity, and so G has an element of order p.

Return to text

Proof of Lemma 6.3:

Since pⁿ and k are coprime, we can use Lemma 6.1 to form the subgroups

P = { x ∈ G | x^(pⁿ) = e }

and

K = { x ∈ G | x^k = e }.

By Lemma 6.1 these two subgroups have only the identity in common, and G ≈ P × K. If p divided |K|, then by Lemma 6.2, K would contain an element of order p. But this element would then be in P as well, which contradicts the fact that only the identity element is in common between P and K. So p does not divide the order of K.

Also note that the order of every element of P is a power of p. Thus, Lemma 6.2 tells us that no other prime other than p divides |P|.

Finally, note that |G| = pⁿ·k = |P|·|K|. Since p does not divide |K|, we have that pⁿ must divide |P|. But no other primes can divide |P|, and so |P| = pⁿ. Hence, |K| = k.

Return to text

Proof of Lemma 6.4:

We will use induction on n. If n = 1, then P is a cyclic group of order p, and hence is generated by non-identity element x in P. We then have X = P, so we can let T = {e}, and hence P ≈ X × T.

Now suppose that the assertion is true for all powers of p less than n. Notice that the order of every element of P is a power of p. Thus, if we let x be an element with the largest order, say m, then the order of all elements in P must divide m. Hence, g^m = e for all elements g in P.

We now let X be the subgroup generated by x. If X = P, then we can again let T = {e} and we are done. If X is not P, we let y be an element of P not in X which has the smallest possible order. Then since the order of y^p is less than the order of y, y^p must be in X. This means that y^p = x^q for some 0 ≤ q < m.

Since y is in P, y^m = e. But

e = y^m = (y^p)^(m/p) = (x^q)^(m/p) = x^{(m q/p)}

Because x is of order m, this can be the identity only if m q/p is a multiple of m. Hence, q is a multiple of p.

If we let k = x^−(q/p)·y then k is not in X because y isn't, and

k^p = (x^−(q/p))^p·y^p = x^−q·y^p = x^−q·x^q = e.

Therefore, we have found an element k of order p that is not in X. If we let K be the group generated by the element k, then X ∩ K = {e}.

Consider the quotient group P/K. What is the order of x K in P/K? We see that

(x K)ⁿ = K ⟺ xⁿ ∈ K ⟺ xⁿ ∈ X ∩ K ⟺ xⁿ = e.

Therefore, the order of x K is the same as the order of x, which is m. Also note that no element of P/K can have an element of higher order since gⁿ = e for all elements g in P.

Now we use the induction. Since the order of P/K is less than the order of P, and x K is an element of maximal order, we have by induction that

P/K ≈ Y × B,

where Y is the subgroup of P/K generated by x K, and B is a subgroup of P/K such that only the identity element K is in the intersection of Y and B.

Let ϕ be the canonical homomorphism from P to P/K given by ϕ(g) = g K. Let T = ϕ^-1(B). Then T is a subgroup of P.

If g is in both X and T, then ϕ(g) is in both Y and B. Since the intersection of Y and B is the identity element, we have ϕ(g) = g K = K. Thus, g is in the subgroup K. But X ∩ K = {e}, so we have

X ∩ T = {e}.

Thus, by the direct product theorem (6.1), we find that X·T ≈ X × T.

We finally need to show that P = X·T. Let u be an element in P, and since P/K ≈ Y × B, we can write ϕ(u) as (x^c·K)·(k K) for some number c, and some k K in B. Then

u ∈ x^c·k·K ⊆ X·T.

Thus, P = X·T, and so P ≈ X × T.

Return to text

Proof of Lemma 6.5:

Since G is expressed as a direct product we can use Proposition 6.2 and find R_q(H) for each factor H in the product, and multiply these numbers together. Since each factor is cyclic, we can use Corollary 2.1. For all of the factors Z ₖ_₁, Z ₖ_₂, … Z ₖ_ₘ, since gcd(k_i, q) = gcd(k_i, p^x) = 1, R_q(H) would be 1. On the other hand, R_q(Z_(pₕ_ⁱ)) is

gcd(p^hⁱ, q) = gcd(p^hⁱ, p^x) = p^Min(hⁱ^{, x)}.

Thus, R_q(G) is the product of the above for factors 1 through n of G, which gives us a grand total of

	⎛	∑	n	Min(hi, x)	⎞
	⎝		i=1		⎠
p					.

Return to text

Proof of Lemma 6.6:

Let us begin by observing the value of the expression

2 Min(h_i, x) − Min(h_i, x − 1) − Min(h_i, x + 1).

When h_i < x, then Min(h_i, x) = Min(h_i, x−1) = Min(h_i, x+1) = h_i, and so the above evaluates to 0. On the other hand, if h_i > x, then the above expression simplifies to

2 x − (x − 1) − (x + 1) = 0.

However, if h_i = x, then Min(h_i, x) = x, Min(h_i, x − 1) = x − 1, and Min(h_i, x + 1) = x. Hence, we have

2 Min(h_i, x) − Min(h_i, x − 1) − Min(h_i, x + 1) = 2 x − (x − 1) − x = 1.

Thus, we see that

2 Min(hi, x) − Min(hi, x − 1) − Min(hi, x + 1) =	⎧	1 if hi = x,
	⎨
	⎩	0 if hi ≠ x.

Thus, if we sum the above expression for i going from 1 to n, we will count the number of terms h_i that are equal to x. Hence this count will be

n
∑	2 Min(hi, x) − Min(hi, x − 1) − Min(hi, x + 1) = 2 ƒ(x) − ƒ(x − 1) − ƒ(x + 1).
i = 1

Return to text

Proof of Theorem 6.2:

We will proceed on induction on the order of the group. If the order of the group is 2, then the theorem is true since the group would be isomorphic to Z₂. Let G be a finite abelian group and suppose the theorem is true for all groups of order less than G. Let p be a prime that divides the order of G. By Lemma 6.3, G ≈ P × K, where P is the subgroup of G containing the elements of order p^m for some m.

Furthermore, if x is an element of maximal order in P, and X is the group generated by x, then by Lemma 6.4, G ≈ X × T × K. Since X will be a nontrivial cyclic group, the orders of T and K will be less than |G|. Thus, by induction, T and K can be written as a direct product of cyclic groups whose orders are powers of primes. Since X is also a cyclic group of order p^r for some r, G can be written as a direct product of cyclic groups whose orders are powers of primes.

We next have to show that this decomposition is unique. We will do this by showing that the number of times Z_(pₓ₎ appears in the decomposition, where p is a prime, is completely determined by the order of the elements in the group G. From Lemma 6.5,

R_pₓ(G) = p^{(∑ Min(h}ⁱ^{, x) )}

where the sum is taken over all i such that p_i = p. Thus, we see that

ƒp(x) =	∑	Min(hi, x) = logp(Rpₓ(G))
	pi = p

will be completely determined by the orders of the elements of G. But then by Lemma 6.6 the number of times that Z_(pₓ₎ appears in the decomposition is given by

2 ƒ_p(x) − ƒ_p(x − 1) − ƒ_p(x + 1).

Hence, the decomposition of G as a direct product of cyclic groups of the form Z_(pₓ₎ is unique.

Return to text

Proof of Corollary 6.2:

By the fundamental theorem of abelian groups (6.2), every abelian group of order p^m must be isomorphic to

Z_(pₕ_¹) × Z_(pₕ_²) × Z_(pₕ_³) × ⋯ × Z_(pₕ_ⁿ )

Also,

p^h¹·p^h²·p^h³ ⋯ p^hⁿ = p^m

Hence h₁ + h₂ + h₃ + ⋯ + h_n = m. Furthermore, the decomposition of the abelian group is unique up to rearrangement of the factors. Thus, there is a one-to-one correspondence between non-isomorphic abelian groups of order p^m and ways m can be written as a sum of positive integers without regard to order.

Return to text

Proof of Corollary 6.3:

We know from the fundamental theorem of abelian groups (6.2) that each such group is isomorphic to a direct product of cyclic groups whose order is a power of a prime. If we collect all factors involving the same primes together, we find that such a group is isomorphic to a direct product of a series of groups of orders

p₁^h¹, p₂^h², p₃^h³, …, and p_n^hⁿ.

We know from Corollary 6.2 that there are exactly P(x) non-isomorphic abelian groups of order p^x. Thus, there are P(h_i) possible groups for the i^th factor in this decomposition. Therefore, there are

P(h₁)·P(h₂)·P(h₃) ⋯ P(h_n)

possible ways of forming a product of groups with orders p₁^h¹, p₂^h², p₃^h³, …, and p_n^hⁿ.

Since the fundamental theorem of abelian groups (6.2) also states that the decomposition is unique up to the rearrangement of the factors, every group thus formed is isomorphically different. So we have exactly P(h₁)·P(h₂)·P(h₃) ⋯ P(h_n) non-isomorphic abelian groups of order m.

Return to text

Proof of Proposition 6.3:

The mapping i(x) = x for all x in G is obviously an automorphism on G, so the set of all automorphisms on G is non-empty. Also, each automorphism is a permutation on the elements of G. Suppose ϕ and ƒ are two automorphisms on G. Then ϕ(ƒ(x)) is a one-to-one and onto mapping from G to G.

Furthermore,

ϕ(ƒ(x·y)) = ϕ(ƒ(x)·ƒ(y)) = ϕ(ƒ(x))·ϕ(ƒ(y)).

So ϕ(ƒ(x)) is a homomorphism on G, hence ϕ·ƒ is an automorphism of G.

Also, since ƒ is one-to-one and onto, ƒ⁻¹ exists on G, and

ƒ(ƒ⁻¹(x)·ƒ⁻¹(y)) = ƒ(ƒ⁻¹(x))·ƒ(ƒ⁻¹(y)) = x·y.

Taking ƒ⁻¹ of both sides of the equation gives us

ƒ⁻¹(x)·ƒ⁻¹(y) = ƒ⁻¹(x·y)

So ƒ⁻¹ is a homomorphism. Hence both ƒ⁻¹ and ϕ·ƒ⁻¹ are automorphisms of G. Therefore by Proposition 2.2, Aut(G) is a subgroup of the group of permutations on the elements of G.

Return to text

Proof of Proposition 6.4:

Consider the mapping

ψ: Z_n^* → Aut(Z_n)

given by ψ(j) = ƒ_j, where ƒ_j(x) = (j·x) mod n. Then given two elements j and k in Z_n^*, we have that

ƒ_j(ƒ_k(x)) = j·(k·x) mod n = (j·k)·x mod n = ƒ_(j·k)(x).

So

ψ(j)·ψ(k) = ƒ_j(ƒ_k) = ƒ_(j·k) = ψ(j·k).

Hence, ψ is a homomorphism from Z_n^* to Aut(Z_n). To see that ψ is one-to-one, note that ƒ_j(1) = j, and so ƒ_j = ƒ_k only if j = k.

To see that ψ is onto, we can use the pigeon-hole principle. If we consider a general automorphism f of Z_n, then f(1) must be a generator of Z_n, since 1 is a generator. But f will be completely determined by knowing f(1). Thus, the number of automorphisms is at most the number of generators of Z_n, which is ϕ(n). Since |Z_n^*| = ϕ(n), we know the function is one-to-one, so it must also be onto.

Return to text

Proof of Proposition 6.5:

First we need to show that Inn(G) is a subgroup. Let ƒ_x(y) = x·y·x^-1 be an inner automorphism. The inverse can be easily found by observing

y ∈ ƒ_x^-1(v) ⟺ x·y·x^-1 = v ⟺ y = x^-1·v·x ⟺ y = ƒ_(x₋₁₎(v),

so the inverse of ƒ_x is also an inner automorphism.

If we consider two inner automorphisms ƒ_x and ƒ_y, then

(ƒ_x·ƒ_y)(v) = ƒ_x(ƒ_y(v)) = x·(y·v·y^-1)·x^-1 = (x·y)·v·(x·y)^-1 = ƒ_(x·y)(v).

Thus the product of two inner automorphisms is also an inner automorphism. So by Proposition 2.2, Inn(G) is a subgroup of Aut(G).

Finally, we need to show that Inn(G) is normal in Aut(G). Let ϕ be any automorphism and let ƒ_x(y) = x·y·x^-1 be an inner automorphism. Then

(ϕ·ƒ_x·ϕ^-1)(v) = ϕ(ƒ_x(ϕ^-1(v))) = ϕ(x·ϕ^-1(v)·x^-1).

Since ϕ is a homomorphism, this will simplify.

ϕ(x·ϕ^-1(v)·x^-1) = ϕ(x)·ϕ(ϕ^-1(v))·ϕ(x^-1) = ϕ(x)·v·[ϕ(x)]^-1 = ƒ_ϕ(x)(v).

So ϕ·ƒ_x·ϕ^-1 is an inner automorphism of G. Therefore, by Proposition 3.4, Inn(G) is a normal subgroup of Aut(G).

Return to text

Proof of Proposition 6.6:

It is clear that the product of two ordered pairs in G is an ordered pair in G. If we let e₁ denote the identity element of K, and e₂ denote the indentity element of H, then

ϕ_e₂(k) = k

since ϕ must map e₂ to the identity automorphism of K. Thus

(k₁, h₁)·(e₁, e₂) = (k₁·ϕ_h₁(e₁), h₁·e₂) = (k₁, h₁) and (e₁, e₂)·(k₂, h₂) = (e₁·ϕ_e₂(k₂), e₂·h₂) = (k₂, h₂).

So (e₁, e₂) acts as the identity element of G.

Next we note that the element (k, h) has an inverse (ϕ_h₋₁(k^-1), h^-1), since

(ϕ_h₋₁(k^-1), h^-1)·(k, h) = (ϕ_h₋₁(k^-1)·ϕ_h₋₁(k), h^-1·h) = (ϕ_h₋₁(k^-1·k), e₂) = (ϕ_h₋₁(e₁), e₂) = (e₁, e₂),

and

(k, h)·(ϕ_h₋₁(k^-1), h^-1) = (k·ϕ_h(ϕ_h₋₁(k^-1)), h·h^-1) = (k·ϕ_e₂(k^-1), e₂) = (k·k^-1, e₂) = (e₁, e₂).

The final thing we need to check is that the multiplication on G is associative. Note that

[(k₁, h₁)·(k₂, h₂)]·(k₃, h₃) = (k₁·ϕ_h₁(k₂), h₁·h₂)·(k₃, h₃) = (k₁·ϕ_h₁(k₂)·ϕ_h₁·h₂(k₃), (h₁·h₂)·h₃),

while

(k₁, h₁)·[(k₂, h₂)·(k₃, h₃)] = (k₁, h₁)·(k₂·ϕ_h₂(k₃), h₂·h₃) = (k₁·ϕ_h₁(k₂·ϕ_h₂(k₃)), h₁·(h₂·h₃)) = (k₁·ϕ_h₁(k₂)·ϕ_h₁(ϕ_h₂(k₃)), (h₁·h₂)·h₃) = (k₁·ϕ_h₁(k₂)·ϕ_h₁·h₂(k₃), (h₁·h₂)·h₃).

Hence the multiplication on G is associative and so G forms a group.

Return to text

Proof of Lemma 6.7:

We will use Proposition 2.2 and observe that

(e₁, h)^-1 = (ϕ_h₋₁(e₁^-1), h^-1) = (e₁, h^-1),

so

(e₁, h₁)·(e₁, h₂)^-1 = (e₁, h₁)·(e₁, h₂^-1) = (e₁·ϕ_h₁(e₁), h₁·h₂^-1) = (e₁, h₁·h₂^-1).

Thus, whenever a and b are in H, a·b^-1 is in H. So H is a subgroup.

The mapping ƒ : G → H given by

ƒ( (k, h) ) = h

is a homomorphism, since

ƒ( (k₁, h₁)·(k₂, h₂) ) = ƒ( (k₁·ϕ_h₁(k₂), h₁·h₂) ) = h₁·h₂ = ƒ( (k₁, h₁) )·ƒ( (k₂, h₂) ).

The kernel of this homomorphism is K, so K is a normal subgroup of G. By restricting the function ƒ to the set H, we find that it is one-to-one and onto. Thus, H ≈ H.

A similar function g : K → K, given by

g(k) = (k, e₂)

can show that K ≈ K. This function is clearly one-to-one and onto, and

g(k₁)·g(k₂) = (k₁, e₂)·(k₂, e₂) = (k₁·ϕ_e₂(k₂), e₂) = (k₁·k₂, e₂) = g(k₁·k₂).

Finally, it is clear that the intersections of the two groups give just {(e₁, e₂)}.

Return to text

Proof of Theorem 6.3:

Note that since H is a subgroup of G, and N is a normal subgroup we have by Lemma 4.3 that N·H is a subgroup of G. We next want to define the homomorphism ϕ. For each h in H, we define

ϕ_h(k) = h·k·h^-1

for all k ∈ N. We first need to show that ϕ_h is an automorphism on N for each h in H, and then we need to show that ϕ itself is a nontrivial homomorphism. Since N is normal, it is clear that this sends elements of N to elements of N. Note that

ϕ_h(k₁·k₂) = h·k₁·k₂·h^-1 = (h·k₁·h^-1)·(h·k₂·h^-1) = ϕ_h(k₁)·ϕ_h(k₂).

So ϕ_h is a homomorphism from N to N. Since

y ∈ ϕ_h^-1(k) ⟺ h·y·h^-1 = k ⟺ y = h^-1·k·h

we see that ϕ_h is a one-to-one and onto function. Thus, ϕ_h is an automorphism of N.

Next, we need to see that ϕ itself is a homomorphism from H to Aut(N). Note that

(ϕ_h₁·ϕ_h₂)(k) = ϕ_h₁(ϕ_h₂(k)) = ϕ_h₁(h₂·k·h₂^-1) = h₁·h₂·k·h₂^-1·h₁^-1 = (h₁·h₂)·k·(h₁·h₂)^-1 = ϕ_h₁·h₂(k).

So ϕ_h₁·ϕ_h₂ = ϕ_h₁·h₂ and we see that ϕ is a homomorphism. In fact, the homomorphism must be nontrivial, because if ϕ_h(k) = k for all h and k, then since ϕ_h(k) = h·k·h^-1 = k, we have that k·h = h·k for all h in H, and k in N. This would indicate that H is a normal subgroup of N·H, which contradicts our original assumption. Thus, ϕ is a nontrivial homomorphism.

We can now proceed in a way similar to how we proved the direct product theorem (6.1). As before, we will begin by showing that every element in N·H can be uniquely written in the form k·h, where k ∈ N and h ∈ H. Suppose that we have

k₁·h₁ = k₂·h₂.

Then k₂^-1·k₁ = h₂·h₁^-1. Since this element is in both N and H, which has just the identity element in the intersection, we must have

k₂^-1·k₁ = h₂·h₁^-1 = e.

Therefore, k₁ = k₂ and h₁ = h₂. Thus, we have shown that every element of N·H is written uniquely as k·h, where k is in N, and h is in H.

We now want to create a mapping

ƒ : N·H → N ⋊_ϕH

defined by

ƒ(x) = (k, h),

where k and h are the unique elements such that k ∈ N, h ∈ H, and x = k·h. The function ƒ is one-to-one since the element (k, h) can only come from k·h. Also, the element maps to (k, h), so ƒ is onto.

The final step is to show that ƒ is a homomorphism. Let x = k₁·h₁, and y = k₂·h₂. Then

x·y = k₁·h₁·k₂·h₂ = (k₁·h₁·k₂·h₁^-1)·(h₁·h₂).

Since N is a normal subgroup, h₁·k₂·h₁^-1 is in N, and so k₁·h₁·k₂·h₁^-1 is in N while h₁·h₂ is in H. Thus,

ƒ(x·y) = ƒ((k₁·h₁·k₂·h₁^-1)·(h₁·h₂)) = (k₁·h₁·k₂·h₁^-1, h₁·h₂) = (k₁·ϕ_h₁(k₂), h₁·h₂) = (k₁, h₁)·(k₂, h₂) = ƒ(x)·ƒ(y).

So ƒ is an isomorphism, and we have N·H ≈ N ⋊_ϕH.

Return to text

Proof of Proposition 6.7:

Let us write G = N ⋊_ϕH, and M = N ⋊_ƒ H. These are two different groups, even though they are both written using ordered pairs. Let us define a mapping

v : G → M

defined by

v( (k, h) ) =(w(k), h).

Because w(k) is one-to-one and onto, certainly v is one-to-one and onto. All we would have to check is that

v( (k₁, h₁) )·v( (k₂, h₂) ) = v( (k₁, h₁) · (k₂, h₂) )

We have that

v( (k₁, h₁) · (k₂, h₂) ) = (w(k₁), h₁)·(w(k₂), h₂) = (w(k₁)·ƒ_h₁(w(k₂)), h₁·h₂) = (w(k₁)·w(ϕ_h₁(w^-1(w(k₂)))), h₁·h₂) = (w(k₁)·w(ϕ_h₁(k₂)), h₁·h₂)

On the other hand,

v( (k₁, h₁) · (k₂, h₂) ) = v( (k₁·ϕ_h₁(k₂), h₁·h₂) ) = (w(k₁·ϕ_h₁(k₂)), h₁·h₂) = (w(k₁)·w(ϕ_h₁(k₂)), h₁·h₂).

Since these are equal, we have an isomorphism.

Return to text

Sage Interactive Problems

6.1 ) Use Sage to define the group Z₂ × Z₆. Show that this group is not isomorphic to Z₁₂.

6.2 ) Define the group S₃ × Z₂ in Sage. Show that this group is not isomorphic to A₄.
(Hint: Count the elements of order 2.)

6.3 ) Use Sage's PartitionsP command to find the number of abelian groups of order 120,000.

For Problems 6.4 through 6.7:Find all of the automorphisms of the following groups.
(Hint: For the non-abelian groups, find the inner automorphisms first.)

6.4 ) S₃.

6.5 ) Z₁₅^*.

6.6 ) D₄.

6.7 ) D₅.

6.8 ) Show that there is only one semi-direct product Z₈^* ⋊ Z₂. Which of the five groups of order 8 is this group isomorphic to?
(Hint: Use Proposition 6.7.)

6.9 ) Use Sage to find the only semi-direct product Z₈^* ⋊ Z₈^*. Is this group isomorphic to any of the three groups of order 16 found by considering Z₈ ⋊ Z₂?

6.10) Use Sage to define the only semi-direct product Z₃ ⋊ Z₄. Show that this group is different than both A₄ and S₃ × Z₂.

6.11) From Problems 6.1, 6.2, 6.10 and section 6.4, we have found six groups of order 12: Z₁₂, Z₂ × Z₆, A₄, D₆, S₃ × Z₂, and Z₃ ⋊ Z₄. Yet the table on page 85 indicates that there are only five non-isomorphic groups of order 12. Which two of these groups are isomorphic? Use Sage to show the isomorphism.