Often a subgroup will depend entirely on a single element of the group; that is, knowing that particular element will allow us to compute any other element in the subgroup.
Suppose that we consider $3 \\in {\\mathbb Z}$ and look at all multiples (both positive and negative) of 3. As a set, this is
It is easy to see that $3 {\\mathbb Z}$ is a subgroup of the integers. This subgroup is completely determined by the element 3 since we can obtain all of the other elements of the group by taking multiples of 3. Every element in the subgroup is “generated” by 3.
If $H = \\{ 2^n : n \\in {\\mathbb Z} \\}\\text{,}$ then $H$ is a subgroup of the multiplicative group of nonzero rational numbers, ${\\mathbb Q}^*\\text{.}$ If $a = 2^m$ and $b = 2^n$ are in $H\\text{,}$ then $ab^{-1} = 2^m 2^{-n} = 2^{m-n}$ is also in $H\\text{.}$ By Proposition 3.31, $H$ is a subgroup of ${\\mathbb Q}^*$ determined by the element 2.
Let $G$ be a group and $a$ be any element in $G\\text{.}$ Then the set
is a subgroup of $G\\text{.}$ Furthermore, $\\langle a \\rangle$ is the smallest subgroup of $G$ that contains $a\\text{.}$
The identity is in $\\langle a \\rangle $ since $a^0 = e\\text{.}$ If $g$ and $h$ are any two elements in $\\langle a \\rangle \\text{,}$ then by the definition of $\\langle a \\rangle$ we can write $g = a^m$ and $h = a^n$ for some integers $m$ and $n\\text{.}$ So $gh = a^m a^n = a^{m+n}$ is again in $\\langle a \\rangle \\text{.}$ Finally, if $g = a^n$ in $\\langle a \\rangle \\text{,}$ then the inverse $g^{-1} = a^{-n}$ is also in $\\langle a \\rangle \\text{.}$ Clearly, any subgroup $H$ of $G$ containing $a$ must contain all the powers of $a$ by closure; hence, $H$ contains $\\langle a \\rangle \\text{.}$ Therefore, $\\langle a \\rangle $ is the smallest subgroup of $G$ containing $a\\text{.}$
If we are using the “+” notation, as in the case of the integers under addition, we write $\\langle a \\rangle = \\{ na : n \\in {\\mathbb Z} \\}\\text{.}$
For $a \\in G\\text{,}$ we call $\\langle a \\rangle $ the cyclic subgroup generated by $a\\text{.}$ If $G$ contains some element $a$ such that $G = \\langle a \\rangle \\text{,}$ then $G$ is a cyclic group. In this case $a$ is a generator of $G\\text{.}$ If $a$ is an element of a group $G\\text{,}$ we define the order of $a$ to be the smallest positive integer $n$ such that $a^n= e\\text{,}$ and we write $|a| = n\\text{.}$ If there is no such integer $n\\text{,}$ we say that the order of $a$ is infinite and write $|a| = \\infty$ to denote the order of $a\\text{.}$
Notice that a cyclic group can have more than a single generator. Both 1 and 5 generate ${\\mathbb Z}_6\\text{;}$ hence, ${\\mathbb Z}_6$ is a cyclic group. Not every element in a cyclic group is necessarily a generator of the group. The order of $2 \\in {\\mathbb Z}_6$ is 3. The cyclic subgroup generated by 2 is $\\langle 2 \\rangle = \\{ 0, 2, 4 \\}\\text{.}$
The groups ${\\mathbb Z}$ and ${\\mathbb Z}_n$ are cyclic groups. The elements 1 and $-1$ are generators for ${\\mathbb Z}\\text{.}$ We can certainly generate ${\\mathbb Z}_n$ with 1 although there may be other generators of ${\\mathbb Z}_n\\text{,}$ as in the case of ${\\mathbb Z}_6\\text{.}$
The group of units, $U(9)\\text{,}$ in ${\\mathbb Z}_9$ is a cyclic group. As a set, $U(9)$ is $\\{ 1, 2, 4, 5, 7, 8 \\}\\text{.}$ The element 2 is a generator for $U(9)$ since
Not every group is a cyclic group. Consider the symmetry group of an equilateral triangle $S_3\\text{.}$ The multiplication table for this group is Table 3.7. The subgroups of $S_3$ are shown in Figure 4.8. Notice that every subgroup is cyclic; however, no single element generates the entire group.
![\"\"](\"images/cyclic-s3-subgroups.svg\")
Every cyclic group is abelian.
Let $G$ be a cyclic group and $a \\in G$ be a generator for $G\\text{.}$ If $g$ and $h$ are in $G\\text{,}$ then they can be written as powers of $a\\text{,}$ say $g = a^r$ and $h = a^s\\text{.}$ Since
$G$ is abelian.
We can ask some interesting questions about cyclic subgroups of a group and subgroups of a cyclic group. If $G$ is a group, which subgroups of $G$ are cyclic? If $G$ is a cyclic group, what type of subgroups does $G$ possess?
Every subgroup of a cyclic group is cyclic.
The main tools used in this proof are the division algorithm and the Principle of Well-Ordering. Let $G$ be a cyclic group generated by $a$ and suppose that $H$ is a subgroup of $G\\text{.}$ If $H = \\{ e \\}\\text{,}$ then trivially $H$ is cyclic. Suppose that $H$ contains some other element $g$ distinct from the identity. Then $g$ can be written as $a^n$ for some integer $n\\text{.}$ Since $H$ is a subgroup, $g^{-1} = a^{-n}$ must also be in $H\\text{.}$ Since either $n$ or $-n$ is positive, we can assume that $H$ contains positive powers of $a$ and $n \\gt 0\\text{.}$ Let $m$ be the smallest natural number such that $a^m \\in H\\text{.}$ Such an $m$ exists by the Principle of Well-Ordering.
We claim that $h = a^m$ is a generator for $H\\text{.}$ We must show that every $h' \\in H$ can be written as a power of $h\\text{.}$ Since $h' \\in H$ and $H$ is a subgroup of $G\\text{,}$ $h' = a^k$ for some integer $k\\text{.}$ Using the division algorithm, we can find numbers $q$ and $r$ such that $k = mq +r$ where $0 \\leq r \\lt m\\text{;}$ hence,
So $a^r = a^k h^{-q}\\text{.}$ Since $a^k$ and $h^{-q}$ are in $H\\text{,}$ $a^r$ must also be in $H\\text{.}$ However, $m$ was the smallest positive number such that $a^m$ was in $H\\text{;}$ consequently, $r=0$ and so $k=mq\\text{.}$ Therefore,
and $H$ is generated by $h\\text{.}$
The subgroups of ${\\mathbb Z}$ are exactly $n{\\mathbb Z}$ for $n = 0, 1, 2,\\ldots\\text{.}$
Let $G$ be a cyclic group of order $n$ and suppose that $a$ is a generator for $G\\text{.}$ Then $a^k=e$ if and only if $n$ divides $k\\text{.}$
First suppose that $a^k=e\\text{.}$ By the division algorithm, $k = nq + r$ where $0 \\leq r \\lt n\\text{;}$ hence,
Since the smallest positive integer $m$ such that $a^m = e$ is $n\\text{,}$ $r= 0\\text{.}$
Conversely, if $n$ divides $k\\text{,}$ then $k=ns$ for some integer $s\\text{.}$ Consequently,
Let $G$ be a cyclic group of order $n$ and suppose that $a \\in G$ is a generator of the group. If $b = a^k\\text{,}$ then the order of $b$ is $n/d\\text{,}$ where $d = \\gcd(k,n)\\text{.}$
We wish to find the smallest integer $m$ such that $e = b^m = a^{km}\\text{.}$ By Proposition 4.12, this is the smallest integer $m$ such that $n$ divides $km$ or, equivalently, $n/d$ divides $m(k/d)\\text{.}$ Since $d$ is the greatest common divisor of $n$ and $k\\text{,}$ $n/d$ and $k/d$ are relatively prime. Hence, for $n/d$ to divide $m(k/d)$ it must divide $m\\text{.}$ The smallest such $m$ is $n/d\\text{.}$
The generators of ${\\mathbb Z}_n$ are the integers $r$ such that $1 \\leq r \\lt n$ and $\\gcd(r,n) = 1\\text{.}$
Let us examine the group ${\\mathbb Z}_{16}\\text{.}$ The numbers 1, 3, 5, 7, 9, 11, 13, and 15 are the elements of ${\\mathbb Z}_{16}$ that are relatively prime to 16. Each of these elements generates ${\\mathbb Z}_{16}\\text{.}$ For example,