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Section14.1Groups Acting on Sets

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Let $X$ be a set and $G$ be a group. A (left) action of $G$ on $X$ is a map $G \\times X \\rightarrow X$ given by $(g,x) \\mapsto gx\\text{,}$ where

  1. $ex = x$ for all $x \\in X\\text{;}$

  2. $(g_1 g_2)x = g_1(g_2 x)$ for all $x \\in X$ and all $g_1, g_2 \\in G\\text{.}$

Under these considerations $X$ is called a $G$-set. Notice that we are not requiring $X$ to be related to $G$ in any way. It is true that every group $G$ acts on every set $X$ by the trivial action $(g,x) \\mapsto x\\text{;}$ however, group actions are more interesting if the set $X$ is somehow related to the group $G\\text{.}$

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Example14.1

Let $G = GL_2( {\\mathbb R} )$ and $X = {\\mathbb R}^2\\text{.}$ Then $G$ acts on $X$ by left multiplication. If $v \\in {\\mathbb R}^2$ and $I$ is the identity matrix, then $Iv = v\\text{.}$ If $A$ and $B$ are $2 \\times 2$ invertible matrices, then $(AB)v = A(Bv)$ since matrix multiplication is associative.

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Example14.2

Let $G = D_4$ be the symmetry group of a square. If $X = \\{ 1, 2, 3, 4 \\}$ is the set of vertices of the square, then we can consider $D_4$ to consist of the following permutations:

\n\\begin{equation*}\n\\{ (1), (13), (24), (1432), (1234), (12)(34), (14)(23), (13)(24) \\}.\n\\end{equation*}\n

The elements of $D_4$ act on $X$ as functions. The permutation $(13)(24)$ acts on vertex 1 by sending it to vertex 3, on vertex 2 by sending it to vertex 4, and so on. It is easy to see that the axioms of a group action are satisfied.

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In general, if $X$ is any set and $G$ is a subgroup of $S_X\\text{,}$ the group of all permutations acting on $X\\text{,}$ then $X$ is a $G$-set under the group action

\n\\begin{equation*}\n(\\sigma, x) \\mapsto \\sigma(x)\n\\end{equation*}\n

for $\\sigma \\in G$ and $x \\in X\\text{.}$

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Example14.3

If we let $X = G\\text{,}$ then every group $G$ acts on itself by the left regular representation; that is, $(g,x) \\mapsto \\lambda_g(x) = gx\\text{,}$ where $\\lambda_g$ is left multiplication:

\n\\begin{gather*}\ne \\cdot x = \\lambda_e x = ex = x\\\\\n(gh) \\cdot x = \\lambda_{gh}x = \\lambda_g \\lambda_h x = \\lambda_g(hx) = g \\cdot ( h \\cdot x).\n\\end{gather*}\n

If $H$ is a subgroup of $G\\text{,}$ then $G$ is an $H$-set under left multiplication by elements of $H\\text{.}$

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Example14.4

Let $G$ be a group and suppose that $X=G\\text{.}$ If $H$ is a subgroup of $G\\text{,}$ then $G$ is an $H$-set under conjugation; that is, we can define an action of $H$ on $G\\text{,}$

\n\\begin{equation*}\nH \\times G \\rightarrow G,\n\\end{equation*}\n

via

\n\\begin{equation*}\n(h,g) \\mapsto hgh^{-1}\n\\end{equation*}\n

for $h \\in H$ and $g \\in G\\text{.}$ Clearly, the first axiom for a group action holds. Observing that

\n\\begin{align*}\n(h_1 h_2, g) & = h_1 h_2 g (h_1 h_2 )^{-1}\\\\\n& = h_1( h_2 g h_2^{-1}) h_1^{-1}\\\\\n& = (h_1, (h_2, g) ),\n\\end{align*}\n

we see that the second condition is also satisfied.

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Example14.5

Let $H$ be a subgroup of $G$ and ${\\mathcal L}_H$ the set of left cosets of $H\\text{.}$ The set ${\\mathcal L}_H$ is a $G$-set under the action

\n\\begin{equation*}\n(g, xH) \\mapsto gxH.\n\\end{equation*}\n

Again, it is easy to see that the first axiom is true. Since $(g g')xH = g( g'x H)\\text{,}$ the second axiom is also true.

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If $G$ acts on a set $X$ and $x, y \\in X\\text{,}$ then $x$ is said to be $G$-equivalent to $y$ if there exists a $g \\in G$ such that $gx =y\\text{.}$ We write $x \\sim_G y$ or $x \\sim y$ if two elements are $G$-equivalent.

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Proposition14.6

Let $X$ be a $G$-set. Then $G$-equivalence is an equivalence relation on $X\\text{.}$

Proof

The relation $\\sim$ is reflexive since $ex = x\\text{.}$ Suppose that $x \\sim y$ for $x, y \\in X\\text{.}$ Then there exists a $g$ such that $gx = y\\text{.}$ In this case $g^{-1}y=x\\text{;}$ hence, $y \\sim x\\text{.}$ To show that the relation is transitive, suppose that $x \\sim y$ and $y \\sim z\\text{.}$ Then there must exist group elements $g$ and $h$ such that $gx = y$ and $hy= z\\text{.}$ So $z = hy = (hg)x\\text{,}$ and $x$ is equivalent to $z\\text{.}$

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If $X$ is a $G$-set, then each partition of $X$ associated with $G$-equivalence is called an orbit of $X$ under $G\\text{.}$ We will denote the orbit that contains an element $x$ of $X$ by ${\\mathcal O}_x\\text{.}$

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Example14.7

Let $G$ be the permutation group defined by

\n\\begin{equation*}\nG =\\{(1), (1 2 3), (1 3 2), (4 5), (1 2 3)(4 5), (1 3 2)(4 5) \\}\n\\end{equation*}\n

and $X = \\{ 1, 2, 3, 4, 5\\}\\text{.}$ Then $X$ is a $G$-set. The orbits are ${\\mathcal O}_1 = {\\mathcal O}_2 = {\\mathcal O}_3 =\\{1, 2, 3\\}$ and ${\\mathcal O}_4 = {\\mathcal O}_5 = \\{4, 5\\}\\text{.}$

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Now suppose that $G$ is a group acting on a set $X$ and let $g$ be an element of $G\\text{.}$ The fixed point set of $g$ in $X\\text{,}$ denoted by $X_g\\text{,}$ is the set of all $x \\in X$ such that $gx = x\\text{.}$ We can also study the group elements $g$ that fix a given $x \\in X\\text{.}$ This set is more than a subset of $G\\text{,}$ it is a subgroup. This subgroup is called the stabilizer subgroup or isotropy subgroup of $x\\text{.}$ We will denote the stabilizer subgroup of $x$ by $G_x\\text{.}$

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Remark14.8

It is important to remember that $X_g \\subset X$ and $G_x \\subset G\\text{.}$

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Example14.9

Let $X = \\{1, 2, 3, 4, 5, 6\\}$ and suppose that $G$ is the permutation group given by the permutations

\n\\begin{equation*}\n\\{ (1), (1 2)(3 4 5 6), (3 5)(4 6), (1 2)( 3 6 5 4) \\}.\n\\end{equation*}\n

Then the fixed point sets of $X$ under the action of $G$ are

\n\\begin{gather*}\nX_{(1)} = X,\\\\\nX_{(3 5)(4 6)} = \\{1,2\\},\\\\\nX_{(1 2)(3 4 5 6)} = X_{(1 2)(3 6 5 4)} = \\emptyset,\n\\end{gather*}\n

and the stabilizer subgroups are

\n\\begin{gather*}\nG_1 = G_2 = \\{(1), (3 5)(4 6) \\},\\\\\nG_3 = G_4 = G_5 = G_6 = \\{(1)\\}.\n\\end{gather*}\n

It is easily seen that $G_x$ is a subgroup of $G$ for each $x \\in X\\text{.}$

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Proposition14.10

Let $G$ be a group acting on a set $X$ and $x \\in X\\text{.}$ The stabilizer group of $x\\text{,}$ $G_x\\text{,}$ is a subgroup of $G\\text{.}$

Proof

Clearly, $e \\in G_x$ since the identity fixes every element in the set $X\\text{.}$ Let $g, h \\in G_x\\text{.}$ Then $gx = x$ and $hx = x\\text{.}$ So $(gh)x = g(hx) = gx = x\\text{;}$ hence, the product of two elements in $G_x$ is also in $G_x\\text{.}$ Finally, if $g \\in G_x\\text{,}$ then $x = ex = (g^{-1}g)x = (g^{-1})gx = g^{-1} x\\text{.}$ So $g^{-1}$ is in $G_x\\text{.}$

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We will denote the number of elements in the fixed point set of an element $g \\in G$ by $|X_g|$ and denote the number of elements in the orbit of $x \\in X$ by $|{\\mathcal O}_x|\\text{.}$ The next theorem demonstrates the relationship between orbits of an element $x \\in X$ and the left cosets of $G_x$ in $G\\text{.}$

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Theorem14.11

Let $G$ be a finite group and $X$ a finite $G$-set. If $x \\in X\\text{,}$ then $|{\\mathcal O}_x| = [G:G_x]\\text{.}$

Proof

We know that $|G|/|G_x|$ is the number of left cosets of $G_x$ in $G$ by Lagrange's Theorem (Theorem 6.10). We will define a bijective map $\\phi$ between the orbit ${\\mathcal O}_x$ of $X$ and the set of left cosets ${\\mathcal L}_{G_x}$ of $G_x$ in $G\\text{.}$ Let $y \\in {\\mathcal O}_x\\text{.}$ Then there exists a $g$ in $G$ such that $g x = y\\text{.}$ Define $\\phi$ by $\\phi( y ) = g G_x\\text{.}$ To show that $\\phi$ is one-to-one, assume that $\\phi(y_1) = \\phi(y_2)\\text{.}$ Then

\n\\begin{equation*}\n\\phi(y_1) = g_1 G_x = g_2 G_x = \\phi(y_2),\n\\end{equation*}\n

where $g_1 x = y_1$ and $g_2 x = y_2\\text{.}$ Since $g_1 G_x = g_2 G_x\\text{,}$ there exists a $g \\in G_x$ such that $g_2 = g_1 g\\text{,}$

\n\\begin{equation*}\ny_2 = g_2 x = g_1 g x = g_1 x = y_1;\n\\end{equation*}\n

consequently, the map $\\phi$ is one-to-one. Finally, we must show that the map $\\phi$ is onto. Let $g G_x$ be a left coset. If $g x = y\\text{,}$ then $\\phi(y) = g G_x\\text{.}$

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