14.1: Groups Acting on Sets

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 $G\text{.}$

Example $14.1$

Let $G = GL_2( {\mathbb R} )$ and $X = {\mathbb R}^2\text{.}$

Solution

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.

Example $14.2$

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:

$$\{ (1), (1 \, 3), (2 \, 4), (1 \, 4 \, 3 \, 2), (1 \, 2 \, 3 \, 4), (1 \, 2)(3 \, 4), (1 \, 4)(2 \, 3), (1 \, 3)(2 \, 4) \}\text{.} \nonumber$$

Solution

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

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

$$(\sigma, x) \mapsto \sigma(x) \nonumber$$

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

Example $14.3$

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:

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

Solution

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

Example $14.4$

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 $G\text{,}$

$$H \times G \rightarrow G\text{,} \nonumber$$

via

$$(h,g) \mapsto hgh^{-1} \nonumber$$ for $h \in H$ and $g \in G\text{.}$

Solution

Clearly, the first axiom for a group action holds. Observing that

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

we see that the second condition is also satisfied.

Example $14.5$

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

$$(g, xH) \mapsto gxH\text{.} \nonumber$$

Solution

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.

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.

Proposition $14.6$

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 $z\text{.}$

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 ${\mathcal O}_x\text{.}$

Example $14.7$

Let $G$ be the permutation group defined by

$$G =\{(1), (1 \, 2 \, 3), (1 \, 3 \, 2), (4 \, 5), (1 \, 2 \, 3)(4 \, 5), (1 \, 3 \, 2)(4 \, 5) \} \nonumber$$

and $X = \{ 1, 2, 3, 4, 5\}\text{.}$

Solution

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

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 $G_x\text{.}$

Remark $14.8$

The ideal gas law is easy to remember and apply in solving problems, as long as you get the proper values a

Example $14.9$

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

$$\{ (1), (1 \, 2)(3 \, 4 \, 5 \, 6), (3 \, 5)(4 \, 6), (1 \, 2)( 3 \, 6 \, 5 \, 4) \}\text{.} \nonumber$$

Solution

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

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

and the stabilizer subgroups are

$$\begin{gather*} G_1 = G_2 = \{(1), (3 \, 5)(4 \, 6) \},\\ G_3 = G_4 = G_5 = G_6 = \{(1)\}\text{.} \end{gather*}$$

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

Proposition $14.10$

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 $G_x\text{.}$

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 $G\text{.}$

Theorem $14.11$

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

$$\phi(y_1) = g_1 G_x = g_2 G_x = \phi(y_2)\text{,} \nonumber$$

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 $g_2 = g_1 g\text{,}$

$$y_2 = g_2 x = g_1 g x = g_1 x = y_1; \nonumber$$

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 $\phi(y) = g G_x\text{.}$