## Section14.1Groups 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 $$G\text{.}$$

###### 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.

###### 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:

\begin{equation*} \{ (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{.} \end{equation*}

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

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

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

###### 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:

\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*}

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

###### 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 $$G\text{,}$$

\begin{equation*} H \times G \rightarrow G\text{,} \end{equation*}

via

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

for $$h \in H$$ and $$g \in G\text{.}$$ 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.

###### 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

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

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.

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

###### Example14.7.

Let $$G$$ be the permutation group defined by

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

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 $${\mathcal O}_4 = {\mathcal O}_5 = \{4, 5\}\text{.}$$

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 $$G_x\text{.}$$

###### Remark14.8.

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

###### Example14.9.

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

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

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 $$x \in X\text{.}$$

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 $$G\text{.}$$

We know that $$|G|/|G_x|$$ is the number of left cosets of $$G_x$$ in $$G$$ by Lagrange's Theorem (TheoremĀ 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

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

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 $$g_2 = g_1 g\text{,}$$

\begin{equation*} y_2 = g_2 x = g_1 g x = g_1 x = y_1; \end{equation*}

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