# Section14.1Groups Acting on Sets¶ permalink

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 $(g,x) \mapsto gx$, where

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

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

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 $(g,x) \mapsto x$; however, group actions are more interesting if the set $X$ is somehow related to the group $G$.

##### Example14.1

Let $G = GL_2( {\mathbb R} )$ and $X = {\mathbb R}^2$. Then $G$ acts on $X$ by left multiplication. If $v \in {\mathbb R}^2$ and $I$ is the identity matrix, then $Iv = v$. 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), (13), (24), (1432), (1234), (12)(34), (14)(23), (13)(24) \}.\end{equation*} 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.

In general, if $X$ is any set and $G$ is a subgroup of $S_X$, the group of all permutations acting on $X$, then $X$ is a $G$-set under the group action \begin{equation*}(\sigma, x) \mapsto \sigma(x)\end{equation*} for $\sigma \in G$ and $x \in X$.

##### Example14.3

If we let $X = G$, then every group $G$ acts on itself by the left regular representation; that is, $(g,x) \mapsto \lambda_g(x) = gx$, 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). \end{gather*} If $H$ is a subgroup of $G$, then $G$ is an $H$-set under left multiplication by elements of $H$.

##### Example14.4

Let $G$ be a group and suppose that $X=G$. If $H$ is a subgroup of $G$, then $G$ is an $H$-set under conjugation; that is, we can define an action of $H$ on $G$, \begin{equation*}H \times G \rightarrow G,\end{equation*} via \begin{equation*}(h,g) \mapsto hgh^{-1}\end{equation*} for $h \in H$ and $g \in G$. 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) ), \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 $H$. The set ${\mathcal L}_H$ is a $G$-set under the action \begin{equation*}(g, xH) \mapsto gxH.\end{equation*} Again, it is easy to see that the first axiom is true. Since $(g g')xH = g( g'x H)$, the second axiom is also true.

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

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

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

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

##### Remark14.8

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

##### 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) \}.\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, \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)\}. \end{gather*} It is easily seen that $G_x$ is a subgroup of $G$ for each $x \in X$.

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 $|{\mathcal O}_x|$. The next theorem demonstrates the relationship between orbits of an element $x \in X$ and the left cosets of $G_x$ in $G$.