Group Actions

Let $K$ be any set and $G$ be a group with binary operation $\cdot$ and identity element $1$ .

Definition 1. A group action on $K$ by $G$ is an induced map $* : G \times K \to K$ such that

$(g \cdot h) * x = g * (h * x)$ for $g,h \in G$ and $x \in K$ ,
$1 * x = x$ for $x \in K$ .

In this case, we say that $G$ acts on $K$ via the map $*$ . If $K = G$ , we say that $G$ acts on itself via the map $*$ .

Denote $\mathrm{Sub}(G) := \{H \in \mathcal P(G) : H \leq G\}$ . For any $H \in \mathrm{Sub}(G)$ , recall that $G/H$ consists of all sets of the form $g H$ . If $K \triangleleft G$ , then $G/H$ forms a subgroup.

Problem 1. Show that $G$ acts on $G/H$ via the map $u * vH = (u \cdot v)H$ .

(Click for Solution)

Solution. We verify Definition 1. Fix $u,v,w \in G$ so that $wH \in G/H$ . Then

$\begin{aligned} (u \cdot v) * wH &= ((u \cdot v) \cdot w)H \\ &= (u \cdot (v \cdot w))H \\ &= u * (v \cdot w)H \\ &= u *(v * wH). \end{aligned}$

Similarly, $1 * wH = (1 \cdot w)H = wH$ .

Remark 1. Setting $H = \{1\}$ , $G$ acts on itself via the map $g * h = g \cdot h$ .

Problem 2. Show that $G$ acts on $\mathrm{Sub}(G)$ via the map $g * H = g H g^{-1}$ .

(Click for Solution)

Solution. We verify Definition 1. Fix $u, v \in G$ and $H \in K$ . Then

$\begin{aligned} (u \cdot v) * H &= (u \cdot v) H (u \cdot v)^{-1} \\ &= (u \cdot v) H (v^{-1} \cdot u^{-1}) \\ &= u \cdot (v H v^{-1}) \cdot u^{-1} \\ &= u \cdot (v * H) \cdot u^{-1} \\ &= u * (v * H). \end{aligned}$

Similarly, $1 * H = 1 H 1^{-1} = H$ .

Remark 2. Setting $H = \{h\}$ and applying similar reasoning, $G$ acts on itself via the map $g * h = g \cdot h \cdot g^{-1}$ .

Problem 3. Given $x , y \in K$ , denote $x \sim y$ to mean that there exists $g \in G$ such that $y = g * x$ . Show that $\sim$ forms an equivalence relation on $K$ .

(Click for Solution)

Solution. We check the three properties of an equivalence relation.

For reflexivity, for any $x \in K$ , $x = 1 * x$ by Definition 1, so that $x \sim x$ , as required.

For symmetry, suppose $x \sim y$ . Then there exists $g \in G$ such that $y = g*x$ . By Definition 1,

$\begin{aligned} x = 1 * x = (g^{-1} \cdot g) * x = g^{-1} * (g * x) = g^{-1} * y \end{aligned}$

implies that $y \sim x$ , as required.

For transitivity, suppose $x \sim y$ and $y \sim z$ . Then there exist $g, h \in G$ such that

$y = g * x,\quad z = h * y.$

By composition and Definition 1,

$z = h * y = h * (g * x) = (h \cdot g) * x$

implies that $x \sim z$ , as required.

Using Problem 3, for each $x \in K$ , define the orbit $\mathcal O_x := [x] \in K/{\sim}$ . It is not hard to check that

$\mathcal O_x = \{g * x : g \in G\} .$

Then $\sim$ partitions $K$ into

$\displaystyle K = \bigsqcup_{\mathcal O_x \in K/{\sim}} \mathcal O_x.$

Problem 4. For any $x \in K$ , show that $G_x \leq G$ , where

$G_x := \{g \in G : g * x = x\}.$

Furthermore, show that for any $g \in G$ ,

$g \cdot G_x \cdot g^{-1} = G_{g * x}.$

Deduce that $|\mathcal O_x| = |G/G_x|$ , i.e., the orbit-stabilizer formula.

(Click for Solution)

Solution. We verify the group axioms for $G_x$ .

For closure, given $g,h \in G_x$ ,

$(g \cdot h)*x = g * (h*x) = g*x = x$

implies that $g \cdot h \in G_x$ . The other three properties follow similarly.

For the conjugate-like claim, fix $g \in G$ . Expand both sets:

$\begin{aligned} g \cdot G_x \cdot g^{-1} &= \{g \cdot h \cdot g^{-1} : h \in G_x\}, \\ G_{g*x} &= \{k \in G : k * (g * x) = g * x\}. \end{aligned}$

We observe that for any $h \in G_x$ ,

$\begin{aligned} (g\cdot h\cdot g^{-1}) * (g*x) &= g *(h * (g^{-1} * (g*x))) \\ &= g *(h * ((g^{-1} \cdot g) *x )) \\ &= g *(h * (1 *x )) \\ &= g *(h * x) \\ &= g * x. \end{aligned}$

We leave it as an exercise to therefore check that equality holds.

Finally, define the map $f : G \to K$ by $f(g) = g * x$ . Then $f(g) = f(h)$ if and only if $h \in g G_x$ , inducing a bijection $\phi : G/G_x \to f(G)$ . By definition,

$f(G) = \{ f(g) : g \in G \} = \{ g * x : g \in G \} = \mathcal O_x.$

Therefore, $|\mathcal O_x| = |G/G_x|$ .

Remark 3. In particular, if $K$ is finite, then

$\displaystyle |K| = \sum_{\mathcal O_x \in K/{\sim}} |G/G_x|.$

Now suppose that $G$ is finite and Abelian. Denote its identity element by $1_G$ .

Problem 5. Let $p$ be a prime number such that $p \mid |G|$ . Show that $G$ contains an element of order $p$ . The subgroup it generates, clearly, also has order $p$ . This result is known as Cauchy’s theorem.

(Click for Solution)

Solution. For each $g \in G$ , denote its order by $o(g)$ . If there exists an element $g$ such that $p \mid o(g)$ , then defining $n := o(g)/p$ , the element $g^n$ is the desired element. Suppose no such element exists. Define

$n_G := \mathrm{lcm}\{o(g) : g \in G\}.$

In particular, $p \nmid n_G$ . We prove by induction on $|G|$ that $|G| \mid n_G^k$ for some integer $k$ . The base case $|G|$ is obvious. For the inductive step, fix $g \in G \backslash \{1_G\}$ . By definition, $\langle g \rangle$ implies that $|\langle g \rangle| \mid n_G$ . By assumption, $p \nmid |\langle g \rangle|$ and $p \mid |G|$ , which implies that $|\langle g \rangle| \neq |G|$ . In particular, for any $x \in G$ , $(x \langle g \rangle)^{n_G} = 1_{G/\langle g \rangle}$ , so that $n_{G/\langle g \rangle} \mid n_G$ .