# Groups: Group Actions

## Fall 2012 #1#algebra/qual/work

Let $$G$$ be a finite group and $$X$$ a set on which $$G$$ acts.

• Let $$x\in X$$ and $$G_x \coloneqq\left\{{g\in G {~\mathrel{\Big\vert}~}g\cdot x = x}\right\}$$. Show that $$G_x$$ is a subgroup of $$G$$.

• Let $$x\in X$$ and $$G\cdot x \coloneqq\left\{{g\cdot x {~\mathrel{\Big\vert}~}g\in G}\right\}$$. Prove that there is a bijection between elements in $$G\cdot x$$ and the left cosets of $$G_x$$ in $$G$$.

## Fall 2015 #2#algebra/qual/work

Let $$G$$ be a finite group, $$H$$ a $$p{\hbox{-}}$$subgroup, and $$P$$ a sylow $$p{\hbox{-}}$$subgroup for $$p$$ a prime. Let $$H$$ act on the left cosets of $$P$$ in $$G$$ by left translation.

Prove that this is an orbit under this action of length 1.

Prove that $$xP$$ is an orbit of length 1 $$\iff H$$ is contained in $$xPx^{-1}$$.

## Spring 2016 #5#algebra/qual/work

Let $$G$$ be a finite group acting on a set $$X$$. For $$x\in X$$, let $$G_x$$ be the stabilizer of $$x$$ and $$G\cdot x$$ be the orbit of $$x$$.

• Prove that there is a bijection between the left cosets $$G/G_x$$ and $$G\cdot x$$.

• Prove that the center of every finite $$p{\hbox{-}}$$group $$G$$ is nontrivial by considering that action of $$G$$ on $$X=G$$ by conjugation.

## Fall 2017 #1#algebra/qual/work

Suppose the group $$G$$ acts on the set $$A$$. Assume this action is faithful (recall that this means that the kernel of the homomorphism from $$G$$ to $$\operatorname{Sym}^*(A)$$ which gives the action is trivial) and transitive (for all $$a, b$$ in $$A$$, there exists $$g$$ in $$G$$ such that $$g \cdot a = b$$.)

• For $$a \in A$$, let $$G_a$$ denote the stabilizer of $$a$$ in $$G$$. Prove that for any $$a \in A$$, \begin{align*} \bigcap_{\sigma\in G} \sigma G_a \sigma^{-1}= \left\{{1}\right\} .\end{align*}

• Suppose that $$G$$ is abelian. Prove that $$|G| = |A|$$. Deduce that every abelian transitive subgroup of $$S_n$$ has order $$n$$.

## Fall 2018 #2#algebra/qual/completed

• Suppose the group $$G$$ acts on the set $$X$$ . Show that the stabilizers of elements in the same orbit are conjugate.

• Let $$G$$ be a finite group and let $$H$$ be a proper subgroup. Show that the union of the conjugates of $$H$$ is strictly smaller than $$G$$, i.e. \begin{align*} \bigcup_{g\in G} gHg^{-1}\subsetneq G \end{align*}

• Suppose $$G$$ is a finite group acting transitively on a set $$S$$ with at least 2 elements. Show that there is an element of $$G$$ with no fixed points in $$S$$.

• Orbit: $$G\cdot x \coloneqq\left\{{g\cdot x {~\mathrel{\Big\vert}~}g\in G}\right\} \subseteq X$$
• Stabilizer: $$G_x \coloneqq\left\{{g\in G{~\mathrel{\Big\vert}~}g\cdot x = x}\right\} \leq G$$
• Orbit-Stabilizer: $$G\cdot x \simeq G/G_x$$.
• $$abc\in H \iff b\in a^{-1}H c^{-1}$$
• Set of orbits for $$G\curvearrowright X$$, notated $$X/G$$.
• Set of fixed points for $$G\curvearrowright X$$, notated $$X^g$$.
• Burnside’s Lemma: $${\left\lvert {X/G} \right\rvert} \cdot {\left\lvert {G} \right\rvert} = \sum_{g\in G} {\left\lvert {X^g} \right\rvert}$$
• Number of orbits equals average number of fixed points.

• Fix $$x$$, then $$y\in {\mathrm{Orb}}(x) \implies g\cdot x = y$$ for some $$g$$, and $$x = g^{-1}\cdot y$$.
• Then \begin{align*} h \in {\operatorname{Stab}}(x) &\iff h\cdot x = x && \text{by being in the stabilizer} \\ &\iff h\cdot (g^{-1}\cdot y) = g^{-1}\cdot y \\ &\iff (g h g^{-1}) \cdot y = y \\ &\iff ghg^{-1}\in G_y && \text{by definition}\\ &\iff h\in g ^{-1} {\operatorname{Stab}}(y) g ,\end{align*} so $${\operatorname{Stab}}(x) = g^{-1}{\operatorname{Stab}}(y) g$$.

Let $$G$$ act on its subgroups by conjugation,

• The orbit $$G\cdot H$$ is the set of all subgroups conjugate to $$H$$, and

• The stabilizer of $$H$$ is $$G_H = N_G(H)$$.

• By orbit-stabilizer, \begin{align*} G\cdot H = [G: G_H] = [G: N_G(H)] .\end{align*}

• Since $${\left\lvert {H} \right\rvert} = n$$, and all of its conjugate also have order $$n$$.

• Note that \begin{align*} H\leq N_G(H) \implies {\left\lvert {H} \right\rvert} \leq {\left\lvert {N_G(H)} \right\rvert} \implies {1\over {\left\lvert {N_G(H)} \right\rvert}} \leq {1\over {\left\lvert {H} \right\rvert}} ,\end{align*}

• Now strictly bound the size of the union by overcounting their intersections at the identity: \begin{align*} {\left\lvert {\bigcup_{g\in G}gHg^{-1}} \right\rvert} &< (\text{Number of Conjugates of } H) \cdot (\text{Size of each conjugate}) \\ & \text{strictly overcounts since they intersect in at least the identity} \\ &= [G: N_G(H)] {\left\lvert {H} \right\rvert} \\ &= {{\left\lvert {G} \right\rvert} \over {\left\lvert {N_G(H)} \right\rvert}} {\left\lvert {H} \right\rvert} \\ & \text{since $G$ is finite} \\ &\leq {{\left\lvert {G} \right\rvert} \over {\left\lvert {H} \right\rvert}} {\left\lvert {H} \right\rvert} \\ &= {\left\lvert {G} \right\rvert} .\end{align*}

• Let $$G\curvearrowright X$$ transitively where $${\left\lvert {X} \right\rvert} \geq 2$$.
• An action is transitive iff there is only one orbit, so $${\left\lvert {X/G} \right\rvert} = 1$$.
• Apply Burnside’s Lemma \begin{align*} 1 = {\left\lvert {X/G} \right\rvert} = \frac{1}{{\left\lvert {G} \right\rvert}} \sum_{g\in G} {\left\lvert { \mathrm{Fix}(g)} \right\rvert} \implies {\left\lvert {G} \right\rvert} = \sum_{g\in G} {\left\lvert { \mathrm{Fix}(g)} \right\rvert} = \mathrm{Fix}(e) + \sum_{\substack{g\in G \\ g\neq e}} {\left\lvert { \mathrm{Fix}(g)} \right\rvert} \end{align*}
• Note that $$\mathrm{Fix}(e) = X$$, since the identity must fix every element, so $${\left\lvert { \mathrm{Fix}(e)} \right\rvert} \geq 2$$.
• If $${\left\lvert { \mathrm{Fix}(g)} \right\rvert} > 0$$ for all $$g\neq e$$, the remaining term is at least $${\left\lvert {G} \right\rvert} -1$$. But then the right-hand side yields is at least $$2 + ({\left\lvert {G} \right\rvert} -1) = {\left\lvert {G} \right\rvert} + 1$$, contradicting the equality.
• So not every $${\left\lvert { \mathrm{Fix}(g)} \right\rvert} > 0$$, and $${\left\lvert { \mathrm{Fix}(g) } \right\rvert} = 0$$ for some $$g$$, which says $$g$$ has no fixed points in $$X$$.
#1 #algebra/qual/work #2 #5 #algebra/qual/completed