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 $X$.