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