Actions

Write \(G/H = \left\{{g_0 H, g_1 H, \cdots, g_N H}\right\}\) for some \(N \coloneqq[G:H]\). Since cosets are equal or disjoint and have equal cardinality, \begin{align*} G = {\textstyle\coprod}_{k \leq N} g_k H \implies {\sharp}G = \sum_{k\leq N} {\sharp}\qty{g_k H} = \sum_{k\leq N} {\sharp}H = N {\sharp}H ,\end{align*} so \({\sharp}G = N {\sharp}H\), \({\sharp}H\) divides \({\sharp}G\) and \(N = [G:H]\) divides \({\sharp}G\).

See https://kconrad.math.uconn.edu/blurbs/grouptheory/cauchypf.pdf.

• Injectivity: \(\Phi(gG_x) = \Phi(hG_x) \iff g\curvearrowright x=h\curvearrowright x \iff gh^{-1}\curvearrowright x = x \iff gh^{-1}\in G_x \iff gG_x = hG_x\).

• Well-definedness: use \(gG_x = hG_x \iff gh^{-1}\in G_x \iff g^{-1}h \curvearrowright x = x\). Then \(g (g^{-1}h) \curvearrowright x = g \curvearrowright x\) on one hand, and on the other \((gg^{-1})h\curvearrowright x = h\curvearrowright x\), so \begin{align*} \Phi(hG_x) \coloneqq h\curvearrowright x = (gg^{-1}) h\curvearrowright x = g(g^{-1}h)\curvearrowright x = g\curvearrowright x = \Phi(gG_x) .\end{align*}

• Surjectivity: equivalent to the action being transitive.

• Fix \(x\in X\) and \(y\in {\mathrm{Orb}}(x)\), so \(g.x=y\) for some \(g\).
• Let \(H_x \coloneqq{\operatorname{Stab}}(x)\) and \(H_y\coloneqq{\operatorname{Stab}}(y)\), the claim is that \(H_x = g^{-1}H_y g\).
• Now just check: \begin{align*} h\in H_x &\iff hx = x \\ &\iff hg^{-1}y = g^{-1}y \\ &\iff ghg^{-1}y = y \\ &\iff ghg^{-1}\in H_y \\ &\iff h\in g^{-1}H_y g ,\end{align*} so \(H_x = g^{-1}H_y g\).

• Define a group action \(\phi: G\curvearrowright G/H \coloneqq\left\{{eH, g_1 H, \cdots, g_{n-1} H}\right\}\) acting on the \(n\) cosets of \(H\) by left-translation \(g.(g_k H) = (gg_k) H\).

• Then use that \(\operatorname{Sym}(G/H) \leq S_n\), so \(\operatorname{im}\phi \leq S_n\) is a subgroup.

• Since \(G\) is simple and \(\ker \phi \leq G\), we have \(\ker \phi = 1, G\). If \(\ker \phi = 1\), \(\phi\) is injective and we’re done.

• Otherwise \(\ker \phi = G\), and acting on \(eH\) yields \(gH = H\) for all \(g\), forcing \(H=G\) and \(n=1\), contradicting that \(H<G\) is proper. \(\contradiction\)

Strategy: form the set \(A \coloneqq\left\{{ (g,x) \in G\times X {~\mathrel{\Big\vert}~}g\curvearrowright x = x }\right\}\) and write/count it in two different ways. Write \({\operatorname{Stab}}(x) = \left\{{g\in G {~\mathrel{\Big\vert}~}gx=x}\right\}\) and \(\mathrm{Fix}(g) = \left\{{x\in X{~\mathrel{\Big\vert}~}gx = x}\right\}\).

First union over \(G\), where the inner set lets \(x\) vary: \begin{align*} A = \coprod_{g_0\in G} \left\{{ (g_0, x) {~\mathrel{\Big\vert}~}g_0 x = x }\right\} \cong \coprod_{g_0\in G} \left\{{g_0}\right\}\times \mathrm{Fix}(g_0) \subseteq G\times X .\end{align*}

Then union over \(X\), where the inner set lets \(g\) vary: \begin{align*} A = \coprod_{x_0\in X} \left\{{ (g, x_0) {~\mathrel{\Big\vert}~}gx_0= x_0 }\right\} \cong \coprod_{x_0\in X} {\operatorname{Stab}}(x_0) \times\left\{{ x_0 }\right\} \subseteq G\times X .\end{align*} Taking cardinalities, and using the fact that \(\left\{{p}\right\} \times A \cong A\) as sets for any set \(A\), we get the following equality \begin{align*} \sum_{g_0\in G} {\sharp} \mathrm{Fix}(g_0) = {\sharp}A = \sum_{x_0\in X} {\sharp}{\operatorname{Stab}}(x_0) .\end{align*}

Now rearrange orbit-stabilizer: \begin{align*} {\mathrm{Orb}}(x_0) = G/{\operatorname{Stab}}(x_0) \implies {\sharp}{\operatorname{Stab}}(x_0) = {\sharp}G/ {\sharp}{\mathrm{Orb}}(x_0) ,\end{align*} and use this to rewrite the RHS: \begin{align*} \sum_{g_0\in G} {\sharp} \mathrm{Fix}(g_0) &= \sum_{x_0\in X} {\sharp}{\operatorname{Stab}}(x_0) \\ &= \sum_{x_0\in X} {{\sharp}G \over {\sharp}{\mathrm{Orb}}(x_0)} \\ &= {\sharp}G \sum_{x_0\in X} {1 \over {\sharp}{\mathrm{Orb}}(x_0)} \\ \\ \implies {1\over {\sharp}G} \sum_{g_0\in G} {\sharp} \mathrm{Fix}(g_0) &= \sum_{x_0\in X} {1\over {\sharp}{\mathrm{Orb}}(x_0)} ,\end{align*} so it suffices to show the right-hand side sum is the number of orbits, \({\sharp}(X/G)\).

Proceed by partitioning the sum up according to which orbit each term comes from: \begin{align*} \sum_{x_0\in X}\qty{1\over {\sharp}{\mathrm{Orb}}(x_0)} &= \sum_{{\mathrm{Orb}}(x_0) \in X/G} \qty{ \sum_{y\in {\mathrm{Orb}}(x_0)} \qty{1\over {\sharp}{\mathrm{Orb}}(x_0)} }\\ &= \sum_{{\mathrm{Orb}}(x_0) \in X/G} \qty{1\over {\sharp}{\mathrm{Orb}}(x_0)}\sum_{y\in {\mathrm{Orb}}(x_0)} 1 \\ &= \sum_{{\mathrm{Orb}}(x_0) \in X/G} \qty{1\over {\sharp}{\mathrm{Orb}}(x_0)} {\sharp}{\mathrm{Orb}}(x_0) \\ &= \sum_{{\mathrm{Orb}}(x_0) \in X/G} 1 \\ &= {\sharp}(X/G) .\end{align*}

• Let \(\phi: G\curvearrowright X\coloneqq\left\{{xH}\right\}\), noting that \({\sharp}X = p\) and \(\operatorname{Sym}(X) \cong S_p\).

• Then \(K\coloneqq\ker \phi\), and importantly \(K \supseteq H\) since \(K\) is the intersection of stabilizers, and contains \({\operatorname{Stab}}(eH) \supseteq H\) since \(gH = H \implies g\in H\).

• Since \(G\) is finite and \(K\leq G\), we have \({\sharp}(G/K)\) dividing \({\sharp}G\), since \begin{align*} [G:K] = {\sharp}(G/K) = {\sharp}G/ {\sharp}K \implies {\sharp}G = {\sharp}(G/K) {\sharp}K .\end{align*}

• Now \begin{align*} G/K \cong K' \leq S_p \implies {\sharp}(G/K)\divides p! .\end{align*}

• So \({\sharp}(G/K)\) divides \(\gcd( {\sharp}G, p!)=p\), using that \(p\) was the minimal prime dividing \({\sharp}G\). This forces \({\sharp}(G/K)\) to be 1 or \(p\).

• If it’s \(p\):

  • Then \(p = [G:K] = [G:H]\) and since \(K\supseteq H\) this forces \(K=H\). Kernels are automatically normal, so we’re done.

• If it’s 1:

  • Then \([G:K] = 1\) and \(K = \ker \phi = G\).
  • Identifying \(\ker \phi = \bigcap_{xH\in G/H} {\operatorname{Stab}}(xH)\), we have \({\operatorname{Stab}}(xH) = xHx^{-1}= G\) for all \(x\), which says \(H\) is normal.