Group Theory: General

• For ease of notation, replace \(H\) in the problem with \(N\) so we remember which one is normal.
• Write \(n\coloneqq{\sharp}N\) and \(m \coloneqq[G:N] = {\sharp}G/N\), where the quotient makes sense since \(N\) is normal.
• Let \(H \leq G\) with \({\sharp}H = n\), we’ll show \(H=N\).
  • Since \({\sharp}H = {\sharp}N\) it suffices to show \(H \subseteq N\).
  • It further suffices to show \(hN = N\) for all \(h\in H\).
• Noting \(\gcd(m, n)=1\), use the division algorithm to write \(1 = ns + mt\) for some \(s,t\in {\mathbf{Z}}\).
• The result follows from a computation: \begin{align*} hN &= h^1 N \\ &= h^{ns + mt}N \\ &= h^{ns} N \cdot h^{mt}N \\ &= \qty{h^n N}^s \cdot \qty{h^t N}^m \\ &= (eN)^s \cdot N \\ &= N ,\end{align*}
  • We’ve used that \(h\in H \implies o(h) \divides {\sharp}H = n\) by Lagrange, so \(h^n = e\).
  • We’ve also used that \({\sharp}G/N = m\), so \((xH)^m = H\) for any \(xH\in G/H\).

• \(H \cap K \leq H \leq G \implies [G: H \cap K] = [G: H] [H : H \cap K]\)
• So it suffices to show \([H: H \cap K] \leq [G: K]\)
• Write \(H/H \cap K = \left\{{ h_1 J, \cdots, h_m J }\right\}\) as distinct cosets where \(J \coloneqq H \cap J\).
• Then \(h_i J\neq h_j J \iff h_i h_j^{-1}\not\in J = H \cap K\).
• \(H\) is a subgroup, so \(h_i h_j^{-1}\in H\) forces this not to be in \(K\).
• But then \(h_i K \neq h_j K\), so these are distinct cosets in \(G/K\).
• So \({\sharp}G/K \geq m\).

• Let \(N{~\trianglelefteq~}P\), then for each conjugacy class \([n_i]\) in \(N\), \(H \cap[g_i] = [g_i]\) or is empty.
• \(G = {\textstyle\coprod}_{i\leq M} [g_i]\) is a disjoint union of conjugacy classes, and the conjugacy classes of \(H\) are of the form \([g_i] \cap H\).
• Then pull out the center \begin{align*} H = \coprod_{i\leq M} [g_i] \cap H = \qty{ Z(G) \cap H } {\textstyle\coprod}\coprod_{i\leq M'} [g_i] .\end{align*}
• Taking cardinalities, \begin{align*} {\sharp}H = {\sharp}\qty{ Z(G) \cap H} + \sum_{i\leq M'} {\sharp}[g_i] .\end{align*}
• \(p\) divides \(H\) since \(H\leq P\) and \(P\) is a \(p{\hbox{-}}\)group.
• Each \({\sharp}[g_i] \geq 2\) since the trivial conjugacy classes appear in the center, forcing \({\sharp}[g_i] \geq p\).
• \(p\) divides \({\sharp}[g_i]\) since \({\sharp}[g_i]\) must divide \({\sharp}P = p^k\)
• So \(p\) must divide the remaining term \(Z(G) \cap H\), which makes it nontrivial.

\todo[inline]{Redo part c}

(DZG): Warning!! I haven’t checked this solution very carefully, and this is kind of a delicate parity argument. Most of the key ideas are borrowed from here.

• Write \(k \coloneqq o(\pi_g)\), then since \(\pi\) is injective, \(k = o(g)\) in \(G\).
• Since \(\pi_g\) as a cycle is obtained from the action of \(g\), we can pick an element \(x_0\) in \(G\), take the orbit under the action, and obtain a cycle of length \(k\) since the order of \(g\) is \(k\). Then continue by taking any \(x_1\) not in the first orbit and taking its orbit. Continuing this way exhausts all group elements and yields a decomposition into disjoint cycles: \begin{align*} \pi_g = (x_0, gx_0, g^2 x_0, \cdots, g^{k-1} x_0) (x_1, gx_1, g^2 x_1, \cdots, g^{k-1} x_1) \cdots (x_m, gx_m, g^2 x_m, \cdots, g^{k-1} x_m) .\end{align*}
• So there are \(m\) orbits all of length exactly \(k\). Proceed by casework.
• If \(k\) is even:
  • This yields \(m\) odd cycles, and thus \(\pi\) has zero (an even number) of even cycles.
  • Thus \(\pi \in \ker \operatorname{sgn}\) and is an even permutation.
• If \(k\) is odd
  • This yields \(m\) even cycles, thus an even number of even cycles iff \(m\) is even
• The claim is that the number of orbit representatives \(m\) is equal to \([G:H] = {\sharp}G/H\) for \(H = \left\langle{ g }\right\rangle\).
  • Proof: define a map \begin{align*} \left\{{ \text{Orbit representatives } x_i }\right\} &\to G/H \\ x &\mapsto xH .\end{align*}
  • This is injective and surjective because \begin{align*} xH = yH &\iff xy^{-1}\in H = \left\langle{ g }\right\rangle \\ &\iff xy^{-1}= g^\ell \\ &\iff x=g^\ell y \\ &\iff y\in {\mathcal{O}}_x ,\end{align*} so \(y\) and \(x\) are in the same orbit and have the same orbit representative.
• We now have \begin{align*} \pi_g \text{ is an even permutation } \iff \begin{cases} k \text{ is odd and } m \text{ is even} & \\ \text{ or } & \\ k \text{ is even} & . \end{cases} \end{align*}
• Everything was an iff, so flip the evens to odds: \begin{align*} \pi_g \text{ is an odd permutation } \iff \begin{cases} k \text{ is even and } m \text{ is odd} & \\ \text{ or } & \\ k \text{ is odd} & . \end{cases} .\end{align*}
• Then just recall that \(k\coloneqq o(\pi_g)\) and \begin{align*} m= [G: \left\langle{ g }\right\rangle] = {\sharp}G / {\sharp}\left\langle{ g }\right\rangle= {\sharp}G / o(g) = {\sharp}G/ o(\pi_g) .\end{align*}

• Let \(H\leq G\) and define \(n\coloneqq[G:H]\).
• Write \(G/H = \left\{{ x_1 H, \cdots, x_n H }\right\}\) for the finitely many cosets.
• Let \(G\) act on \(G/H\) by left translation, so \(g\cdot xH \coloneqq gxH\).. Call the action \(\psi: G\to \operatorname{Sym}(G/H)\).
• Then \({\operatorname{Stab}}(xH) = xHx^{-1}\) is a subgroup conjugate to \(H\), and \(K\coloneqq\ker \psi = \bigcap_{i=1}^n xHx^{-1}\) is the intersection of all conjugates of \(H\).
• Kernels are normal, so \(K{~\trianglelefteq~}G\), and \(K\subseteq xHx^{-1}\) for all \(x\), meaning \(K\) is contained in every conjugate of \(H\).
• The index \([G:K]\) is finite since \(G/K \cong \operatorname{im}\psi\) by the first isomorphism theorem, and \({\sharp}\operatorname{im}\psi \leq {\sharp}\operatorname{Sym}(G/H) = {\sharp}S_n = n! < \infty\).

• Use that \(N{~\trianglelefteq~}G \iff N = {\textstyle\coprod}' [n_i]\) is a disjoint union of (full) conjugacy classes.
• Take cardinalities: \begin{align*} p = {\sharp}N = \sum_{i=1}^m {\sharp}[n_i] = 1 + \sum_{i=2}^m [n_i] .\end{align*}
• The size of each conjugacy class divides the size of \(H\) by orbit-stabilizer, so \({\sharp}[n_i] \divides p\) for each \(i\).
• But the entire second term must sum to \(p-1\) for this equality to hold, which forces \({\sharp}[n_i] = 1\) (and incidentally \(m=p-1\))
• Then \([n_i] = \left\{{ n_i }\right\} \iff n_i \in Z(G)\), and this holds for all \(i\), so \(N \subseteq Z(G)\).

• Idea: take a semidirect product involving \(C_9\) and \(C_7\). We’ll need some facts: \(\mathop{\mathrm{Hom}}(C_m, C_n) \cong C_d\) where \(d = \gcd(m, n)\), and \(\mathop{\mathrm{Aut}}(C_m)\cong C_m^{\times}\) which has order \(\phi(m)\) (since one needs to send generators to generators), which can be explicitly calculated based on the prime factorization of \(m\).

• Some calculations we’ll need:

  • \(\mathop{\mathrm{Aut}}(C_9) \cong C_9^{\times}\cong C_{\phi(9)} \cong C_6\), using that \(\phi(p^k) = p^{k-1}(p-1)\).
  • \(\mathop{\mathrm{Aut}}(C_7) \cong C_7^{\times}\cong C_{\phi(7)}\cong C_6\) using that \(\phi(p) = p-1\).

• To get a nonabelian group, we need a nontrivial semidirect product, so look at \(\mathop{\mathrm{Hom}}(G, \mathop{\mathrm{Aut}}(H))\) in the two possible combinations.

  • \(\mathop{\mathrm{Hom}}(C_7, \mathop{\mathrm{Aut}}(C_9)) \cong \mathop{\mathrm{Hom}}(C_7, C_6) \cong C_1 \coloneqq\left\{{e}\right\}\) using that \(\mathop{\mathrm{Hom}}(C_m, C_n) \cong C_{d}\) for \(d = \gcd(m, n)\). So there are no nontrivial homs here, so only the direct product is possible.
  • \(\mathop{\mathrm{Hom}}(C_9, \mathop{\mathrm{Aut}}(C_7)) \cong \mathop{\mathrm{Hom}}(C_9, C_6) \cong C_3\), so use this!
  • Note that we don’t have to consider possibilities for \(C_3\times C_3\), since including this as a factor would yield no elements of order 9.

• So take \(G\coloneqq C_7 \rtimes_\psi C_9\) for some \(\psi: C_9 \to \mathop{\mathrm{Aut}}(C_7)\), and we can take the presentation \begin{align*} G = \left\langle{x, y{~\mathrel{\Big\vert}~}x^7, y^9, yxy^{-1}= \psi(x)}\right\rangle .\end{align*}

• It now suffices to find a nontrivial \(\psi: C_7\to C_7\). Writing it multiplicatively as \(C_7 = \left\langle{x{~\mathrel{\Big\vert}~}x^7}\right\rangle\), any map that sends \(x\) to a generator will do. It suffices to choose any \(k\) coprime to \(7\), and then take \(\psi(x) \coloneqq x^k\), which will be another generator.

• So take

\begin{align*} G = \left\langle{x, y{~\mathrel{\Big\vert}~}x^7, y^9, yxy^{-1}= x^2}\right\rangle .\end{align*}

Since \(Z(G)\) is a subgroup of \(G\) and \(|G| = p^3\), by Lagrange’s theorem, \(|Z(G)| \in \{1, p, p^2, p^3\}\).

Since we stipulated that \(G\) is nonabelian, \(|Z(G)| \ne p^3\). Also, since \(G\) is a \(p\)-group, it has nontrivial center, so \(|Z(G)| \ne 1\). Finally, by the \(G/Z(G)\) theorem, \(|Z(G)| \ne p^2\): if \(|Z(G)| = p^2\), then \(|G/Z(G)| = p\) and so \(G/Z(G)\) would be cyclic, meaning that \(G\) is abelian. Hence, \(|Z(G)| = p\).

Then, since \(|Z(G)| = p\), we have that \(|G/Z(G)| = p^2\), and so \(G/Z(G)\) is abelian. Thus, \([G, G] \in Z(G)\). Since \(|Z(G)| = p\), then \(|[G,G]| \in \{ 1, p\}\) again by Lagrange’s theorem. If \(|[G,G]| = p\) then \([G,G] = Z(G)\) and we are done. And, indeed, we must have \(|[G,G]| = p\), because \(G\) is nonabelian and so \(|[G,G]| \ne 1\).

• Suppose \(G = \left\langle{ a, b}\right\rangle\) with \(a^2 = b^2 = e\), satisfying some unknown relations.

• Consider \(ab\). Since \(G\) is finite, this has finite order, so \((ab)^n = e\) for some \(n\geq 2\).

• Note \(\left\langle{ab, b}\right\rangle \subseteq \left\langle{a, b}\right\rangle\), since any finite word in \(ab, b\) is also a finite word in \(a, b\).

• Since \((ab)b = ab^2 = a\), we have \(\left\langle{ab, b}\right\rangle \subseteq \left\langle{a, b}\right\rangle\), so \(\left\langle{ab, b}\right\rangle = \left\langle{a, b}\right\rangle\).

• Write \(D_{2n} = F(r, s) / \ker \pi\) for \(\pi: F(r, s)\to D_{2n}\) the canonical presentation map.

• Define \begin{align*} \psi: F(r, s) &\to G \\ r &\mapsto ab \\ t &\mapsto b .\end{align*}

• This is clearly surjective since it hits all generators.

• We’ll show that \(ab, a\) satisfy all of the relations defining \(D_{2n}\), which factors \(\psi\) through \(\ker \pi\), yielding a surjection \(\tilde \psi: D_{2n} \twoheadrightarrow G\).

  • \((ab)^n = e\) by construction, \(b^2 = e\) by assumption, and \begin{align*} b (ab) b^{-1}= babb^{-1}= ba = b^{-1}a^{-1}= (ab)^{-1} ,\end{align*} corresponding to the relation \(srs^{-1}= r^{-1}\). Here we’ve used that \(o(a) = o(b) = 2\) implies \(a=a^{-1}, b=b^{-1}\).

• Surjectivity of \(\tilde \psi\) yields \(2n = {\sharp}D_{2n} \geq {\sharp}G\).

• The claim is that \({\sharp}G \geq 2n\), which forces \({\sharp}G = 2n\). Then \(\tilde \psi\) will be a surjective group morphism between groups of the same order, and thus an isomorphism.

  • We have \(\left\langle{ ab }\right\rangle\leq G\), so \(n\divides {\sharp}G\).
  • Since \(b\not\in \left\langle{ ab }\right\rangle\), this forces \({\sharp}G > n\), so \({\sharp}G \geq 2n\).

Remark: see a more direct proof in Theorem 2.1 and Theorem 1.1 here

• Write \({\sharp}G = p^2 q\)

• Cases: first assume \(p>q\), then do \(q<p\).

• In any case, we have \begin{align*} n_p \divides q &\implies n_p \in \left\{{ 1,q }\right\} \\ \\ n_q \divides p^2 &\implies n_q \in \left\{{ 1, p, p^2}\right\} .\end{align*}

• If \(n_p=1\) or \(n_q=1\), we’re done, so suppose otherwise.

• Case 1: \(:p>q\).

  • Using that \([n_p]_p \equiv 1\), consider reducing elements in \(\left\{{1, q}\right\} \operatorname{mod}p\).
  • Since \(q<p\), we just have \(q\operatorname{mod}p = q\), and as long as \(q\neq 1\) we have \(q\not\equiv 1\operatorname{mod}p\). But since \(n_p\neq 1\) and \(n_p\neq q\), this is a contradiction. \(\contradiction\)

• Case 2: \(p< q\):

  • Using that \([n_q]_q \equiv 1\), consider reducing \(\left\{{1, p, p^2}\right\}\operatorname{mod}q\).

  • Since now \(p<q\), we have \(p\operatorname{mod}q = p\) itself, so \(p\operatorname{mod}q \neq 1\) and we can rule it out.

  • The remaining possibility is \(n_q = p^2\).

  • Supposing that \(n_p \neq 1\), we have \(n_p=q\), so we can count \begin{align*} \text{Elements from Sylow } q: n_q( {\sharp}S_q - 1) &= p^2(q-1) + 1 ,\end{align*} where we’ve used that distinct Sylow \(q\)s can only intersect at the identity, and although Sylow \(p\)s can intersect trivially, they can also intersect in a subgroup of size \(p\).

  • Suppose all Sylow \(p\)s intersect trivially, we get at least \begin{align*} \text{Elements from Sylow } p: n_p( {\sharp}S_p - 1) &= q(p^2-1) .\end{align*} Then we get a count of how many elements the Sylow \(p\)s and \(q\)s contribute: ` \begin{align*} q(p^2-1) + p^2(q-1) + 1 = p^2q - q + p^2q - p^2 + 1 = p^2q + (p^2-1)(q-1)

    p^2q = {\sharp}G ,\end{align*} `{=html} provided \((p^2-1)(q-1) \neq 0\), which is fine for \(p\geq 2\) since this is at least \((2^2-1)(3-2) = 3\) (since \(p<q\) and \(q=3\) is the next smallest prime). \(\contradiction\)

  • Otherwise, we get two Sylow \(p\)s intersecting nontrivially, which must be in a subgroup of order at least \(p\) since the intersection is a subgroup of both. In this case, just considering these two subgroups, we get \begin{align*} \text{Elements from Sylow } p: n_p( {\sharp}S_p - 1) &> p^2 + p^2 - p = 2p^2-p -1 .\end{align*} Then a count: \begin{align*} p^2(q-1) + (2p^2-p - 1) + 1 &= p^2 q- p^2 + 2p^2 -p \\ &= p^2 q + p^2 -p \\ &= p^2q + p(p-1) \\ &> p^2q = {\sharp}G ,\end{align*} a contradiction since this inequality is strict provided \(p\geq 2\). \(\contradiction\)