Groups: Sylow Theory

• We have \(g_j \subseteq Z(g_k)\) for all \(k\) by assumption.
• If we can show \(Z(g_k) = G\) for all \(k\), then \(g_k \in Z(G)\) for all \(k\).
  • Then each conjugacy class is size 1, and since \(G = {\textstyle\coprod}_{i=1}^n [g_i] = {\textstyle\coprod}_{i=1}^n \left\{{g_i}\right\}\), every \(g\in G\) is some \(g_i\). So \(G \subseteq Z(G)\), forcing \(G\) to be abelian.
• If we can show \(G \subseteq \bigcup_{h\in H} h Z(g_k) h^{-1}\) for some \(k\), this forces \(Z(g_k) = G\) and \(g_k \in Z(G)\).
  • If we can do this for all \(k\), we’re done!
• Since \(g\in G\) is in some conjugacy class, write \(g=hg_j h^{-1}\) for some \(h\in G\) and some \(1\leq j\leq n\).
• Now use \(g_j \in Z(g_k)\) for all \(k\): \begin{align*} g\in G &\implies g = hg_j h^{-1}&& \text{for some } h\in H \\ g_j \in Z(g_k) \forall k &\implies g\in hZ(g_k)h^{-1}&&\text{for some }h, \, \forall 1\leq k \leq n \\ &\implies g\in \bigcup_{h\in G} h Z(g_k) h^{-1} &&\forall 1\leq k \leq n \\ .\end{align*}
  • Note that it’s necessary to get rid of the \(h\) dependence, since now now every \(g\in G\) is in \(\bigcup_{h\in G} hZ(g_k)h^{-1}\).
• Now \begin{align*} G \subseteq \bigcup_{h\in G} hZ(g_k) \subseteq G \,\,\forall k \implies Z(g_k) = G\,\, \forall k ,\end{align*} and we’re done.

Part a: write \(36 = 2^2\cdot 3^2\), then the distinct groups correspond to all combinations of integer partitions of the exponents \(2\) and \(2\):

• \((2, 2) \leadsto ({\mathbf{Z}}/4{\mathbf{Z}}) \times ({\mathbf{Z}}/9{\mathbf{Z}})\)
• \((1+1, 2) \leadsto ({\mathbf{Z}}/2{\mathbf{Z}})^2 \times ({\mathbf{Z}}/9{\mathbf{Z}})\)
• \((2, 1+1) \leadsto {\mathbf{Z}}/4{\mathbf{Z}}\times ({\mathbf{Z}}/3{\mathbf{Z}})^2\)
• \((1+1, 1+1) \leadsto ({\mathbf{Z}}/2{\mathbf{Z}})^2 \times ({\mathbf{Z}}/3{\mathbf{Z}})^2\)

Part b: let \(H_2 \leq G\) be the unique Sylow \(2{\hbox{-}}\)subgroup, so \({\sharp}H_2 = 2^2\). Then \(H_2{~\trianglelefteq~}G\), so \(G/H\) is a group of size \([G:H] = 3^2\). Note that by pure numeric observation (and the hint), it is necessary to find a normal subgroup \(N{~\trianglelefteq~}G\) of size 3 and thus index 12, since \({\sharp}A_4 = 4!/2 = 12 = 36/3\). Since normal subgroups are kernels of homomorphisms, we can look for a morphism \(\phi: G\to S_4\) where \(\ker(\phi)\) is order 3, anticipating applying the first isomorphism theorem to get \(G/\ker(\phi) \cong \operatorname{im}(\phi) \leq S_4\), where we hope that \(\operatorname{im}(\phi) \cong A_4\). So we look for an action of \(G\) on a set with 4 elements.

Applying the Sylow theorems, we have \(n_3 \divides 2^2 \implies n_3\in\left\{{1,2,4}\right\}\) and \(n_3\equiv 1\operatorname{mod}4\implies n_3\in \left\{{1, 4}\right\}\). Note that \(n_3=1\) implies the Sylow \(3{\hbox{-}}\)subgroup is normal, in which case \(G \cong H_2 \times H_3\) splits into a product of groups of order \(2^2, 3^2\), and thus \(G\) is abelian since any group of order \(p^2\) is abelian for any prime \(p\) and products of abelian groups are abelian. So we can conclude that \(n_3 = 4\) and consider the action of \(G\) on the set of four Sylow \(3{\hbox{-}}\)subgroups of \(G\) to get a map \(\phi: G\to S_4\).

It is a theorem that this action is transitive, and so \(\operatorname{im}(\phi)\) is a transitive subgroup of \(S_4\). By standard facts in Galois theory, the only such subgroups are \(S_4, A_4, D_4, C_2^2, C_4\), and so \({\sharp}\operatorname{im}(\phi) \in \left\{{24 ,12, 8, 4}\right\}\). By the hint it suffices to show that \({\sharp}\operatorname{im}(\phi) = 12\).

• \({\sharp}\operatorname{im}(\phi)\neq 24\), since then \(\phi\) surjects and \(G/\ker(\phi) \cong S_4\), but the RHS has order 24 and the LHS order \(36/k\) for some \(k\in {\mathbf{Z}}\), which is impossible.
• \({\sharp}\operatorname{im}(\phi) \neq 8\) since then \(36/k = 8\) for some \(k\), again impossible.
• \({\sharp}\operatorname{im}(\phi) \neq 4\): if so, \({\sharp}\ker(\phi) = 9\) so \(\ker(\phi)\) is a Sylow \(3{\hbox{-}}\)subgroup. Since kernels are normal and all Sylows are conjugate, this would force \(n_3 = 1\), a contradiction.

Summarizing, to produce \(N\), consider \(G\curvearrowright{\operatorname{Syl}}_3(G)\) to get a map \(\phi: G\to S_4\). Then \(\operatorname{im}(\phi)\leq S_4\) must have order 12 with kernel \(N\coloneqq\ker(\phi)\) of order 3, and since the only subgroup of order 12 in \(S_4\) is \(A_4\), we have \(G/N\cong A_4\).

Part 3: Since \(N{~\trianglelefteq~}G\), we need to show:

• \(H \cap N = \left\{{e}\right\}\),
• \(NH = G\),
• \(H{~\trianglelefteq~}G\)

The first two conditions will give a semidirect product, and the third will show that the semidirect product is actually direct.

Note that \(L \coloneqq H \cap N\) is a subgroup of both \(H\) and \(N\), so if \(L\neq \left\{{e}\right\}\) then \({\sharp}L = 3\) and \(L\) is a subgroup of both \(N\) and \(A_4\). So \(L = N\) and thus \(N \subseteq A_4\) is a cyclic subgroup of order 3 generated by (say) \(n\); however since \(2\divides {\sharp}A_4 = 12\), Cauchy’s theorem produces an element \(m\) of order two. But then the subgroup \(\left\langle{n, m}\right\rangle\) is an order 6 subgroup of \(A_4\), which by the hint does not exist, so by contradiction we must have \(L \coloneqq N \cap H = \left\{{e}\right\}\).

That \(NH = G\) now follows by counting elements: \(NH \leq G\) is a subgroup, and \({\sharp}NH = ({\sharp}N)\cdot({\sharp}H) = 3\cdot 12 = {\sharp}G\).

Finally, to see that \(H{~\trianglelefteq~}G\), we’ll instead show that the semidirect product is trivial directly. By the first two conditions above, we already have \(G\cong N\rtimes_\phi H\) for some \(\phi: H\to \mathop{\mathrm{Aut}}(N)\). Since \(N\cong C_3\) is cyclic, \(\mathop{\mathrm{Aut}}(N)\cong N^{\times}\) which is of order \(\phi(3) = 2\), so \(\phi\) is a group morphism from a group of order 12 to one of order 2. If \(\phi\) is nontrivial, \({\sharp}\ker \phi = 12/2 = 6\) is a subgroup of order 6 in \(H\cong A_4\), contradicting the hint and forcing \(\phi\) to be trivial and \(G\cong H\times N\).

Part d If \(G\coloneqq D_{18}\) has a normal Sylow \(2{\hbox{-}}\)subgroup, by part 2 \(\exists N{~\trianglelefteq~}G\) with \(G/N\cong A_4\), and (claim) since \(G\) has a subgroup \(H\cong A_4\) we must have \(G = N\times H\), a product of groups of orders 3 and 12 respectively. Somehow this is a contradiction??

That \(H\cong A_4 \leq G\) exists: unclear, maybe even not true. Not sure what the intended approach is, so here’s an alternative.

A general fact: for \(D_{2n}\), for any odd prime \(p\divides 2n\), the Sylow \(p{\hbox{-}}\)subgroup \(H_p\) is cyclic and normal since \(D_{2n}\cong C_n \rtimes C_2\) and \(p > 2\) implies \(H_p\) descends to a Sylow \(p{\hbox{-}}\)subgroup of \(C_n\) and all subgroups of cyclic groups are abelian and cyclic. Here \(n=18\), so take \(p=3\) to get \(H_{3} {~\trianglelefteq~}D_{18}\) a normal subgroup of order 9. If \(n_2=1\), so there is one single normal Sylow \(2{\hbox{-}}\)subgroup, then \(H_2, H_3\) are both normal and we get \(D_{18} \cong H_2 \times H_3\) as a direct product. But \(H_2, H_3\) are abelian and \(D_{18}\) is nonabelian, so this is a contradiction.

\todo[inline]{Part (b) not finished!}