D+F 4.5.24

Prove that if $G$ is a group of order 231 then $Z(G)$ contains a Sylow 11-subgroup of $G$. Also, show that there is a normal Sylow 7-subgroup in $G$.

$\textit{Proof}$ : Begin by showing there is a normal Sylow 7-subgroup in $G$: By Sylow's Theorem (iii), $$n_p \equiv 1(modp)$$ and $$n_p \mid m$$ when $|G| = p^am$.
With $p=7$, we can write $|G|=231=7 \cdot 33$. The possibilities for $n_7$ are 1,8,15,22, and 29. Only the number 1 divides 33, thus $$n_7=1$$ Since Sylow p-subgroups are conjugate to each other, the Sylow subgroup of order 7 (231 has only 1 factor of 7) is normal in $G$ by corollary 20 p.142.

Now, to show that $Z(G)$ contains a Sylow-11 subgroup we initially proceed as above. The possibilities for $n_{11}$ are 1,12, and 23. But, only the number 1 also divides 21. Therefore, there is a unique Sylow 11 subgroup, $H$ in $G$ that is also normal. By Corollary 15 p.134, $$N_G(H)/C_G(H) \lesssim Aut(H) \quad (\star)$$ Since $H \lhd G$, $N_G(H)=G$. Also, $Aut(H)=Aut(\mathbb{Z}_{11})$ since $\mathbb{Z}_{11}$ is the only group of order 11. So $(\star)$ above becomes $$G/C_G(H) \lesssim Aut(\mathbb{Z}_{11}) \cong \mathbb{Z}_{11}^* \cong \mathbb{Z}_{10} $$ In terms of group orders, this means that $$|G/C_G(H)| \large{\mid} 10$$ So, $$\frac{231}{|C_G(H)|} \mid 10 \Longrightarrow \frac{231}{|C_G(H)|}=1,2,5,10 \Longrightarrow |C_G(H)|=231$$ So, $H$ commutes with all 231 of the elements of $G$. In other words, $$C_G(H) \leq Z(G) \quad \blacksquare$$