Exercise 4.5.25 (If $|G| = 385$, $Z(G)$ contains a Sylow $7$-subgroup of $G$.)

Question

Prove that if $G$ is a group of order $385$ then $Z(G)$ contains a Sylow $7$-subgroup of $G$ and a Sylow $11$-subgroup is normal in $G$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Here $|G| = 385 = 5 \cdot 7 \cdot 11$. By Sylow's Theorem,
\begin{align*}
&n_{11} \mid 5 \cdot 7 = 35,\qquad n_{11} \equiv 1 \pmod {11},\
&n_7 \mid 5 \cdot 11 = 55,\qquad n_7 \equiv 1 \pmod 7.
\end{align*}
Therefore
$$n_{11} = 1,\qquad n_7 = 1,$$
so $G$ contains a normal Sylow $11$-subgroup $Q$ and a normal Sylow $7$ subgroup $P$.

As in Exercise 24, since $P \unlhd G$, $G$ acts on conjugation on $P$, so there is a homomorphism
$$\varphi : G 	o \mathrm{Aut}(P).$$
$P$ is cyclic of order $7$, thus $|\mathrm{Aut}(P)| = 6$, and $|G| = 5 \cdot 7 \cdot 11$, so $\mathrm{g.c.d.}(|\mathrm{Aut}(P)| , |G|) = 1$. This shows that $\varphi$ is trivial: for all $g$ in $G$ and for all $x \in P$, $gx = xg$, thus
$$P \leq Z(G).$$
This proves that $Z(G)$ contains a Sylow $7$-subgroup of $G$.
\end{proof}

Exercise 4.5.25 (If $|G| = 385$, $Z(G)$ contains a Sylow $7$-subgroup of $G$.)

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Exercise 4.5.25 (If $|G| = 385$, $Z(G)$ contains a Sylow $7$-subgroup of $G$.)

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