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Let $G$ be a group of order $385$. Prove that $Z(G)$, the center of $G$, contains an element of order $7$.

I used the Sylow theorem and realized that there are one Sylow $11$-subgroup, which is normal in $G$, and also one Sylow $7$-subgroup, which is normal as well.

I tried to solve it with the commutator of $G$ and concluded that $G'$ is trivial. Does it make sense? Anyway I will be glad for a clue.

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Every group of order 385 contains a central Sylow 7-subgroup and a normal Sylow 11-subgroup. Indeed, note that $385 = 5 \cdot 7 \cdot 11$. Now Sylow’s Theorem forces $n_7 = n_{11} = 1$, so that $G$ has a unique, hence normal Sylow $7$- and a normal Sylow $11$-subgroup. Let $P_7$ denote the (unique) Sylow $7$-subgroup.

We have $N_G(P_7) = G$. Hence we obtain $G/C_G(P_7) \leq \mathsf{Aut}(P_7)$. Moreover, $|\mathsf{Aut}(P_7)| = 6 = 2 \cdot 3$. Thus $|G/C_G(P_7)| = 1$, so that $G/C_G(P_7) = 1$, hence $C_G(P_7) = G$. Thus $P_7 \leq Z(G)$.

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  • $\begingroup$ can you please explain how you got this $G/C_G(P_7)≤𝖠𝗎𝗍(P_7). Moreover, |𝖠𝗎𝗍(P_7)|=6=2⋅3. Thus |G/C_G(P_7)|=1$ $\endgroup$ – user2993422 Sep 9 '15 at 0:02
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    $\begingroup$ By the normalizer-centralizer theorem we obtain $G/C_G(P_7) \leq \mathsf{Aut}(P_7)$, e.g., see here. $\endgroup$ – Dietrich Burde Sep 9 '15 at 7:45
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    $\begingroup$ Note that $Aut(\mathbb{Z}/p)\simeq (\mathbb{Z}/p)^{\times}$ has order $p-1$. $\endgroup$ – Dietrich Burde Sep 9 '15 at 7:53
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From this link, a 7-Sylow subgroup of $G$ intersects the center of $G$ nontrivially.

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  • $\begingroup$ I know that this is old, but I'm having trouble proving that the Sylow 7-subgroup is not contained in the centralizer of a noncentral element. Could you elaborate on this a little bit? $\endgroup$ – coldnumber Aug 3 '16 at 13:18

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