3
$\begingroup$

I can't make sense of this line in a proof:

If there is such $G$ we may assume that $n_3(G) = 10$ and $n_5(G) = 36$. By (NC) there is an element $x$ of order $15$.

The (NC) theorem is :

Let $G$ be a group and $H \le G$ a subgroup. Then $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\operatorname{Aut}(H)$.

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    Please note for further reference the editing I have done. – 2017-02-21

1 Answers 1

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Let $P$ be a 5-Sylow subgroup of $G$, then $n_5(G)$ equals the index of the normalizer $N := N_G(P)$, so $|N| = 15$. We have to show that $N$ is cyclic (indeed, any group of order 15 is cyclic) which can be shown by the theorem you mentioned: If $N$ is not cyclic, then $N$ also cannot be abelian, so $C_G(N) = P$. Now $N/C_G(P) \cong C_3$ is isomorphic to a subgroup of $\mathrm{Aut}(P) \cong C_4$ which is a contradiction.