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I am trying to understand the proof of the statement that if $|G| = 60$ and $G$ has more than one Sylow $5$-subgroup, then $G$ is simple. This statement and its proof can found in Page 145 of Dummit and Foote's Abstract Algebra.

The proof is by contradiction assuming that $|G| = 60$ and $n_5>1$ but there exists $H$ a normal subgroup of $G$ with $H \neq 1$ or $G$. By Sylow's theorem the only possibility for $n_5$ is $6$. Let $P \in Syl_5(G)$ so that $|N_G(P)| = 10$ since its index is $n_5$.

In the second paragraph, it is assumed that $5$ divides $|H|$ and a contradiction is obtained.

I could not understand the beginning sentence of the third paragraph. Why is "If $|H| = 6, 12$, $H$ has a normal, HENCE CHARACTERISTIC, Sylow subgroup, which is therefore also normal in $G$". Why is the statement true, especially why a normal subgroup of $H$ is also characteristic in $H$ so that it is normal in $G$ (using the assumption that $H$ is normal in $G$).

Shaun
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Vanya
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  • See here for your last question. You mean this post, right? – Dietrich Burde Sep 21 '23 at 09:47
  • The questions posed in the second link are different from mine. I want to know why it suffices to consider |H| = 6, or 12, and not all the divisors, 2, 3, 4, 6, 12 of 12. Secondly, why a normal subgroup is automatically characteristic? – Vanya Sep 21 '23 at 09:56
  • If a Sylow $p$-subgroup $P$ of a group $G$ is normal in $G$, then $P$ is the unique Sylow $p$-subgroup of $G$, and so it is characteristic in $G$. – Derek Holt Sep 21 '23 at 10:02
  • Thank you. I understood. Can you please clarify why it is sufficient to consider $|H| = 6, 12$ and not all the divisors of $12$ as orders of $H$ – Vanya Sep 21 '23 at 10:05
  • Well if $|H|=2,3$ or $4$ then it is a Sylow subgroup of itself so it obviously has a normal Sylow subgroup. – Derek Holt Sep 21 '23 at 10:12
  • Thank you. I understood. – Vanya Sep 21 '23 at 10:19

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