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I wanted to prove that the intersection of all Sylow $p$-subgroups of a finite group G is a normal subgroup of $G$.

Can someone enlighten me how is this implication possible:

If an automorphism $\sigma$ of $G$ maps every Sylow $p$-subgroup to a Sylow $p$-subgroup, then the image of the intersection of all Sylow $p$-subgroups under $\sigma$ is itself?

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Let $N$ be the intersection of all $p$-Sylow subgroups of $G$. If $P$ is a $p$-Sylow subgroup then $N = \cap_{g\in G}gPg^{-1}$. Thus we have for all $x\in G$ that $$ xNx^{-1} = \bigcap_{g\in G}(xg)P(xg)^{-1} = \bigcap_{g\in G}gPg^{-1} = N, $$ so N is normal.

For the question on $\sigma$ see the accepted answer here:

Intersection of all $p$-Sylow subgroups is normal

So in fact, this works and $N$ is a characteristic subgroup and in particular normal.

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  • $\begingroup$ I already see it now. Thanks! $\endgroup$
    – the inner beauty
    Commented Apr 20, 2022 at 9:49
  • $\begingroup$ $N = \cap_{g\in G}gPg^{-1}$ is by Sylow II, right? $\endgroup$
    – user1007416
    Commented May 15, 2022 at 20:05

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