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If $P$ is a normal Sylow p-subgroup of a finite group $G$ and $f:G\to G$ is an endomorphism, then $f(P)$ is a subgroup of $P$.

I have a solution but it does not use that fact that $P$ is Sylow p-subgroup of $G$. Any help is greatly appreciated. Thanks.

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    $\begingroup$ What is your solution? Perhaps it requires being a Sylow $p$-subgroup, but you don't see where it requires it (or it could be wrong). $\endgroup$
    – Clayton
    Feb 12, 2013 at 12:34

2 Answers 2

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$f(P)$ is a $p$-group (for the first isomorphism theorem), so it is contained is some $p$-Sylow (for the Sylow's theorem). But since $P$ is a normal $p$-Sylow, it is the unique $p$-Sylow (this again for the Sylow's theorem); so $f(P)\subseteq P$.

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  • $\begingroup$ If $P$ is a normal Sylow p-subgroup of $G$, does it always follow that it is the unique Sylow p-subgroup of $G$? $\endgroup$ Feb 12, 2013 at 13:04
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    $\begingroup$ Yes, any two Syslow p-subgroups are conjugate. $\endgroup$ Feb 12, 2013 at 13:56
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I have discovered this answer thru Galoisfan's idea:
Let $y\in P$. Since $P$ is a Sylow p-subgroup, every element of $P$ has order a power$(\ge0)$ of some fixed prime $p$, i.e, $y^{p^k}=e$. Thus, $f(y^{p^k})=f(e)$ implies that $[f(y)]^{p^k}=e$, where $f(y)\in f(P)$. Hence, every element of $f(P)$ has order a power of $p$ and so $f(P)$ is a $p$-subgroup of $G$. By the Second Sylow Theorem, there exists $x\in G$ such that $f(P)$ is a subgroup of $xPx^{-1}$. But since $P$ is normal in $G$, $xPx^{-1}=P$. Therefore, $f(P)$ is a subgroup of $P$.

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    $\begingroup$ Yes, that's exactly what Galoisfan meant, but it is always better to write and understand such ideas in one's own words - as you just did. $\endgroup$ Feb 12, 2013 at 13:58
  • $\begingroup$ I am glad I figured it out sir Hagen von Eitzen. Thanks. $\endgroup$ Feb 12, 2013 at 14:04

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