I have earlier this year in January tried some questions in Group theory but couldn't post them due to my illness.

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

Let $x, y \in f(P)$.

I have prove that $f(P )$ is a subgroup. But why is it subgroup of $P$ I am unable to prove it? Can you please tell what property of normal p -subgroups should I use?


  • 2
    $\begingroup$ By Sylow's theorem, $P$ is the only Sylow $p$-subgroup of $G$. Now $f(P)$ is a $p$-subgroup, so it is contained in some Sylow $p$-subgroup. But since there is only one, we have $f(P)\le P$. $\endgroup$ Sep 2, 2021 at 5:58
  • $\begingroup$ Why are you taking $x,y\in f(P)$? To prove that $f(P)\leq P$, you would normally just need to show that any one element of $f(P)$ is in $P$. $\endgroup$ Sep 2, 2021 at 15:29
  • $\begingroup$ @BrauerSuzuki Why P should be only sylow p-subgroup of G? As far as I know, the theorems implies converse ie if sylow subgroup is unique then it is normal.So, I think your argument is not valid $\endgroup$
    – Avenger
    Sep 19, 2021 at 5:53
  • 2
    $\begingroup$ All Sylow $p$-subgroups are conjugate. So there is only one if and only if it's normal. $\endgroup$ Sep 19, 2021 at 6:05
  • $\begingroup$ @BrauerSuzuki Thanks! $\endgroup$
    – Avenger
    Sep 19, 2021 at 6:22

1 Answer 1


Notice that $P=xPx^{-1}$ for any $x\in G$ ($P$ is normal in $G$). So, if we find an $x\in G$ such that $f(P)<xPx^{-1}$, then we are done. To obtain this, we want to use the second Sylow theorem. Therefore, to use the theorem, we need to show that $f(P)$ is a $p$-subgroup of $G$, i.e. the order of each $f(a)\in f(P)$, $a\in P$, is a power of $p$.

Notice that $|f(a)|$ divides $|a|$, where $|a|$ is a power of $p$ ($P$ is a $p$-subgroup). Then $|f(a)|$ is a power of $p$.


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