# If f is a endomorphism and P is a normal Sylow p-subgroup then...

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?

thanks

• 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 $f(P)\le P$. Sep 2, 2021 at 5:58
• 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 $P$. Sep 2, 2021 at 15:29
• @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 Sep 19, 2021 at 5:53
• All Sylow $p$-subgroups are conjugate. So there is only one if and only if it's normal. Sep 19, 2021 at 6:05
• @BrauerSuzuki Thanks! Sep 19, 2021 at 6:22

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 $$f(P), then we are done. To obtain this, we want to use the second Sylow theorem. Therefore, to use the theorem, we need to show that $$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$$.