I'm stuck on this herstein exercise for a long time.
Let $P$ is a $p$-Sylow subgroup of $G$ and order of $a$ is a prime power then if $a\in N(P)$ prove $a\in P$
I was doing like this but stuck in middle, consider $H'=\cup_{i=1}^{\infty} a^iP$. So if $n$ is smallest such that $a^n\in P$ we get $|P'|=n|P|$ also $|P'| \mid |N(P)|$ so if we can show $n\neq 1 \implies p|n$ we would be done, but I'm stuck here. Please help me someone with this, or give any new hint/solution
On
As Derek suggested, we need to assume that the order of $a$ is a power of $p$; we can find a counterexample if not for any abelian group whose order is not a prime power. A somewhat simpler proof than the other answer is to first note that $P$ is a normal Sylow $p$-subgroup of $N(P)$ because by definition it is normal and it is of maximal order with respect to having order a power of $p$. Every subgroup of $N(P)$ whose order is a power of $p$ is therefore contained in $P$. This includes the cyclic subgroup generated by $a$, hence $a\in P$.
An idea in hints:
== Show that $\;\left(\left|N_G(P)/P\right|\,,\,\,p\right)=1\;$ . Further Hint: use Cauchy theorem to show that otherwise there exists $\;g\in G\;\;s.t.\;\;\left|\langle gP\rangle\right|=p\;$ , and use the Correspondence theorem to show there exists a $\;p$ - subgroup subgroup of $\;G\;$ containing $\;P\;$ , which cannot be.
== Show that we can assume $\;ord(a)=p\;$ and assume $\;a\in N_H(P)\setminus P\;$ , and get an immediate contradiction to the first point above.