Element of prime power in normalizer Let $K$ be Sylow $p$-subgroup of a finite group $G$; than prove that if $x \in N_G(K)$ and the order of $x$ is power of $p$, than $x \in K$.
My try ,
Let $K$ be a Sylow $p$-subgroup,
$o(K) = p^k$ and since $x$ is in $N_G(K)$ then  $ xKx^-1 = K$.
Now $p^a / p^k $ where $1\le a \le k$ and also $p^a/ o(N_G(K))$ this implies that the $N_G(K)$ has a subgroup whose order is $p^a$ after this how to conclude...
 A: It's not a trivial claim but neither that hard to prove:
Claim 1: With the question's notation, $\;[N(K):K]=\left|N(K)/K\right|\;$ is coprime with $\;p\;$ .
Proof: Otherwise there's an element $\;xP\;,\;\;x\in N(K)\;$ of order $\;p\;$ (why?), and thus $\;\langle xP\rangle\le N(K)/K\;$ is of order $\;p\;$ . By the correspondence theorem, there exists a subgroup $\;P\le X\le N(H)\;$ s.t. $\;X/K=\langle xP\rangle\;$ .
But then both $\;K\;$ and $\;X/K\;$ are $\;p\,-$ groups, which makes $\;X\;$ a $\;p\,-$ group containing $\;P\;$, contradicting the maximality of $\;P\;$ as $\;p\,$ subgroup of $\;G\;\;\;\;\;\square$ .
Now to our original problem: suppose the order of $\;x\;$ is $\;p\;$ (otherwise simply take $\;x^{p^{a-1}}\;$...) . If $\;x\notin K\;$ , then $\;xP\;$ is an element of order $\;p\;$ in $\;N(K)/K\;$ , contradicting claim 1 above. Q.E.D.
A: Let $x \in N_G(K)$ have order a power of $p$, so the subgroup $\langle x \rangle$ has $p$-power order. Since $K \unlhd N_G(K)$, the set $K\langle x \rangle$ is actually a subgroup of $N_G(K)$. And $|K\langle x \rangle|=\frac{|K| \cdot |\langle x \rangle|}{|K \cap \langle x \rangle|}$. which is a $p$-power. But $K$ is the Sylow subgroup and hence the largest $p$-subgroup of $N_G(K)$. So $K\langle x \rangle=K$, which implies $x \in K$.
