Show that if $a^{-1}Pa=P$ with $P$ a Sylow $p$-subgroup and $a$ of order $p^m$ then $a\in P$ 
If $P$ is a Sylow $p$ subgroup and $a\in G$ is of order $p^m$ for some $m$, show that if $a^{-1}Pa=P$ then $a\in P$.

Attempt:
Since $P$ is a Sylow $p$ subgroup so is $a^{-1}Pa$ and hence they are conjugates so $a^{-1}Pa=g^{-1}Pg$ for some $g$ .But 
$a^{-1}Pa=P\implies g^{-1}Pg=P$.
I feel I am getting nowhere.Please help me someone.
 A: $\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$If $H, K \le G$, we have

$H K = \{ h k : h \in H, k \in K \}$ is a subgroup of $G$ iff $K H \subseteq H K$.

Let $H = \langle a \rangle$, $K = P$ of order $p^{n}$, say. For $h = a^{i} \in H$ and $k \in P$ we have
$$
k h = k a^{i} = a^{i} a^{-i} k a^{i} \in a^{i} K \subseteq H K.
$$
Hence $H K$ is a subgroup of $G$, and
$$
p^{n} = \Size{K} \text{ divides } \Size{H K} = \frac{\Size{H} \Size{K}}{\Size{H \cap K}} \text{ divides } p^{n + m}.
$$
It follows that $K = H K$, so that $a \in K$.
A: Suppose $a^{-1}Pa=P$ for $a\notin P$
Then $a\in N_G(P)$, but then $p$ divides $|N_G(P)/P|$ (since $aP$ has order dividing $p^m$).
Hence $|N_G(P)|=|N_G(P)/P||P|$ is divisible by $p|P|$ contradicting the fact that $P$ is a Sylow $p$ subgroup.
A: If a $p$-element $a$ normalizes the Sylow $p$-subgroup $P$, then $a \in N_G(P)$. But $P$, being a Sylow $p$-subgroup of its normalizer, is the only Sylow $p$-subgroup of $N_G(P)$ (it is by definition normal in its normalizer!). Hence the $p$-subgroup $\langle a \rangle \subseteq P$, in particular $a \in P$.
