Let $G$ a finite group such that $\lvert G \rvert=pm$, with $p$ a prime and $\gcd(p,m)=1$. $G$ has an unique Sylow $p$-subgroup $P$. Prove $P\lhd G$. I just made this exercise, left as homework, and I'm almost sure that I did something wrong, or at least that there's a better way to solve it. Here it goes:

Let $G$ a finite group such that $\lvert G \rvert=pm$, with $p$ a prime and $\gcd(p,m)=1$. Suppose that $G$ has an unique Sylow $p$-subgroup $P$. Prove that $P\trianglelefteq G$.

My try:

Proof: The Sylow $p$-subgroups $P$ of $G$ are, in our case, the subgroups of $G$ with order $p$. There's only one, so $\lvert P \rvert=p$. We know that, if $N_G(P)$ denotes the normaliser of $P$ in $G$, verifies this: $P\trianglelefteq N_G(P)$, because $P\leq G$. Also we know that the normaliser $P$ is exactly the centralizer of $P$, that is a subgroup of $G$. Then:
\begin{equation}
P\trianglelefteq N_G(P)=G_P\leq G \implies P \trianglelefteq G.
\end{equation}

We started Sylow today, so I used some lemmas we proved this week to make this proof, but it seems to me that there's something wrong. 
I would like my proof to be criticized. Thank you.
 A: Any automorphism, $\varphi$, of $G$ sends $P$ to another group of order $p$. But there is only one, so the image $\varphi(P)=P$, since it is the only one. So $P$ is characteristic, hence normal--since conjugation is just applying the automorphism $g\mapsto x^{-1}gx$.
I'm not sure how your proof achieves the proof of normality, if you want that approach to be more effective, you might want to add more details to make it clearer to the reader how that actually proves the result you're asked to show.
Another simple way of showing this--if you don't want to suss out the details of your proof more explicitly to make it work--is to note that $x^{-1}Px$ is a subgroup of $G$ of order $p$ (it is a bijective homomorphism) and apply uniqueness--if you haven't learned the word "characteristic" yet, this is the low-tech way of doing the proof I started my answer with.

If you cannot see the bijective homomorphism property, it is fairly easy:
$1$) Bijection:  Consider the map $\varphi_x(g)=x^{-1}gx$, then the (two-sided) inverse is the map $\varphi_{x^{-1}}: g\mapsto xgx^{-1}$, so it is a bijection.
$2$) Homomorphism:  Consider
$$\varphi_x(gh)=x^{-1}ghx=x^{-1}g(xx^{-1})hx$$
$$=(x^{-1}gx)(x^{-1}hx)=\varphi_x(g)\varphi_x(h).$$
