Characteristic subgroup 
Suppose that $|G|=pm$, where $p\nmid m$ and p is a prime. If $H$ is normal subgroup of order $p$ in $G$, prove that $H$ is characteristic.

I looked up the solution manual:

Since $H$ is of order $p$ and is normal in $G$, if $\phi$ is an automorphism of $G$ and if $\phi (H)\neq H$, then $H\phi (H)$ is  a subgroup of G and is of order $p^2$. Thus $p^2=|H\phi (H)|$ must divide $|G|=pm$, by Lagrange's Theorem. Thus $p^2|pm$, and so $p|m$, contrary to assumption. Thus $H$ is characteristic.

What I didn't understandwas that if $H$ is characteristic, then $H=\phi (H)$, therefore $|H\phi (H)|=\frac {|H||\phi (H)|}{|H\bigcap \phi (H)|}=p$. 
On the other hand, If $|H\phi (H)|=p^2$, then since $|H\phi (H)|=\frac {|H||\phi (H)|}{|H\bigcap \phi (H)|}=p$, $|H\bigcap \phi (H)|=1$, therefore, $H$ is not characteristic, since if it were, then $H\bigcap \phi (H)=\phi (H)\neq (e)$.
 So how can you say that $|H\phi (H)|=p^2$?
 A: @pxc3110 You are perfectly right, the solution manual assumes $|\phi(H)H|=p^2$ and this is not obvious since $\phi(H) \cap H$ can contain more elements than only the identity. Let us do it differently, observe that $H \subseteq H\phi(H) \subseteq G$ and that index$[\phi(H)H:H]$ divides index$]G:H]=m$. As you observed, index$[\phi(H)H:H]$ is a $p$-power. But $p$ does not divide $m$. Hence index$[\phi(H)H:H]=1$, that is $\phi(H)H=H$, which implies that $\phi(H) \subseteq H$. The same argument goes for $\phi^{-1}$, hence $\phi(H)=H$.
A: Perhaps it is easier to phrase things this way : $H$ is a $p-$Sylow subgroup of $G$. Since $H$ is normal, it is the only $p-$Sylow subgroup of $G$. Hence, if $\varphi$ is any automorphism of $G$, $\varphi(H)$ is also a $p$-Sylow subgroup of $G$, whence $\varphi(H) = H$.
Hence, $H$ is characteristic.
A: In a proof by contradiction, we assume nonsense. Indeed, it turns out to be nonsense that $H\phi(H)$ has order $p^2$. So if you start there, you will reach some absurd statement, which is what you did.
The logic has to go through like this: assume H is not characteristic. Then $H\cap\phi(H)$ is trivial, so $H\phi(H)$ has order $p^2$, which is a contradiction. That proves that H is characteristic.
