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 $|H\phi (H)|=p^2$?