Homomorphic image of a Sylow p-subgroup is Sylow p-subgroup.

Let $G$ be finite group and $\phi$ : $G \to H$ be a group epimorphism. If $P$ is a Sylow p-subgroup of $G$ then $\phi(P)$ is a Sylow p-subgroup of $H$.

It looks very simple but I'm stuck here.

Let $\lvert P\rvert=p^{k}$. Clearly $\lvert \phi(P)\rvert=p^{n}$ where $n\leq k$. Suppose $\phi(P)$ is not a maximal p-subgroup in $H$ and let $Q$ be a maximal p-subgroup of $H$ containing $\phi(P)$. Then $\lvert Q\rvert=p^{m}$ where $n<m$ and $\phi^{-1}(Q)$ is strictly larger than $P$.

For the rest of the proof, what can I do? Thanks.

Solution 1:

Hint: by one of the isomorphism theorems $H = \phi(G) \cong G/ker(\phi)$. Write $N=ker(\phi)$. The image of $P$ under $\phi$ is $PN/N$. And this latter group $\cong P/P \cap N$. Can you show from here that $PN/N \in Syl_p(G/N)$?