If Q is a p-Sylow-Group of H there is a p-Sylow-Group P of G with $\phi(P)=Q$ while $\phi:G\rightarrow H$ epimorphism

Question

Let G be a finite group and $\phi: G \rightarrow H$ a group-epimorphism. Proof: If $Q\in Syl_p(H)$ there is a $P\in Syl_p(G)$ with $Q=\phi(P)$.

Answer 1

Since $\phi$ is an epimorphism, $[H:Q] = [G:\phi^{-1}(Q)]$ is something coprime to $p$. This shows that any Sylow-$p$-subgroup of $\phi^{-1}(Q)$ is a Sylow-$p$-subgroup of $G$. Let $P$ be a Sylow-$p$-subgroup of $\phi^{-1}(Q)$. We want to show that $\phi(P) = Q$.

Suppose $\phi^{-1}(Q)$ has size $np^k$, where $n$ is coprime to $p$, and suppose $Q$ has size $p^\ell$ with $\ell < k$ since $\phi$ is an epimorphism.

The kernel $K$ of the restriction $\phi : \phi^{-1}(Q)\rightarrow Q$ has size $np^{k-\ell}$. Since the group generated by $K$ and $P$ is a subgroup of $\phi^{-1}(Q)$ of size divisible by $np^{k-\ell}$ and $p^k$, we conclude that it's all of $\phi^{-1}(Q)$. This shows that $\phi(P) = Q$.

Answer 2

Pick $R\in\operatorname{Syl}_p(\phi^{-1}(Q))$ and the $P\in\operatorname{Syl}_p(G)$ such that it contains the $p$ subgroup $R<G$.

Answer 3

Since it is a epimorphism, we can think the canonical map $\phi: G \to G/N$; where $N$ is normal in $G$.

Notice that if $H\in Syl_p(N)$ then $H=P\cap N$ for some $P\in Syl_p(G)$ and you can say that $\phi(P)\in Syl_p(G/N)$ as $p \nmid [G:P]\implies p\nmid [\bar G: \bar P]$.

Now, if $Q\in Syl_p(\bar G)$ then $\phi( P)^{\bar g}=Q \implies \phi(P^g)=Q$ so we are done.