# Inverse image of Sylow $p$-subgroup is Sylow $p$-subgroup

I am trying to prove or disprove the following statement:

Let $H$ and $G$ be finite groups, $p$ a prime number, $\psi: H \to G$ an injective homomorphism and $S$ a Sylow $p$-subgroup of $G$ with $\psi^{-1} ( S ) \neq 1$. Then $\psi^{-1}(S)$ is a Sylow $p$-subgroup of $H$.

It is easy to show that $\psi^{-1}(S)$ is a $p$-group. So in order to prove the statement, we would need to show that $\psi^{-1}(S)$ is a maximal $p$-group. But I am not sure whether injectivity is enough to prove this statment (and I assume it isn't). But I cannot think of a counterexample.

For a counterexample we would need a group $H$ of order $ap^m$ und $G$ of order $bp^n$ with $a\,|\,b$, $p\,\not |\, a,b$, $m\leq n$ and a subgroup $S \leq G$ with $|S|=p^n$, but $|\psi^{-1}(S)| < p^m$.

• It may be helpful to identify $H$ with a subgroup of $G$. Then you have to prove that a Sylow $p$-subgroup of $G$ intersected with $H$ gives you a Sylow $p$-subgroup of $H$. Can you conclude? – Crostul Jan 15 '16 at 11:06

This is not true in general. For instance let $G=S_5$, $H=\{\sigma \in S_5 \mid \sigma(5)=5\}$ and $\psi$ the canonical inclusion $H \to G$.
Then the subgroup $S$ of $G$ generated by $(1,2)(3,5)$, $(1,3)(2,5)$ and $(1,2)$ is a $2$-Sylow of $G$ (it has cardinality 8), but $\psi^{-1}(S)=S \cap H = \langle(1,2)\rangle$ has cardinality 2 and hence is not a $2$-Sylow of $H$.