1
$\begingroup$

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 $|\psi^{-1}(S)| < p^m$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Crostul Jan 15 '16 at 11:06
2
$\begingroup$

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 $H$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.