If $f:G\to H$ is a surjective homomorphism of finite groups, then $f$ sends Sylow $p$-subgroups to Sylow $p$-subgroups.

Here's what I have. Suppose $\vert G \vert=p^km$ with $(p,m)=1$. Let $P\in \text{Syl}_p(G)$. Then $\vert P \vert =p^k$. By the 1st Isomorphism Theorem, $\vert H\vert =p^km/ \vert \ker (f)\vert=p^ln$ for some $l\le m$ and $n\bigg\vert m$, with $(p,n)=1$. How can I show $f(P)$ has order $p^l$? Is there a better approach to this?


1 Answer 1


Let $P$ be a $p$-Sylow in $G$. Of course, $f(P)$ is a $p$-subgroup of $H$. We have to show that the index of $f(P)$ in $H$ is prime to $p$. Let $\bar P$ be the subgroup $f^{-1}(f(P))$ of $G$. Since $P\subseteq \bar P$, the index of $\bar P$ in $G$ is also prime to $p$. Since $f$ is surjective, the induced map $$ \bar f\colon G/\bar P\rightarrow H/f(P) $$ is a bijection. It follows that the index of $f(P)$ in $H$ is prime to $p$.

  • $\begingroup$ Why $f(P)$ is a p group? How did you show that $\endgroup$
    – Napu
    Apr 19, 2021 at 19:19

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