## Image and Preimage of Sylow p-subgroups under Epimorphism

Suppose G and H are p-groups, and $\phi:G\to H$ is a surjective homomorphism.

Then for any Sylow p-subgroup P of G, $\phi(P)$ is a Sylow p-subgroup of H.

Conversely, for any Sylow p-subgroup Q of H, $Q=\phi(P)$ for some Sylow p-subgroup P of G.

Proof:

By the First Isomorphism Theorem, $G/\ker\phi\cong\phi(G)=H$. Write $N=\ker\phi$. Then $\phi(P)=\{pN:p\in P\}=PN/N$.

Since $P$ is a Sylow $p$-subgroup of $G$, $[G:P]$ is relatively prime to $p$. Thus, $[G:PN]=[G:P]/[PN:P]$ is also relatively prime to $p$.

Then $\displaystyle [H:\phi(P)]=[G/N:PN/N]=[G:PN]$ is also relatively prime to $p$. Since $\phi(P)\cong P/\ker\phi|_P$, $\phi(P)$ is a $p$-group, so $\phi(P)$ is a Sylow $p$-subgroup of $H$.

Part 2: Let $Q$ be a Sylow $p$-subgroup of $H\cong G/N$. Then by Correspondence Theorem, $Q\cong K/N$ for some subgroup $K$ with $N\subseteq K\subseteq G$.

Then, $[G:K]=[H:Q]$ is relatively prime to $p$, so $K$ contains a Sylow $p$-subgroup $P$.

Consider $P/N\cong\phi(P)\subseteq Q\cong K/N$. By previous part, $\phi(P)$ is a Sylow $p$-subgroup of $H$, so $\phi(P)=Q$.