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.

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