If $\theta :G\rightarrow H$ is a surjective homomorphism, then $\theta(\Phi(G))\leq\Phi(H)$.

Question

This is a claim when I try to solve another problem related to the Fratinni group of a p-group, and I saw an answer Frattini subgroup of a finite elementary abelian $p$-group is trivial. I am stuck at the second step where the author claims that If $\theta :G\rightarrow H$ is a surjective homomorphism, then $\theta(\Phi(G))\leq\Phi(H)$, where $\Phi(G)$ and $\Phi(H)$ are the Frattini subgroup of $G$ and $H$. Any help is appreciated.

Answer 1

Welcome to MSE!

This follows from the correspondence theorem.

Let $h \not \in \Phi(H)$. Then for some maximal subgroup $M \leq H$, we must have $h \not \in M$ (since $\Phi$ is the intersection of the maximal subgroups). But then any preimage $\theta^{-1} h \not \in \theta^{-1} M$, which is a maximal subgroup by correspondence. So $\theta^{-1}h \not \in \Phi(G)$.

Taking contrapositives gives the desired claim: $\theta [ \Phi(G) ] \leq \Phi(H)$.

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I hope this helps ^_^