Let $G$ be a group and let $M \le G$ be a maximal subgroup of $G$. Let $f:G \to H$ be a surjective homomorphism such that $M$ doesn't contain $kerf$.
The task is to prove that $f(M) = H$.
My attempt: I say the $\{s_1, …, s_n \}$ is the generating set of $M$. I assume by contradiction that $f(M) \ne H$ so there exists $e \ne h \in H$ such that $h \notin f(M)$. Since $f$ is surjective, there exists $e \ne g \in G$ such that $f(g) = h$, which implies $ g \notin M$. The I want to look at the subgroup that is generated by $\{s_1, …, s_n, g \}$ which is bigger then $M$ and show that it is not $G$ which means it is a contradiction. However, I couldn't manage to prove it is not $G$, and I am stuck here.
Another direction at which I thought about is to somehow use the correspondence theorem, since $kerf$ is a normal subgroup of $G$, but I am not sure how can I use it here.
Help would be appreciated.
Best Answer
Consider $L=f^{-1}(f(M))$; it is a subgroup which contains $M$ and which is different of $M$ since there exists $x\in kerf$ which is not in $M$ and $x\in L$. This implies that $L=G$ since $M$ is maximal, we deduce that $f(L)=f(M)=H$.