If $M$ is a maximal subgroup and $f$ a surjective homomorphism then $f(M) = H$ 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.
 A: 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 $f(L)=f(M)=H$.
A: We are given the existence of
$k \in \ker f \setminus M; \tag 1$
let
$\langle M, k \rangle \le G \tag 2$
be the smallest, with respect to set inclusion, subgroup of $G$ containing both $M$ and $k$; $\langle M, k \rangle$ is evidently the subgroup of $G$ generated by the elements of $M$ and $k$, in the sense that every $p \in \langle M, k \rangle$ may be written as a product
$p = \displaystyle \prod_1^n p_i, \; n \in \Bbb N, \tag 3$
where either $p_i \in M$ or $p_i = k^m$, $m \in \Bbb Z$, $1 \le i \le n$.  Another way to describe $\langle M, k \rangle$ is as the set of products
$p = \displaystyle \prod_1^n p_i, \; p_i \in M \cup \langle k \rangle, \; 1 \le i \le   n; \tag 4$
that is, the $p \in \langle M, k \rangle$ are generated by taking arbitrary finite products from the set $M \cup \langle k \rangle$.  It is easy to see that the set $\langle M, k \rangle$ of all such $p$ is closed under the multiplication given on $G$, that it contains the identity $e$ of $G$ (just take every $p_i = e_G$, the identity of $G$), and that 
$p^{-1} = \displaystyle \prod_0^{n - 1} p_{n - i}^{-1} \in \langle M, k \rangle; \tag 5$
it follows then by virtue of (1) that $\langle M, k \rangle$ is itself a group and that
$M \subsetneq \langle M, k \rangle \le G. \tag 6$
Now since $M$ is a maximal subgroup of $G$, we in fact must have
$\langle M, k \rangle = G, \tag 7$
since it properly contains $M$ (6).  Thus every $g \in G$ may be expressed as a product of the form (3), whence
$f(g) = f \left ( \displaystyle \prod_1^n p_i \right ) = \displaystyle \prod_1^n f(p_i) \in f(M), \tag 8$
since $p_i \in M$ or $p_i \in \langle k \rangle$, $1 \le i \le n$; in the former case, $f(p_i) \in M$; in the latter, $f(p_i) = e_H$ the identity of $H$, since $\langle k \rangle \le \ker f$; since $f(g) \in f(M)$ for every $g \in G$, we conclude that
$H = f(G) \subset f(M) \subset f(G) = H, \tag 9$
and so
$f(M) = H \tag{10}$
as required.  $OE\Delta$.
