If $M$ is a maximal subgroup and $f$ a surjective homomorphism then $f(M) = H$

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 $f(L)=f(M)=H$.

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. $OE\Delta$.