Maximal subgroups and isomorphism Let $G$ and $H$ be two groups and $\phi:G\rightarrow H$ be a homomorphism.
Suppose that $M$ and $N$ are maximal subgroups of $G$ and $H$, respectively.
Now, are subgroups $\phi(M)$ and $\phi^{-1}(N)$ maximal subgroups of $N$ and $M$ respectively?
 A: The answer is no. Take for example $G=H=\mathbb{Z}/4\mathbb{Z}$ and $\phi:\mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z}:x \mapsto 4x$. There is only one choice of maximal subgroup, $M=N=2\mathbb{Z}/\mathbb{Z}$, but $\phi(M) = 4\mathbb{Z}/4\mathbb{Z} \neq N$ and $\phi^{-1}(N) =\mathbb{Z}/4\mathbb{Z} \neq M$.
Classification question
In the comments, a classification question was asked:
Suppose $\phi:G\to H$ had the property that for every maximal subgroup $M$ of $G$, $\phi(M)$ is a maximal subgroup of $H$. For finite groups, such a $\phi$ cannot be very much different from an isomorphism unless it is a trivial map with either $G$ or $H$ very special.
If $G$ has a maximal subgroup $M$ containing the kernel of $\phi$, then $\phi(M) < \phi(G)$ cannot be maximal in $H$ unless $\phi(G) = H$. Hence in such a case $\phi$ must be surjective. Then if $M$ is any maximal subgroup, the kernel $\ker(\phi)$ must be contained in $M$, since
otherwise $\phi(M) = \phi(M+\ker(\phi)) = \phi(G) = H$ is not maximal in $H$.
If $\phi$ is not surjective, then the kernel of $\phi$ is not contained in any maximal subgroup of $G$. If $G$ is finite (noetherian), then the kernel must be all of $G$, so that $\phi:G \to H:x \mapsto 1_H$ is trivial. If additionally $G$ is not the identity, then $\phi(M) = 1$ must be maximal in $H$, so $H$ is cyclic of prime order. If $G$ is the identity, then again $\phi(G)=1$ but $H$ can be anything, since $G$ has no maximal subgroups.
Proposition: If $G$ is finite, then $\phi:G \to H$ takes maximal subgroups of $G$ to maximal subgroups of $H$ iff one of the following cases occurs:


*

*$\phi(G)=H$ and $\ker(\phi) \leq \Phi(G)$ [ “$G$ is a Frattini extension of $H$” ]

*$H$ is cyclic of prime order and $\phi(G)=1$

*$|G|=1$


Now if $G$ is not finite, one can get other weird examples such as $\phi:\mathbb{Q} \to \mathbb{Q}:x \mapsto 0$. Here $\phi$ is not surjective, and yet every maximal subgroup of $\mathbb{Q}$ is mapped to a maximal subgroup of $\mathbb{Q}$.
A similar example takes $\phi:\mathbb{Q} \times \mathbb{Z}/2\mathbb{Z} \to \mathbb{Q} \times \mathbb{Z}/2\mathbb{Z}: (x,y) \mapsto (x,0)$. The kernel is $0\times \mathbb{Z}/2\mathbb{Z}$. The Frattini subgroup of $G$ is $\Phi(G)=\mathbb{Q}\times 0$, so there is a maximal subgroup supplementing the kernel. However, there is no maximal subgroup containing the kernel. More directly, the only maximal subgroup of $G$ is $M=\mathbb{Q} \times 0$ and $\phi(M)=M$ is indeed a maximal subgroup of $H$. Note that $\phi$ is not surjective, and $\ker(\phi)$ is not contained in the Frattini.
