Extend isomorphism of subgroups to homomorphism of groups Given two finite groups $G_1$ ang $G_2$ with respective subgroups $H_1$ and $H_2$ satisfying $H_1\cong H_2$ via the isomorphism $\phi$, is it always possible to extend $\phi$ to an homomorphism $\bar\phi:G_1\to G_2$? i.e. is it possible to construct a morphism $\bar\phi:G_1\to G_2$ with $\bar\phi(x)=\phi(x)$ for all $x\in H_1$?
 A: Take the two alternating groups $A_n,\  A_m$ with $m,n \ge5,\, m > n$. Both of them contain elements of order $5$, and hence we have subgroups $H_1\subset A_m$ and $H_2\subset A_n$ of order $5$ in them that are isomorphic (e.g., they could be the subgroups generated by $5$-cycles).
But we cannot  find a homomorphism $A_m\to A_n$ extending this isomorphism of $H_1$ with $H_2$: it cannot be trivial as the image has to contain $H_2$, and it cannot  be injective as $A_m$ has order bigger than $A_n$. There are no other normal subgroups ($A_m$ is  simple), hence  there is no candidate for the kernel for any intended extended homomorphism.
A: No. Let $G_1$ be any simple nonabelian finite group and $H_1$ a proper nontrivial subgroup, and then define $G_2=H_2\cong H_1$. Our putative map $\tau:G_1\to G_2$ cannot be the trivial map as it restricts to the isomorphism $H_1\to H_2$, but $\tau$ must have nontrivial kernel since $|G_1|>|G_2|$. Kernels are normal subgroups and this contradicts the hypothesis that $G_1$ is simple.
A: Here is another counter example with $G_2$ abelian, and $G_1$ not simple. Take $G_2$ as  Klein's 4-group and  $G_1$ to be $S_n$ with $n>4$.
As we can find in $G_1$ two transpositions $\tau, \tau'$ operating on disjoint subsets the subgroup generated by them is ismorphic to the Klein's group, $G_2$.
As $S_n$ does not have a normal subgroup of index 4, the isomorphism of these two Klein's groups cannot be extended to whole of $S_n$ as a homomorphism.
