prove that two subgroups are isomorphic

Assume that $$G$$ is a group, and that $$N$$ is a maximal normal subgroup of $$G$$. This means that if $$H$$ is a subgroup of $$G$$ such that $$N\subsetneq{H}$$, then $$H=G.$$ Assume that $$H_1$$ and $$H_2$$ are two non-trivial subgroups of $$G$$ (i.e. $$H_1\neq{\{1\}}\neq{H_2}$$) such that $${H_1}\cap{N}=\{1\}={H_2}\cap{N}.$$ Then, it must be the case that $$H_1$$ is isomorphic to $$H_2$$.

This is an exercise in a graduate level algebra textbook. I cannot think of a way to approach this problem!

• Hint. Think about the second isomorphism theorem. – Arturo Magidin Jul 2 '19 at 4:27

Since $$H_1N\ \text{and}\ H_2N$$ are subgroups of $$G$$ such that $$N \leqslant H_1N, H_2N \Rightarrow H_1N\ \text{and}\ H_2N$$ are equal to $$G.$$ (Since $$N$$ is maximal.)
Now from the special case of the second isomorphism theorem, we have for $$N$$ a normal subgroup and $$H$$ an arbitrary subgroup with $$HN =G$$, $$G/N \cong H/(H \cap N)$$. Therefore, $$G/N \cong H_1/(H_1 \cap N) = H_1$$ and similarly $$G/N \cong H_2/(H_2 \cap N) = H_2$$.
Hence, $$H_1 \cong H_2$$.
Since $$N$$ is a maximal normal subgroup of $$G$$, the quotient group $$G/N$$ can have no proper subgroups. This is because under the projection homomorphism $${\pi}:G\rightarrow{G/N}$$ given by $$g\mapsto{gN}$$, the inverse image of any non-trivial subgroup of $$G/N$$ would be a subgroup of $$G$$ that contains $$N$$. I claim (inspired by your proofs!) that any subgroup $$H$$ of $$G$$ for which $$H\cap{N}=\{1\}$$ is isomorphic to $$G/N$$. (Hence, two such subgroups must be isomorphic to each other.) To prove this, consider the map $${\phi}:H\rightarrow{G/N}$$ defined to be $${\phi}(h)=hN$$. This is an injective group homomorphism (since $$H\cap{N}=\{1\}$$), whose image $${\phi}(H)$$ must be a non-trivial subgroup of $$G/N$$. But as $$N$$ is maximal, this only such is $$G/N$$ itself, proving that $$\phi$$ is surjective and hence an isomorphism.