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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!

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    $\begingroup$ Hint. Think about the second isomorphism theorem. $\endgroup$ – Arturo Magidin Jul 2 '19 at 4:27
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As stated in the comment we use second isomorphism theorem,

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, $H_1 \cong H_2$.

Hope this helps.

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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.

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