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Given finitely generated (not necessarily abelian) group $G$ and normal subgroup $N\triangleleft G$, then $G/N\cong H\leq G$. Is this statement correct?

The reason why I am suspicious this to be true is because I know for cyclic groups, this is known to be true (edit: it is not). And as long as $N$ is cyclic, it's easy to see that so too can this result be extended to the finitely generated case (I'm very confident in this, at least). When considered the cases in which $N$ is not cyclic, however, it becomes a little bit less obvious.

My expectation is that this claim may not be true judging by the fact that I haven't been able to find any sources online confirming the statement. But nonetheless, I appreciate any and all help I can get. Thanks all!

J.G.131
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As pointed out, the very first counterexample that comes to mind is $(G, N) = (\mathbb{Z}, 2\mathbb{Z})$. I assume what you claimed to be true is that all quotients of finite cyclic groups can be realized as subgroups. Moreover, you can use the classification of finite abelian groups to show that any quotient of a finite abelian group is also isomorphic to some subgroup. Perhaps the intuition to have here is that finite abelian groups (more generally, modules over PIDs) are quite close to vector spaces in terms of their good behavior?

Anyways, the claim does not hold true for arbitrary finite groups either. The smallest counterexample is $(G, N) = (Q_8, \langle -1 \rangle)$, where $Q_8$ denotes the quaternion group of order $8$. Indeed, the quotient $G/N$ is isomorphic to the Klein-$4$ group, which has three elements of order $2$, but $G$ only has one. (You can see that this is the smallest counterexample since the only other non-abelian groups up to isomorphism of order at most $8$ are $S_3$ and $D_8$, both of which have the property you mention).

Haran
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