I have a group isomorphism $G/\mathbb{Z}\to \mathbb{Z}^{(*n)}$, can I conclude from this fact that $G$ is isomorphic to $\mathbb{Z}^{(*(n+1))}$, the free product of (n+1) copies of $\mathbb{Z}$.
2026-04-01 10:23:59.1775039039
Do isomorphisms of quotients give isomorphisms of groups?
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No, since perhaps $G$ is $\Bbb Z\times\Bbb Z^{(*n)}$.