This is supposed to be well-known and easy to establish. If $G$ is Hopfian, then any surjective endomorphism $f:G \rightarrow G$ must be an automorphism, hence a unit in the endomorphism ring, hence correspond to a unit of $\mathbb{Z}.$ So, $f$ corresponds to $\pm 1.$ That is, $f$ is $\pm 1_G$, $1_G$ being the identity function on $G.$ Is this the correct approach to take?
2026-02-23 13:37:49.1771853869
If $G$ is a group and $\mathrm{End}(G) \cong \mathbb{Z},$ why is $G$ Hopfian?
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