# Residually finite and finitely many homomorphisms to any finite group implies Hopfian

## Statement

Suppose $G$ is a residually finite group and is also a group having finitely many homomorphisms to any finite group. Then, $G$ is a Hopfian group.

## Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

Given: A group $G$ that is resudally finite and has finitely many homomorphisms to any fixed finite group. A surjective endomorphism $\varphi$ of $G$.

To prove: $\varphi$ is an automorphism.

Proof: We prove this by contradiction.

ASSUMPTION THAT WILL LEAD TO CONTRADICTION: We assume that $\varphi$ is not an automorphism. Since it is surjective, it must fail to be injective, so that it must have a nontrivial kernel. Let $g$ be a non-identity element in the kernel of $\varphi$.

Step no. Assertion/construction Given data/assumptions used Previous steps used Explanation
1 There is a surjective homomorphism $\alpha:G \to K$ for some finite group $K$ such that $\alpha(g)$ is not the identity element $G$ is residually finite, $g$ is a non-identity element [SHOW MORE]
2 All the homomorphisms $\alpha \circ \varphi^n$, with $n$ varying over the natural numbers, are pairwise distinct homomorphisms from $G$ to $K$ $\varphi$ is surjective, $g$ is in the kernel of $\varphi$ Step (1) [SHOW MORE]
3 There are only finitely many homomorphisms from $G$ to $K$ $G$ has only finitely many homomorphisms to any finite group Step (1) (specifically, that $K$ is finite) Given direct.
4 We have the required contradiction. Steps (2), (3) [SHOW MORE]