Residually finite not implies Hopfian

This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., residually finite group) need not satisfy the second group property (i.e., Hopfian group)
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Statement

A residually finite group need not be a Hopfian group.

Definitions used

Term Definition used
residually finite group every non-identity element is outside of some normal subgroup of finite index, i.e., there is a homomorphism to a finite group where it has a non-identity image.
Hopfian group every surjective endomorphism is an automorphism.

Proof

Consider an infinite-dimensional vector space over a field, or more generally, any infinite external direct power or restricted external direct power of a nontrivial finite group

1. This is residually finite: For any non-identity element, find any coordinate where it takes a non-identity value and consider the normal subgroup of those elements that are the identity at that coordinate. (Equivalently, the quotient map here is projection on that coordinate).
2. This is not Hopfian: Suppose $I$ is the indexing set. Pick $i \in I$ and consider a bijection $I \setminus \{ i \} \to I$. By coordinate shifting, this induces an isomorphism from the quotient by the $i^{th}$ coordinate subgroup to the whole group, and hence, a surjective endomorphism of the whole group that is not injective. Explicitly, if $H_j, j \in I$ are all copies of a nontrivial finite group $H$, and $G$ is the product of the $H_j$s, then a bijection $I \setminus \{ i \} \to I$ induces an isomorphism $G/H_i \to G$ which thus gives a surjective endomorphism from $G$ to $G$ with nontrivial kernel.