Finitely generated and residually finite implies Hopfian

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., finitely generated residually finite group) must also satisfy the second group property (i.e., Hopfian group)
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Statement

Any finitely generated residually finite group (i.e., a group that is both finitely generated and residually finite) is a Hopfian group.

Definitions used

Term	Definition used here
Finitely generated group	has a finite generating set
Residually finite group	every non-identity element is outside some normal subgroup of finite index. In particular, there is a surjective homomorphism to a finite group such that the given non-identity element is not in the kernel of the homomorphism.
Hopfian group	every surjective endomorphism of the group is an automorphism

Facts used

Proof

Proof from given facts

The proof follows directly by piecing together facts (1) and (2).

Hands-on proof

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Given: A finitely generated group $G$ that is also residually finite. A surjective homomorphism $\varphi :G\to G$ .

To prove: $\varphi$ is an automorphism of $G$ .

Proof: We prove this by contradiction.

ASSUMPTION THAT WILL LEAD TO CONTRADICTION: Suppose $\varphi$ is not an automorphism of $G$ . Then, since it is surjective, it must fail to be injective, so there exists an non-identity element $g$ in its kernel.

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] Since $G$ is residually finite, and $g$ is a non-identity element, there is a normal subgroup $N$ of finite index in $G$ such that $g\notin N$ . Let $K=G/N$ and $\alpha :G\to K$ be the quotient map.
2	All the homomorphisms $\alpha \circ \varphi ^{n}$ , for $n$ varying over positive integers, are pairwise distinct homomorphisms from $G$ to $K$	$\varphi$ is surjective, $g$ is in the kernel of $\varphi$	Step (1)	[SHOW MORE] Suppose $m<n$ . We need to show that $\alpha \circ \varphi ^{m}\neq \alpha \circ \varphi ^{n}$ . Since $\varphi$ is surjective, $\varphi ^{m}$ is also surjective. In particular, there is $h\in G$ such that $\varphi ^{m}(h)=g$ . Since $n>m$ and $g$ is in the kernel of $\varphi$ , $\varphi ^{n}(g)$ is the identity element. Thus, $\alpha \circ \varphi ^{m}$ sends $h$ to a non-identity element of $K$ but $\alpha \circ \varphi ^{n}$ sends $h$ to the identity element of $K$ . Hence, these two maps are unequal.
3	There are only finitely many homomorphisms from $G$ to $K$	$G$ is finitely generated	Step (1) (specifically, that $K$ is finite)	[SHOW MORE] A homomorphism of groups is completely specified by where the generators go. Since $G$ has a finite generating set and $K$ is finite, there are only finitely many possibilities for a homomorphism, bounded by $\|K\|^{s}$ where $s$ is the minimum size of generating set for $G$ .
4	We have the required contradiction		Steps (2), (3)	<toggledisplay>Step (2) gives infinitely many pairwise distinct homomorphisms from $G$ to $K$ , Step (3) asserts that there are only finitely many.