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.

Related facts

Applications

• Finitely generated and residually finite implies Hopfian

Similar facts

• Finitely many homomorphisms to any finite group implies every subgroup of finite index has finitely many automorphic subgroups

Opposite facts

• Finitely many homomorphisms to any finite group not implies Hopfian

Proof

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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] 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 $\alpha \circ \varphi ^{n}$, 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] 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 $K$	$G$ 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] Step (2) constructs infinitely many pairwise distinct homomorphisms from $G$ to $K$. Step (3) shows there can be only finitely many. This gives a contradiction.