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
Hopfian group	every surjective endomorphism of the group is an automorphism

Proof

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. Suppose $\varphi$ is not an automorphism of $G$. Then, since it is surjective, there exists an non-identity element $g$ in its kernel.

1. There is a surjective homomorphism $\alpha :G\to K$ for some finite group $K$ such that $\alpha (g)$ is not the 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 $\alpha \circ \varphi ^{n}$ are distinct homomorphisms from $G$ to $K$: [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 $K$: [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 from (2) and (3).

External links

• MathOverflow discussion