# 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

## Proof

### Proof from given facts

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

### Hands-on 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 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]
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]
3 There are only finitely many homomorphisms from $G$ to $K$ $G$ is finitely generated Step (1) (specifically, that $K$ is finite) [SHOW MORE]
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.