Residually finite and finitely many homomorphisms to any finite group implies Hopfian

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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.

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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]
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]
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]