Residually finite and finitely many homomorphisms to any finite group implies Hopfian
Suppose is a residually finite group and is also a group having finitely many homomorphisms to any finite group. Then, is a Hopfian group.
- Finitely many homomorphisms to any finite group implies every subgroup of finite index has finitely many automorphic subgroups
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 group that is resudally finite and has finitely many homomorphisms to any fixed finite group. A surjective endomorphism of .
To prove: is an automorphism.
Proof: We prove this by contradiction.
ASSUMPTION THAT WILL LEAD TO CONTRADICTION: We assume that is not an automorphism. Since it is surjective, it must fail to be injective, so that it must have a nontrivial kernel. Let be a non-identity element in the kernel of .
|Step no.||Assertion/construction||Given data/assumptions used||Previous steps used||Explanation|
|1||There is a surjective homomorphism for some finite group such that is not the identity element||is residually finite, is a non-identity element||[SHOW MORE]|
|2||All the homomorphisms , with varying over the natural numbers, are pairwise distinct homomorphisms from to||is surjective, is in the kernel of||Step (1)||[SHOW MORE]|
|3||There are only finitely many homomorphisms from to||has only finitely many homomorphisms to any finite group||Step (1) (specifically, that is finite)||Given direct.|
|4||We have the required contradiction.||Steps (2), (3)||[SHOW MORE]|