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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Get more facts about finitely generated residually finite group|Get more facts about Hopfian group
|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|
- Finitely generated implies finitely many homomorphisms to any finite group
- Residually finite and finitely many homomorphisms to any finite group implies Hopfian
Proof from given facts
The proof follows directly by piecing together facts (1) and (2).
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Given: A finitely generated group that is also residually finite. A surjective homomorphism .
To prove: is an automorphism of .
Proof: We prove this by contradiction.
ASSUMPTION THAT WILL LEAD TO CONTRADICTION: Suppose is not an automorphism of . Then, since it is surjective, it must fail to be injective, so there exists an non-identity element in its kernel.
|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 , for varying over positive integers, 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||is finitely generated||Step (1) (specifically, that is finite)||[SHOW MORE]|
|4||We have the required contradiction||Steps (2), (3)||<toggledisplay>Step (2) gives infinitely many pairwise distinct homomorphisms from to , Step (3) asserts that there are only finitely many.|