Local finiteness is extension-closed

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This article gives the statement, and possibly proof, of a group property (i.e., locally finite group) satisfying a group metaproperty (i.e., extension-closed group property)
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Statement

Statement with symbols

Suppose G is a group with a normal subgroup H such that both H and G/H are locally finite groups: any finitely generated subgroup of H is finite, and any finitely generated subgroup of G/H is finite. Then, G is itself locally finite.

Related facts

Facts used

  1. Schreier's lemma: In its factual form, this states that any subgroup of finite index in a finitely generated group is again finitely generated.
  2. Finiteness is extension-closed
  3. First isomorphism theorem

Proof

Given: A group G with a normal subgroup H such that both H and G/H are locally finite. Let \varphi:G \to G/H be the quotient map.A finite subset A of G with K = \langle A \rangle.

To prove: K is finite.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 \varphi(K) is a finite subgroup of G/H G/H is locally finite Since A is finite, \varphi(A) is finite. By local finiteness of G/H, \langle \varphi(A) \rangle is a finite subgroup of G/H. We have \langle \varphi(A) \rangle = \varphi(\langle A \rangle) = \varphi(K), so \varphi(K) is finite.
2 K \cap H is a normal subgroup of finite index in K Fact (3) We have \varphi(K) \cong K/(K \cap \ker \varphi) = K/(K \cap H). Since the left side is finite, we conclude that K \cap H has finite index in K.
3 K \cap H is finitely generated Fact (1) A is finite Step (2) A is finite, so K = \langle A \rangle is finitely generated, and by step (2), K \cap H has finite index in K. Thus, by fact (1), K \cap H is also a finitely generated group.
4 K \cap H is finite H is locally finite Step (3) K \cap H is a finitely generated subgroup of H. Local finiteness of H forces K \cap H to be finite.
5 K is finite Fact (2) Steps (2), (4) By step (4), K \cap H is finite, and by step (2), H/(H \cap N) is finite. Thus, by fact (2), K is finite.
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