Local finiteness is extension-closed

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.

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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