local finiteness is closed under extension, proof that

Let $G$ be a group and $N$ a normal subgroup of $G$ such that $N$ and $G/N$ are both locally finite. We aim to show that $G$ is locally finite. Let $F$ be a finite subset of $G$ . It suffices to show that $F$ is contained in a finite subgroup of $G$ .

Let $R$ be a set of coset representatives of $N$ in $G$ , chosen so that $1\in R$ . Let $r\colon G/N\to R$ be the function mapping cosets to their representatives, and let $s\colon G\to N$ be defined by $s(x)=r(xN)^{-1}x$ for all $x\in G$ . Let $\pi\colon G\to G/N$ be the canonical projection. Note that for any $x\in G$ we have $x=r(xN)s(x)$ .

Put $A=r({\left\langle\pi(F)\right\rangle})$ , which is finite as $G/N$ is locally finite. Let $B=s(F\cup AA\cup A^{-1})$ , let $C=B\cup B^{-1}$ and let

D=\{a^{-1}ca\mid a\in A\hbox{ and }c\in C\}\subseteq N.

Put $H={\left\langle D\right\rangle}$ , which is finite as $N$ is locally finite. Note that $1\in A\subseteq R$ and $1\in B\subseteq C\subseteq D\subseteq H\leq N$ .

For any $a_{1},a_{2}\in A$ we have $a_{1}a_{2}=r(a_{1}a_{2}N)s(a_{1}a_{2})\in AB$ . Note that $D^{-1}=D$ , and so every element of $H$ is a product of elements of $D$ . So any element of the form $a^{-1}ha$ , where $a\in A$ and $h\in H$ , is a product of elements of the form $a^{-1}a_{1}^{-1}ca_{1}a$ for $a_{1}\in A$ and $c\in C$ ; but $a_{1}a=a_{2}b$ for some $a_{2}\in A$ and $b\in B$ , so $a^{-1}ha$ is a product of elements of the form $b^{-1}a_{2}^{-1}ca_{2}b=b^{-1}(a_{2}^{-1}ca_{2})b\in CDB\subseteq H$ , and therefore $a^{-1}ha\in H$ .

We claim that $AH\leq G$ . Let $a_{1},a_{2}\in A$ and $h_{1},h_{2}\in H$ . We have $(a_{1}h_{1})(a_{2}h_{2})=a_{1}a_{2}(a_{2}^{-1}h_{1}a_{2})h_{2}$ . But, by the previous paragraph, $a_{1}a_{2}\in AB$ and $a_{2}^{-1}h_{1}a_{2}\in H$ , so $a_{1}a_{2}(a_{2}^{-1}h_{1}a_{2})h_{2}\in ABHH\subseteq AH$ . Thus $AHAH\subseteq AH$ . Also, $(a_{1}h_{1})^{-1}=h_{1}^{-1}a_{1}^{-1}\in Ha_{1}^{-1}$ . But $a_{1}^{-1}=r(a_{1}^{-1}N)s(a_{1}^{-1})\in AB$ , so $Ha_{1}^{-1}\subseteq HAB\subseteq AHAH\subseteq AH$ . Thus $(AH)^{-1}\subseteq AH$ . It follows that $A H$ is a subgroup of $G$ , and it is clearly finite.

For any $x\in F$ we have $x=r(xN)s(x)\in AB$ . So $F\subseteq AH$ , which completes the proof.

Title	local finiteness is closed under extension, proof that
Canonical name	LocalFinitenessIsClosedUnderExtensionProofThat
Date of creation	2013-03-22 15:36:53
Last modified on	2013-03-22 15:36:53
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	6
Author	yark (2760)
Entry type	Proof
Classification	msc 20F50
Related topic	LocallyFiniteGroup