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.