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 $A=r(\genby{\pi(F)})$ , 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 $H=\genby{D}$ , which is finite as $N$ is locally finite. Note that $1\in A\subseteq R$ and $1\in B\subseteq C\subseteq D\subseteq H\subgroup N$ .

For any $a_1,a_2\in A$ we have $a_1a_2=r(a_1a_2N)s(a_1a_2)\in AB$ . 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 $a^{-1}a_1^{-1}ca_1a$ for $a_1\in A$ and $c\in C$ ; but $a_1a=a_2b$ for some $a_2\in A$ and $b\in B$ , so $a^{-1}ha$ is a product of elements of the form $b^{-1}a_2^{-1}ca_2b=b^{-1}(a_2^{-1}ca_2)b\in CDB\subseteq H$ , and therefore $a^{-1}ha\in H$ .

We claim that $AH\subgroup G$ . Let $a_1,a_2\in A$ and $h_1,h_2\in H$ . We have $(a_1h_1)(a_2h_2)=a_1a_2(a_2^{-1}h_1a_2)h_2$ . But, by the previous paragraph, $a_1a_2\in AB$ and $a_2^{-1}h_1a_2\in H$ , so $a_1a_2(a_2^{-1}h_1a_2)h_2\in ABHH\subseteq AH$ . Thus $AHAH\subseteq AH$ . Also, $(a_1h_1)^{-1}=h_1^{-1}a_1^{-1}\in Ha_1^{-1}$ . 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 $AH$ 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.