# 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 $AH$ 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.

Title local finiteness is closed under extension   , proof that LocalFinitenessIsClosedUnderExtensionProofThat 2013-03-22 15:36:53 2013-03-22 15:36:53 yark (2760) yark (2760) 6 yark (2760) Proof msc 20F50 LocallyFiniteGroup