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

Put $A=r(\u27e8\pi (F)\u27e9)$, 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

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

Put $H=\u27e8D\u27e9$, which is finite as $N$ is locally finite. Note that $1\in A\subseteq R$ and $1\in B\subseteq C\subseteq D\subseteq H\le N$.

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 ${a}^{-1}{a}_{1}^{-1}c{a}_{1}a$
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
${b}^{-1}{a}_{2}^{-1}c{a}_{2}b={b}^{-1}({a}_{2}^{-1}c{a}_{2})b\in CDB\subseteq H$,
and therefore ${a}^{-1}ha\in H$.

We claim that $AH\le G$.
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, ${({a}_{1}{h}_{1})}^{-1}={h}_{1}^{-1}{a}_{1}^{-1}\in H{a}_{1}^{-1}$.
But ${a}_{1}^{-1}=r({a}_{1}^{-1}N)s({a}_{1}^{-1})\in AB$,
so $H{a}_{1}^{-1}\subseteq HAB\subseteq AHAH\subseteq AH$.
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 |
---|---|

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 |