# Finite subgroup of an infinite Group

• Dec 12th 2011, 09:26 PM
Amer
Finite subgroup of an infinite Group
The only subgroup of finite order of a an infinite group is the identity
My proof

Let G be a group with infinite order
let H be a subgroup and let $x \in H$ and $x \ne 1$
$\{x , x^2 , x^3 , x^4 , ... \} \subseteq H$
which make H with an infinite order
am I rite ?
is there any counter example
• Dec 13th 2011, 12:14 AM
FernandoRevilla
Re: Finite subgroup of an infinite Group
Quote:

Originally Posted by Amer
The only subgroup of finite order of a an infinite group is the identity

Counterexample: $G=\mathbb{R}-\{0\}$ with the usual product and $H=\{1,-1\}$ .
• Dec 13th 2011, 04:16 AM
Drexel28
Re: Finite subgroup of an infinite Group
More generally, take any field $k$ then if there exists non-zero $x\in k$ with $k^n=1$ (i.e. a non-trivial $n^{\text{th}}$ root of unity) then the group $\mu_n$ of $n^{\text{th}}$-roots of unity form a finite subgroup of $k^\times$. This clearly generalizes Dr. Revilla's example (so that, for example, the $3$-roots of unity in $\mathbb{C}$).

But, perhaps even more of a trivial counterexample. Take any infinite group $G$ and consider $G\times\mathbb{Z}_2$....(Evilgrin)
• Dec 13th 2011, 04:26 AM
Amer
Re: Finite subgroup of an infinite Group
Thanks