Finite subgroup of an infinite Group

• Dec 12th 2011, 08:26 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 $\displaystyle x \in H$ and $\displaystyle x \ne 1$
$\displaystyle \{x , x^2 , x^3 , x^4 , ... \} \subseteq H$
which make H with an infinite order
am I rite ?
is there any counter example
• Dec 12th 2011, 11:14 PM
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: $\displaystyle G=\mathbb{R}-\{0\}$ with the usual product and $\displaystyle H=\{1,-1\}$ .
• Dec 13th 2011, 03:16 AM
Drexel28
Re: Finite subgroup of an infinite Group
More generally, take any field $\displaystyle k$ then if there exists non-zero $\displaystyle x\in k$ with $\displaystyle k^n=1$ (i.e. a non-trivial $\displaystyle n^{\text{th}}$ root of unity) then the group $\displaystyle \mu_n$ of $\displaystyle n^{\text{th}}$-roots of unity form a finite subgroup of $\displaystyle k^\times$. This clearly generalizes Dr. Revilla's example (so that, for example, the $\displaystyle 3$-roots of unity in $\displaystyle \mathbb{C}$).

But, perhaps even more of a trivial counterexample. Take any infinite group $\displaystyle G$ and consider $\displaystyle G\times\mathbb{Z}_2$....(Evilgrin)
• Dec 13th 2011, 03:26 AM
Amer
Re: Finite subgroup of an infinite Group
Thanks