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

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 $\displaystyle H=\{1,-1\}$ .

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)

Re: Finite subgroup of an infinite Group