# Thread: Finite subgroup of an infinite Group

1. ## 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

2. ## Re: Finite subgroup of an infinite Group

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

3. ## 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 $G\times\mathbb{Z}_2$....

Thanks

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# example of infinite group which has finite subgroup

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