Infinite non abelian group with finite order elements I'm looking for an example of an infinite non abelian group in which every element is of finite order.
I know the examples for abelian ones.
 A: Consider $\mathbb Z_2^{\mathbb N} \times S_3$. (or simply $S_3^\mathbb N$)
Alternatively (no pun intended) consider $A_\infty$
A: This is a variation on the answer of Carry on Smiling, but if you already know an example of an infinite abelian group $G$ in which every element has finite order, and if $H$ is a finite nonabelian group, then $G \times H$ is an infinite nonabelian group in which every element has finite order. For instance $G = (\mathbb{Z}/2\mathbb{Z})^\mathbb{N}$ and $H = S_3$...
A: The Grigorchuk group gives an even more extreme example. It is an infinite group that is finitely generated, and all of whose elements have finite order! It is non-Abelian, as you require.
It can be defined as a subgroup of the automorphism group of a rooted binary tree. The whole automorphism group itself is uncountable, but the Grigorchuk group is, of course, countable since it is finitely generated.
Here is an excellent survey article on the Grigorchuk group (which I highly recommend, based on personal experience): The Grigorchuk Group — Katie Waddle.
A: Let $F$ be an infinite field of characteristic $p$, an odd prime. Consider the group 
$$G=\begin{Bmatrix} \begin{bmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{bmatrix}\colon a,b,c\in F\end{Bmatrix}.$$
Then $G$ is an infinite non-abelian group, and by elementary linear algebra, it can be shown that every non-identity element of this group has order $p$.
