3
$\begingroup$

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.

$\endgroup$
8
  • 8
    $\begingroup$ How about the permutations of $\mathbb N$ which fix all but finitely many numbers? $\endgroup$
    – lulu
    Jun 3, 2016 at 14:12
  • $\begingroup$ I think this could work. $\endgroup$
    – Non-Being
    Jun 3, 2016 at 14:13
  • $\begingroup$ To be clear: I understood the question to seek examples in which every element had finite order. If you only want some elements to have finite order then it's a lot easier. $\endgroup$
    – lulu
    Jun 3, 2016 at 14:16
  • $\begingroup$ @m_t I don't see how every element will be of finite order $\endgroup$
    – Non-Being
    Jun 3, 2016 at 14:22
  • 1
    $\begingroup$ @Non-Being Edit your question to make it say what you mean, please. $\endgroup$
    – M. Vinay
    Jun 3, 2016 at 14:26

4 Answers 4

7
$\begingroup$

Consider $\mathbb Z_2^{\mathbb N} \times S_3$. (or simply $S_3^\mathbb N$)

Alternatively (no pun intended) consider $A_\infty$

$\endgroup$
2
  • $\begingroup$ $\mathbb Z_2$ is the group with two elements $\{0,1\}$, with addition $\bmod 2$. And $\mathbb Z_2^\mathbb N$ is the product of $\mathbb Z_2$ with itself $\mathbb N$ times. (formally it is the functions $f:\mathbb N\rightarrow \mathbb Z_2$ with coordinate-wise operations). $\endgroup$
    – Asinomás
    Jun 3, 2016 at 14:28
  • $\begingroup$ @Non-Being It is the infinite direct product of (copies of) $\mathbb Z_2$s. You can also think of it as the group of all infinitely long binary strings under the bitwise XOR (en.wikipedia.org/wiki/Exclusive_or#Bitwise_operation) operation. All elements are of order $2$. This group is Abelian, but when you take its direct product with $S_3$, you get a non-Abelian group, with any non-trivial element having order $2$, $3$, or $6$. $\endgroup$
    – M. Vinay
    Jun 3, 2016 at 14:35
6
$\begingroup$

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 $H = S_3$...

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
0
3
$\begingroup$

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 $p$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .