Give an example of a nonabelian group in which a product of elements of finite order can have infinite order.

Question

So, I let a,b be elements in such a group. So |a|=n and |b|=m, n and m are finite. But |ab| needs to be infinite, but since |ab|=lcm(n,m), how can that be possible?

Answer 1

Take the free group on letters $a,b$ with the sole relations that $a^2=b^2=1$. Now $ab$ has infinite order, since $$abababab\cdots ab \neq 1$$

Answer 2

Matrices, invertible ones are a group under multiplication and here is a good example

Answer 3

Consider group of all permutations of $\mathbb{Z}$; permutation group on $\geq 3$ letters is always non-abelian.

Consider $\sigma(x)=-x+1$ and $\tau(x)=-x+2$, the two permutations of $\mathbb{Z}$.

• Show that $\tau\circ\tau=\sigma\circ\sigma=I$, hence order of $\sigma,\tau$ is $2$.

• Find $\sigma\circ \tau$.

• Is it of finite order?

Answer 4

Compose two reflections on the plane and you get a rotation. Reflections have order 2. The rotation can have arbitrary order, including infinite.