Find Group $G$ with Infinite Order Element: Ablien Not Allowed

[alexmahone]

Find a group $G$ that contains elements $a$ and $b$ such that $a^2=e$, $b^2=e$, but the order of the element $ab$ is infinite.

My attempt:

Clearly $G$ cannot be abelian. So I looked at two commonly known non-abelian groups, namely

(i) The group of symmetries of the equilateral triangle
(ii) 2 by 2 matrices

Neither of these seem to work. Any help would be much appreciated, guys.

 

[I like Serena]

Find a group $G$ that contains elements $a$ and $b$ such that $a^2=e$, $b^2=e$, but the order of the element $ab$ is infinite.

My attempt:

Clearly $G$ cannot be abelian. So I looked at two commonly known non-abelian groups, namely

(i) The group of symmetries of the equilateral triangle
(ii) 2 by 2 matrices

Neither of these seem to work. Any help would be much appreciated, guys.

2x2 matrices work... which types will satisfy the conditions?

 

[alexmahone]

2x2 matrices work... which types will satisfy the conditions?

I looked at $A=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ and $B=\begin{bmatrix}-1&0\\0&1\end{bmatrix}$.
$A^2=B^2=I$
However, $AB=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ and it turns out that $(AB)^4=I$.

 

[I like Serena]

More generally any reflection has order 2.
Which matrix has an infinite order?

 

[alexmahone]

More generally any reflection has order 2.
Which matrix has an infinite order?

I guess the zero matrix has infinite order but it wouldn't be part of a group anyway because it has no multiplicative inverse.

 

[I like Serena]

I guess the zero matrix has infinite order but it wouldn't be part of a group anyway because it has no multiplicative inverse.

How about a rotation?
What kinds of rotations will do and which ones won't?

 

[alexmahone]

How about a rotation?
What kinds of rotations will do and which ones won't?

For a rotation by $\theta$ to have infinite order, I guess $n\theta\neq 2k\pi$ or $\displaystyle\theta\neq \frac{2k\pi}{n}$. So I guess something like $\theta=\sqrt{2}$ will have infinite order. Is this what you mean?

 

[I like Serena]

For a rotation by $\theta$ to have infinite order, I guess $n\theta\neq 2k\pi$ or $\displaystyle\theta\neq \frac{2k\pi}{n}$. So I guess something like $\theta=\sqrt{2}$ will have infinite order. Is this what you mean?

More or less.
Indeed, a rotation with $\theta=\sqrt{2}$ has infinite order.

More specifically, with $\frac{\theta}{2\pi} \in \mathbb Q$ we have finite order.
That's because with such $\theta$ we can always find a multiple that will be a multiple of $2\pi$.
And with $\frac{\theta}{2\pi} \in \mathbb R \backslash \mathbb Q$ we have infinite order, because we're excluding all such multiples.
(Can we prove it properly?)Back to the problem at hand, consider that the composition of 2 reflections with different axes is a rotation.

 

[johng1]

The infinite dihedral group $D_\infty$ (https://en.wikipedia.org/wiki/Infinite_dihedral_group) provides an easy example.
$D_\infty=<r,s\,|\,s^2=1,\,s^{-1}rs=r^{-1}>$
Let $a=rs$ and $b=s$. Then $a^2=rsrs=r(s^{-1}rs)=rr^{-1}=1$ and $ab=r$ has infinite order.