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

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Discussion Overview

The discussion revolves around finding a group \( G \) that contains elements \( a \) and \( b \) such that both \( a^2 = e \) and \( b^2 = e \), while the element \( ab \) has infinite order. The scope includes group theory concepts, particularly focusing on non-abelian groups and the properties of specific matrix representations and rotations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants propose that the group cannot be abelian, leading to the exploration of known non-abelian groups such as the symmetries of the equilateral triangle and 2x2 matrices.
  • One participant suggests specific 2x2 matrices \( A \) and \( B \) and calculates that \( (AB)^4 = I \), indicating that this combination does not yield an infinite order element.
  • Another participant notes that reflections generally have order 2 and questions which matrices would have infinite order.
  • There is a discussion about rotations, with one participant suggesting that a rotation by \( \theta \) will have infinite order if \( n\theta \neq 2k\pi \) or \( \theta \neq \frac{2k\pi}{n} \), using \( \theta = \sqrt{2} \) as an example.
  • A later reply clarifies that rotations with \( \frac{\theta}{2\pi} \in \mathbb{Q} \) have finite order, while those with \( \frac{\theta}{2\pi} \in \mathbb{R} \backslash \mathbb{Q} \) have infinite order, suggesting a proof may be needed.
  • Another participant introduces the infinite dihedral group \( D_\infty \) as an example, providing a specific construction with elements \( a = rs \) and \( b = s \) that satisfies the conditions of the problem.

Areas of Agreement / Disagreement

Participants express differing views on the types of matrices and rotations that can yield an infinite order element. While some agree on the properties of reflections and rotations, the discussion remains unresolved regarding the specific examples that meet the initial criteria.

Contextual Notes

Limitations include the need for further exploration of specific matrix types and rotations, as well as the potential for additional examples beyond those discussed. The mathematical steps and conditions for infinite order elements are not fully resolved.

alexmahone
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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.
 
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Alexmahone said:
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?
 
I like Serena said:
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 $(AB)^4=I$.
 
More generally any reflection has order 2.
Which matrix has an infinite order?
 
I like Serena said:
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.
 
Alexmahone said:
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?
 
I like Serena said:
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?
 
Alexmahone said:
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.
 

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