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.

  • 2
    Think of linear transformations of $\mathbb{R}^2$. Then we can imagine involutions $a,b$ as each flipping the plane 180 degrees over some line. If the angle between the two lines were just so... – basket Oct 2 '16 at 17:14

Consider $GL_2(R)$ $$ A=\begin{pmatrix}-1 & 1\\0&1\end{pmatrix}, B = \begin{pmatrix}-1&0\\0&1\end{pmatrix} $$ $$ A^2=B^2=I, (AB)^n = \begin{pmatrix}1&n\\0&1\end{pmatrix} $$

  • Thanks a lot! But what is the motivation for the solution, i.e. how did you arrive at it? – Thomas Oct 3 '16 at 3:07
  • @Thomas, honestly, I just knew it. To get insides - you may try to use the same diagonals and 0 as lower left element in matrices replacing upper right element by x in A and by y in B, then do calculations and watch upper right element depends on n,x,y – hOff Oct 3 '16 at 23:40

The simplest example is probably the free group on two elements $a, b$ with the only relations $a^2 = e$, $b^2 = e$. By the way, a finite group like symmetries of a triangle cannot have an element of infinite order.

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