Let $G$ be a finite group and $g,h \in G$ both have order 2. Determine the possible orders of $gh$.

So I first think of this in terms of symmetric groups. Obviously $g$ and $h$ could be the same transposition in $S_n$ and thus $gh$ is the identity (order 1). They could also be two disjoint transpositions and have order 2. And if they are two non-disjoint transpositions they would have order 3.

For order 4 or higher I couldn't really construct more examples, but I think that is mainly because I am limiting myself to the symmetric groups. Perhaps there are finite groups that are more complex that would help me arrive at an answer faster... Any thoughts or hints?

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    $\begingroup$ In the dihedral group of order $2n$, you can have the product of two elements of order $2$ be $n$. $\endgroup$ – Arturo Magidin Oct 23 at 2:18
  • $\begingroup$ In a sense there are no finite groups that are more complex then symmetric groups: en.wikipedia.org/wiki/Cayley%27s_theorem. Of course, not every element of a symmetric group is a transposition. $\endgroup$ – Carsten S Oct 23 at 11:36

Consider the dihedral group $D_{2n} = \langle a,b\mid a^n = b^2 = 1 = (ab)^2\rangle$ of order $2n$. We have also $D_{2n} = \langle ab,b\rangle$ with both $ab,b$ elements of order $2$, and $a = (ab)b$ has order $n$. This holds for all $n\in\mathbb{N}^*$, $n\ge 3$. Therefore, every positive integer can be a possible order (it is easy to find examples for $n = 1,2$).

  • $\begingroup$ Thanks for this response, everything else makes sense but could you explain why $(ab)^2=1$? $\endgroup$ – CharlieCornell Oct 23 at 2:34
  • $\begingroup$ @CharlieCornell It is from the definition of dihedral groups: $bab = a^{-1}$. $\endgroup$ – Hongyi Huang Oct 23 at 2:35

Keith Conrad gives this example (4.4) in one of his expository papers:

Let $$ A = \begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix}, B = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}.$$ Then you can check that $$ A^2 = I, B^2 = I, AB = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.$$

Notice that $$ (AB)^m = \begin{pmatrix} 1 & m \\ 0 & 1 \end{pmatrix}.$$ So if we look at everything mod $m$, then $A$ and $B$ have order $2$ and $AB$ has order $m$. The group here is $\operatorname{GL}_2(\mathbf{Z}/m)$. The one exception is when $m = 2$ in which case $B \equiv I \pmod 2$ (which isn't order $2$). So this provides an example for all $m \ge 3$.

And, of course, if you don't look mod $m$, this provides an example where $A$ and $B$ have order $2$ and their product has infinite order.


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