# Order of the products of two order-2 elements in a finite group

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?

• In the dihedral group of order $2n$, you can have the product of two elements of order $2$ be $n$. – Arturo Magidin Oct 23 at 2:18
• 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. – 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 $$n = 1,2$$).
• Thanks for this response, everything else makes sense but could you explain why $(ab)^2=1$? – CharlieCornell Oct 23 at 2:34
• @CharlieCornell It is from the definition of dihedral groups: $bab = a^{-1}$. – Hongyi Huang Oct 23 at 2:35
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.