Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find and example of two elements $a,b$ in a finite group $G$ such that $|a| = |b| = 2, a \ne b$ and $|ab|$ is odd. Any ideas as to how I would go about finding it?


share|cite|improve this question
For this to happen it is necessary that $a$ and $b$ do not commute for otherwise $(ab)^2=abab=aabb=a^2b^2=1$. Have you tried the smallest non-abelian group that you know? – Jyrki Lahtonen Feb 19 '13 at 21:14
Any ideas as to how I would go about finding [blah]? The most general answer to this question, which covers a vast expanse of territory in many forms of endeavor: pick stuff repeatedly and test it; explore. – anon Feb 19 '13 at 21:20
Get to know as many examples as you can swallow and digest. You undoubtedly saw the group had to be nonabelian, so you should try every one of the (I hope several) nonabelian groups you know. – Lubin Feb 19 '13 at 21:56
up vote 11 down vote accepted

In $S_3$ the multiplication of $(a,b)$ and $(b,c)$ is $(a,b,c)$

share|cite|improve this answer

There is a very general example you should know about, that of dihedral groups. A dihedral group has order $2n$, for any $n \ge 2$, and it is generated by two elements of order $2$, whose product has order $n$.

Probably the simplest way to see these groups is as a group of bijective maps on $\mathbf{Z}_{n}$, $$ a : x \mapsto -x, \qquad b: x \mapsto -x-1. $$ (Coefficients are written as integers, but meant to be in $\mathbf{Z}_{n}$.) These are clearly elements of order $2$, while because of $$ a \circ b(x) = (a(b(x)) = -(-x - 1) = x + 1 $$ $ab = a \circ b$ has order $n$.

(Geometrically, such a group is the group of congruences of a regular $n$-gon.)

If you do the same thing over $\mathbf{Z}$, you find that $ab$ has infinite order.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.