Let $G$ be a group of order 6. Suppose that $a,b\in G$ with $a$ of order 3 and $b$ of order 2. Show that either $G$ is cyclic or $ab\not=ba$.

Question

Let $G$ be a group of order 6. Suppose that $a,b\in G$ with $a$ of order 3 and $b$ of order 2. Show that either $G$ is cyclic or $ab\not=ba$.

Attempt at Answer: Suppose $ab=ba$. Then $o(ab)=o(a)o(b)$ since $o(a)$ and $o(b)$ are relatively prime (o being shorthand for 'order'). From here, I'd guess that I want to show that $G$ is NOT cyclic for $ab=ba$. Then I'd show the converse.

The problem is that I don't get how to get from here to there.

Any help/hints would be greatly appreciated.

Answer 1

You've essentially finished the proof already. If $ab = ba$ then $o(ab) = o(a) o(b) = 2 \cdot 3 = 6$ and $G$ is cyclic. If $ab \ne ba$ then $G$ can't be cyclic, since it's not even abelian.

Answer 2

Just a supplement to T.Bongers' proof:

We know that if $ab = ba$, then $o(ab) = o(a)o(b)=6$. What that means then is that the set $$ \langle ab \rangle =\{e,ab,(ab)^2,\dots,(ab)^5\} $$ Is the (cyclic) subgroup of $G$ generated by $ab$ ($e$ here is the identity). Now, here's the punchline: $\langle ab \rangle$ is a subset of $G$ that has as many elements as $G$. It follows that $\langle ab \rangle = G$, which means that $G$ has to be cyclic.

This should prove a very useful addition to your bag of tricks. Things get a bit trickier, however, if $G$ isn't finite.