Prove that if $o(a)=m, o(b)=n$ and $gcd(m,n)=1$ then $o(ab)=mn$

Question

Prove that if $o(a)=m, o(b)=n$ and $gcd(m,n)=1$ then $o(ab)=mn$

I am asking this because I saw someone asked a similar question and was told that the statement is wrong.

I managed to show quite quickly that $o(ab)|mn$, but no matter what I tried I was not able to show the equality.

Is there a mistake in the statement?

This is missing context. If the group is abelian, sure. If not? There's not a lot you can say about the order of $ab$ if $a,b$ don't commute. — lulu, May 15 '19 at 01:17
Compare with https://math.stackexchange.com/questions/67180/order-of-product-of-two-elements-in-a-group — Jens Schwaiger, May 15 '19 at 01:33
$S_3$ has elements of order $2$ and $3$ but no element of order $6$. — lhf, May 15 '19 at 02:04

score 1 · Answer 1 · answered May 15 '19 at 01:45

If $a,b$ commute, then you certainly have $(ab)^{mn}=e$ so $o(ab)|mn$. Now, suppose $o(ab)=r$, so $$a^rb^r=e$$ Raising to the $m$ in both sides, gives you $b^{rm}=e$, so $n|rm$. Since $\gcd(m,n)=1$ you actually have that $n|r$.

Similarly, (by raising to the $n$ in both sides), you can prove that $m|r$.

So, you have $n|r$ and $m|r$. Since $\gcd(m,n)=1$, then you have $mn|r$, i.e. $mn|o(ab)$.

Prove that if $o(a)=m, o(b)=n$ and $gcd(m,n)=1$ then $o(ab)=mn$

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