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I'm stuck with this question:

$G$ is a group. $a, b \in G$ such that $ab = ba$. Show that if $a$ has order $m$, $b$ has order $n$, and $gcd(a,b)=1$, then the order of $ab$ is $mn$.

To help us solve the problem, we've been told (as a hint) to consider $(ab)^{mN}$ and $(ab)^{nN}$.

Here's what I've done so far:

$(ab)^{mn} = (ab) \cdot (ab) \cdot (ab) ... \cdot (ab)$ (mn times)

Since $ab = ba$ and since $G$ is associative:

$(ab)^{mn} = a^{mn} \cdot b^{mn} = e^n \cdot e^n = e$ where $e$ is the identity element.

But I haven't shown that $mn$ is the least positive integer for which this is possible, and I haven't used the hint (which I would like to).

Any help with this problem would be much appreciated.

Jack

Jack
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    This question already has many answers on this site, however, suppose that the order of $ab$ is $k$, then you already have that $k \mid mn$. Note that $1 = (ab)^{kn} = a^{kn}$, so $m \mid kn$. Analogously, you have that $n \mid km$. Use that $m,n$ are coprime to prove that $m \mid k, n \mid k$ and hence $mn \mid k$. Therefore $k = mn$. – Student Mar 23 '17 at 11:27
  • Sorry for the duplicate. I thought it was quite a "niche" question but clearly not. Thanks for your help! – Jack Mar 23 '17 at 11:33
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    No problem. Normally you should get some suggestions when you fill in the title of your question, so it is always worth it to give those a closer look :) btw, I did not saw that you had a hint. The $N$ in the hint is the $k$ is have used (but I guess you figured that one out :) ) – Student Mar 23 '17 at 11:35
  • What does $\mathrm{gcd}(a,b)$ mean ? These are group elements. – Antoine Mar 23 '17 at 11:50
  • Yesterday the same duplicate, it seems. – Dietrich Burde Mar 23 '17 at 15:11

1 Answers1

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Use some number theory:

(1). If ord($a)=m$ then for any integer $r$ we have $a^r=1 \implies m|r.$

Proof: If $m\not | \;r$ then $r=mx+r'$ for some integers $x,r'$ with $0<r'<m.$ But then $1=a^r=(a^m)^xa^{r'}=a^{r'}$, a contradiction, since the def'n of ord($a)$ requires that $a^{r'}\ne 1$ when $0<r'<$ord($a).$

(2). For your Q, let ord($ab)=s.$ Then $$1=((ab)^s)^n=a^{sn}(b^n)^s=a^{sn}.$$ By (1) we have $m=ord(a)|sn,$ but since $\gcd (m,n)=1$, we must have $m|s$.

Interchanging $a,b$ and interchanging $m,n$ we also obtain $n|s.$

Now since $m|s$ and $n|s$ and $\gcd(m,n)=1$ we must have $$mn|s.$$ But also we have $(ab)^{mn}=1$ so by (1) we have $$s=\text{ord}(ab)|mn.$$ Since $mn|s$ and $s|mn,$ we have $s=mn.$

DanielWainfleet
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