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This a question from Algebra by Artin. Let a,b be elements of an abelian group of orders m,n respectively. What can you say about the order of their product ab?

Here is my attempt:

$a^m=1=b^n$

$a^mb^n=1$

$b^n=a^{-m}$

$b^{n-1}=a^{-(m-1)}$
(Not sure if i can assume above step)

So:

$aa^{m-1}bb^{n-1}=aa^{m-1}a^{-(m-1)}b=1$

$ab=1$

So the order of the product $ab$ is $1$

Tom Gatward
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  • No, this is not true. from $b^n=a^{-m}$ you can not conclude $b^{n-1}=a^{-(m-1)}$ because $a^{-(m-1)}=a^{-m+1}=b^{n+1}\ne b^{n-1}$ in general. I have a feeling the order should be the least common multiple of $m$ and $n$. – Jesko Hüttenhain Dec 04 '14 at 09:11
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    Something went wrong: The only element of order $1$ in any group is the identity. So this is implying that the product of any two non-identity elements in any abelian group is the identity. – Kaj Hansen Dec 04 '14 at 09:11
  • @JeskoHüttenhain showed where you made a mistake. The question is about the product $ab$ so consider $c=ab$ and look for $n$ such that $c^n=1$. The abelian character of the group will be very useful at this point. – Tom-Tom Dec 04 '14 at 09:13

1 Answers1

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Suppose

$$ord(a)=m\;,\;\;ord(b)=n\;,\;\;\text{gcd}\,(m,n)=d\implies xm+yn=d\;,\;\;x,y\in\Bbb Z\;,\;\;m,n,d\in\Bbb N$$

Now, define $\;c:=ab\;$ :

$$c^{mn/d}\stackrel{\text{why?}}=a^{mn/d}\;b^{mn/d}=\left(a^m\right)^{n/d}\left(b^n\right)^{m/d}=1\cdot1=1$$

(Observe that $\;\frac nd\,,\,\,\frac md\in\Bbb Z\;$)

But of course

$$\frac{mn}d=\text{lcm}\,(m,n)\;\;,\;\;\;\text{and we're done}$$

Timbuc
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    I think you have only shown that the order of $c=ab$ divides $\operatorname{lcm}(m,n)$. This does not prove that they are equal. – Jesko Hüttenhain Dec 04 '14 at 11:50
  • I know @Jesko. In the most general case, the above is the most we can say about the order of $;ab;$. It isn't true that it always equals the lowest common multiple of the elements' orders. – Timbuc Dec 04 '14 at 12:00