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Possible Duplicate:
Order of elements in abelian groups

Let $G$ be an abelian group and suppose that $G$ has elements of orders $m$ and $n$, respectively. Prove that $G$ has an element whose order is the least common multiple of $m$ and $n$.

I've attempted this problem for quite some time, but didn't seem to get anywhere.

First, let $a$ and $b$ be the elements whose orders are $m$ and $n$, respectively. I guessed that we can find the element of order $lcm(m,n)$ explicitly, instead of simply proving its existence. Furthermore, I also guessed that the element can be expressed in the form $a^kb^l$, because the statement must also hold when $G$ is generated by $a$ and $b$.

Then I let $k$ be the smallest positive integer such that $a^k$ is a power of $b$, say $a^k=b^l$. Then I proved that $l$ is also the smallest positive integer such that $b^l$ is a power of $a$, and that $ml=nk$. I'm not sure whether it's correct though.

Then I tried to find the order of ab. I can prove that the order is divisible by $\frac{lcm(m,n)}{\gcd(m,n)}$, but I can't prove whether it is equal to $lcm(m,n)$. Apparently, taking any $a^ib^j$ won't be any better. And now, I'm at wits end.

Please tell me whether I'm on the right path. If not, please give me some adequate hints so I can work on it.

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    Please do not use titles entirely composed of $\TeX$.2011-11-03
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    Sorry. Is that inappropriate?2011-11-03
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    It's often a good idea, when you've failed to prove something, to try to disprove it. For example, taking $ab$ doesn't work in general, since we could have $b=a^{-1} \neq id$. $a$ itself (which is of the form $a^i b^j$) works in this case.2011-11-03
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    @adjwhrwe, it is quite inconvenient.2011-11-03
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    @adjwhrwe yours first guesses were correct, infact there is an element of order $lcm(n,m)$ of the form $a^kb^l$. Because you're so close to the solution I'll give you something I hope will help you: "Try to do first the case of $gcd(n,m)=1$".2011-11-03
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    Thanks a lot guys! I've managed to prove it.2011-11-03
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    In addition to the duplicate, see also [this one](http://math.stackexchange.com/questions/41303/examples-and-further-results-about-the-order-of-the-product-of-two-elements-in-a).2011-11-03

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