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Let $(G,*)$ be an abelian group with two elements a and b of order p,q co-prime. What is the order of a*b ?

At most pq as $(a*b)^{pq}=a^{pq}*b^{pq}=1_G$

I guess it is exactly pq but I do not see why ?

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    True but the answer gives only a hint that did not help me – 2017-01-08
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    Actually there are, in addition, many complete solutions of this question at MSE, e.g. [here](http://math.stackexchange.com/questions/424097/order-of-products-of-elements-in-a-finite-abelian-group), and the linked questions there (and in the related questions, too). – 2017-01-08

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notice that $(ab)^n=a^n b^n$, so if this is equal to $1$ then $a^n=(b^n)^{-1}$. This would imply that $a^n$ is contained in $\langle b \rangle $ and $\langle a \rangle$. The intersection of these two subgroups is $\{e\}$ by Lagrange's theorem.

So we need $a^n=b^n=e$. So $p$ and $q$ must both divide $n$, hence $\text{lcm}(p,q)=pq$ must divide $n$. So the order of $ab$ is $pq$.

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    why Lagrange imply the intersection to be {e} ? – 2017-01-08
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    p/#G and q/#G with p q coprime – 2017-01-08
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    because the order of the subgroup must divide both $p$ and $q$. – 2017-01-08
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    I got it thanks – 2017-01-08
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    No, use the fact that $\langle a \rangle \cap \langle b \rangle $ is a subgroup of both $\langle a \rangle $ and $\langle b \rangle$. Using Lagrang'es theorem twice tells you that $|\langle a \rangle \cap \langle b\rangle |$ is a divisor of both $p$ and $q$. – 2017-01-08
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    Great.${}{}{}{}{}$ – 2017-01-08