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 – Joachim Jan 08 '17 at 19:55
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1Actually there are, in addition, many complete solutions of this question at MSE, e.g. here, and the linked questions there (and in the related questions, too). – Dietrich Burde Jan 08 '17 at 19:58
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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 $pq$.
Asinomás
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1No, 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 $q$. – Asinomás Jan 08 '17 at 19:48
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