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How we can prove in an abelian group if an element a has order k and an element b has order j and if k and j are relatively prime then the element a*b has the order kj?

Alexander Gruber
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kalpeshmpopat
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1 Answers1

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$(ab)^{kj}=a^{kj}b^{kj}=(a^k)^j(b^j)^k=1^j1^k=1$. So it remains to prove that if $(ab)^n=1$ then $kj|n$.

Assume $(ab)^n=1$, then $a^n=b^{-n}$. Choose $p,q$ such that $pk+qj=1$, which is possible since $k,j$ are relatively prime. Then $1=a^{npk}=b^{-npk}=b^{-n(1-qj)}=b^{-n}b^{nqj}=b^{-n}$, hence $b^n=1$, which implies $j|n$. Similarly, $k|n$, so $kj|n$.

abc
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    One can give a slightly different proof that if $(ab)^n=1$ then $k$ divides $n$. For $((ab)^n)^j=1$. But $b^{nj}=1$, so $a^{nj}=1$. Thus $k$ divides $nj$, and because $k$ and $j$ are relatively prime, it follows that $k$ divides $n$. – André Nicolas Mar 12 '13 at 06:39