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Suppose that $a$ and $b$ are elements of finite order in a group such that $ab=ba$ and $\langle a \rangle \cap \langle b \rangle = \{e\} $. Prove that $|ab|$ is the least common multiple of $|a|$ and $|b|$,

Help please.., any hint?

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Hint: Note that $(ab)^n=a^nb^n$ because $ab=ba$. Thus $(ab)^n=e$ implies $a^n=b^{-n}$, which is in both $\langle a\rangle$ and $\langle b\rangle$.

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is this correct? let $|a|=m ;|b|=n ; k= lcm\left( m,n \right)=pm = qn$ so $a^m=2$ and $b^n=e$ let $|ab|=x$ then $\left( ab \right)^k = a^k b^k = a^\left( pm \right) b^\left(qn \right) = e^p e^q = e $ so $ x|k$ since $ a^k b^k = e $ then $a^k = b^\left(-k\right)$ and $b^k = a^\left(-k\right)$ so $\left( a^k \right) ^n = \left( b^{-k} \right) ^n = \left( b^n \right) ^{-k} = \left( e \right) ^{-k} = e$ since $ |a| = m $ so $ m|kn$ similarly $\left( b^k \right) ^m = \left( a^{-k} \right) ^m = \left( a^m \right) ^{-k} = \left( e \right) ^{-k} = e$ $ |b| = n $ so $ n|km$ is this correct? –  chihiroasleaf Jan 9 '13 at 10:58
    
@BabakSorouh can you give me hint what is the next step? thanks –  chihiroasleaf Jan 9 '13 at 12:01
    
See the answer above. $a^n=b^{-n}$; so $a^n\in\langle a\rangle\cap\langle b\rangle=e$ as well as $b^{-n}\in\langle a\rangle\cap\langle b\rangle=e$. This means that if $n$ be the order of $ab$, then $a^n=e,b^{-n}=e$. But $|a|=k$ and $|b|=s$, so; $k|n$ and $s|n$ and then $t=\text{lcm}(k,s)|n$. Conversely, $$k\mid t\to t=k_1k,\;\; s\mid t\to t=s_1s$$ for some integers $k_1,s_1$. Hence $(ab)^t=a^tb^t$ (why?) and so $(ab)^t=a^{k_1k}b^{s_1s}=e$. It means that $n\mid t$. It is done. You asked for a hint!. ;-) –  B. S. Jan 9 '13 at 13:22
    
ahh..., I see..., thanks for your help... :) –  chihiroasleaf Jan 9 '13 at 13:38
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