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Given a group $G$, $a,b\in G$, $ab=ba$, $o(a)=n$, $o(b)=m$.

If $\gcd(n,m)\ne 1$, and $(a)\cap (b) = \{e\}$, prove that $o(ab)=\operatorname{lcm}(n,m)$.

P.S. $(a)$ denotes the cyclic group generated by the element $a$.

I’m a little confused... Aren’t the conditions after “if” contradictory?

Thanks for your help down there. But...can someone show me how to prove this?

Any help would be appreciated.

the_fox
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Lihua
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  • Why are they contradictory? How does $\gcd(n,m)\neq 1$ and $\langle a\rangle\cap \langle b\rangle = {e}$ contradict each other? – Arturo Magidin Mar 07 '19 at 05:23
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    There's no contradiction, think, for example, of the group $G=C_2\times C_3$ ($C_n$=the cyclic group of order $n$) and take $a=(x,1)$ and $b=(1,y)$, with $x\in C_2$ and $y\in C_3$ non-trivial – kneidell Mar 07 '19 at 05:25
  • Suppose a is an operation to rotate something clockwise 90 degrees, b is 60 degrees. then $a^2=b^3$, meaning these two cyclic groups have more than one element in common. I just can’t figure out an example where these two conditions are not contradictory.. – Lihua Mar 07 '19 at 05:30
  • Well, what if $a$ and $b$ are rotating different things, and each of them leaves the other thing unchanged? Also, look at the xample that kneidell gave. – Arturo Magidin Mar 07 '19 at 05:36
  • @ArturoMagidin yeah that makes sense. But still I need a proof of this problem. QAQ – Lihua Mar 07 '19 at 05:44
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    @kneidell thank u so much – Lihua Mar 07 '19 at 05:45
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    Let $k=\operatorname{lcm}(n, m).$ Can you show $(ab)^k=1$? If so, then all you have left to do is show that for $r \lt k, (ab)^r \neq 1$. – Robert Shore Mar 07 '19 at 07:41
  • I can't find this exact question, but this is very similar – Robert Chamberlain Mar 07 '19 at 08:33

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