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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\}$.

Is this true or false: 'the orders of $a$ and $b$ are relatively prime'.

Help, please...

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the condition on <a>\cap <b> is missing. –  Ittay Weiss Jan 9 '13 at 9:45
    
what do you mean? :) $ \langle a \rangle $ mean subgroup generated by a –  chihiroasleaf Jan 9 '13 at 9:46
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what do you mean by "ab=ba and <a>\cap <b>"? I understand what ab=ba means but what else are you trying to say? –  Ittay Weiss Jan 9 '13 at 9:48
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ahh..., sorry... it's not complete... $ \langle a \rangle \cap \langle b \rangle = e $ it's what I mean... –  chihiroasleaf Jan 9 '13 at 9:49
    
please edit your post then (and you mean <a>\cap <b>={e}) –  Ittay Weiss Jan 9 '13 at 9:51

1 Answer 1

Consider the Klein Group: $$V=\{e,a,b,c\}$$ where in $$a^2=b^2=c^2=1$$ and $$ab=ba=c,bc=cb=a,ca=ac=b$$ We have $|a|=|b|=|c|=2$ and we can see that $$\langle a\rangle=\{e,a\}\\ \langle b\rangle=\{e,b\}\\ \langle c\rangle=\{e,c\}$$ Now, check if your claim is true or not.

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so.., it's false, right? –  chihiroasleaf Jan 9 '13 at 10:13
    
@chihiroasleaf: Yes. ;-) –  Babak S. Jan 9 '13 at 10:14
    
ok.., thank you... :) –  chihiroasleaf Jan 9 '13 at 10:15
    
@chihiroasleaf: I did nothing for you. I hope I could help you. Welcome. :-) –  Babak S. Jan 9 '13 at 10:19
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This example generalises. If $G$ is the direct product of the cyclic groups $A=\langle a\rangle$ and $B=\langle b\rangle$, $G=A\times B$, then $ab=ba$ and $A\cap B=1$. In the example here, $A$ and $B$ are both cyclic of order two. –  user1729 Jan 9 '13 at 11:38

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