Assume two natural numbers $m,n$ are coprime (this means $(m,n)=1$) such that for each $a ,b \in G$ we have $a^m b^m = b^m a^m$ and $a^n b^n = b^n a^n$. Then $G$ is an abelian group.
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What are $m$ and $n$ ? What have you tried? – Jun 06 '15 at 10:00
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where are the your comment you use a^m b^m = b^m a^m and a^n b^n = b^n a^n?? – dibak chopra Jun 06 '15 at 10:22
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m and n are natural numbers that (m,n)=1 – dibak chopra Jun 06 '15 at 10:23
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m and n are co-prime means that (m,n)=1 – dibak chopra Jun 06 '15 at 11:32
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@dibak chopra - This is a very good question +1 from me and check my solution here http://math.stackexchange.com/questions/326702/how-to-prove-that-a-group-with-some-properties-is-abelian/327766#327766 – Nicky Hekster Jun 07 '15 at 12:58
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@Nicky Hekster i read your solution but i confused in those answers and comments.why we have (b^um a^um)^vn=b^um+vn a^um+vn?? – dibak chopra Jun 08 '15 at 19:35
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@Dibak - I cannot find that in my proof. Aren't you looking at the wrong proof, namely that below mine of vonbrand?? – Nicky Hekster Jun 08 '15 at 21:05
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Hint: $(m,n)=1$ then there exist $c,d\in \mathbb{Z}$ s.t. $cm+dn=1$. Thus $$gh=g^{cm+dn}h^{cm+dn}=g^{cn}g^{dm}h^{cm}h^{dn}=\cdots=h^{cm+dn}g^{cm+dn}=hg.$$ because $$(g^mh^n)^{cm}=g^m(h^ng^m)^{cm-1}h^n=g^m(h^ng^m)^{cm}g^{-m}h^{-n}h^n=g^m((h^ng^m)^{c})^mg^{-m}=(h^ng^m)^{cm}$$ Similarly, $(g^mh^n)^{dn}=(h^ng^m)^{dn}$ and so, $h^ng^m=g^mh^n$.
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yes.thus i ask farhad about ... and where are the answer use a^m b^m = b^m a^m and a^n b^n = b^n a^n – dibak chopra Jun 06 '15 at 10:47
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