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 December 17th, 2014, 11:06 PM #1 Newbie   Joined: Dec 2014 From: all of world is my home Posts: 2 Thanks: 0 abstract algebra q2 hello i have a question from abstract algebra if G is abelian and finite and with o(G) degree and n is integral number and (n,o(G))=1 then prove that: for all g in G exists one x in G ; g=x^n thanks very much for your time Last edited by greg1313; December 18th, 2014 at 09:58 AM.
 January 25th, 2015, 05:08 AM #2 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra Let $\mathrm o(G)=m$. $\gcd(n,m)=1$ $\implies$ there exist integers $r,s$ such that $rn+sm=1$. Hence $g=g^{rn+sm}=g^{rn}g^{sm}=g^{rn}=x^n$ where $x=g^r$. If $n=$ then $x=g$ is not unique. If $n>1$ then suppose $g=x^n=y^n$. Then $e=x^n(y^{-1})^n=(xy^{-1})^n$ as $G$ is Abelian. If $xy^{-1}\ne e$ then $n$ must divide $m=\mathrm o(G)$, which is impossible as $\gcd(n,m)=1$ and $n>1$. Hence $xy^{-1}=e$ so $x$ is unique. Last edited by Olinguito; January 25th, 2015 at 05:15 AM.

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