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 Abstract Algebra Abstract Algebra Math Forum

 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. Tags abstract, algebra Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post micle Algebra 1 June 17th, 2013 09:06 AM ustus Abstract Algebra 4 October 14th, 2012 12:00 PM MastersMath12 Abstract Algebra 1 September 24th, 2012 11:07 PM forcesofodin Abstract Algebra 10 April 5th, 2010 10:31 PM micle Abstract Algebra 0 December 31st, 1969 04:00 PM

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