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 July 3rd, 2011, 12:50 PM #1 Newbie   Joined: Jul 2011 Posts: 2 Thanks: 0 Isomorphism Hi everyone! I have an abelian group $G$ which is a finit group of order n and a function $\varphi(a)=a^m$ and I want to prove that if $mcd(m,n)=1$ then $\varphi$ is an isomorphism. Someone told me first prove that $\varphi$ is an homomorphism and then prove that $ma+nb= 1$ for some $a\in\mathbb{Z}$ but I don't mind how to do that. So...some suggestions? Thanks!
 July 3rd, 2011, 01:11 PM #2 Newbie   Joined: Jul 2011 Posts: 2 Thanks: 0 Re: Isomorphism I forgot to said that $\varphi:G\rightarrow G$
 July 8th, 2011, 02:58 AM #3 Member   Joined: Jul 2011 From: Trieste but ever Naples in my heart! Italy, UE. Posts: 62 Thanks: 0 Have you noted that being $G$ an abelian groups than $\varphi$ is a his endomorphism?
 July 9th, 2011, 04:18 PM #4 Senior Member   Joined: Oct 2009 Posts: 105 Thanks: 0 Re: Isomorphism For the Isomorphism part, you have if $a^m= b^m$ then $a= b$ This is what you want to prove. Use the fact that any element (other than the identy) raised to the mth power, is not the identity. The rest of the Isomorphism part should follow after that.
 July 10th, 2011, 12:57 PM #5 Senior Member   Joined: Jul 2011 Posts: 227 Thanks: 0 Re: Isomorphism If you can prove your homomorphism is a surjection and a bijection that will implicate it's a bijection and so an isomorphism. And isomorphism i: G-G is called an automorphism.
July 11th, 2011, 02:24 AM   #6
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Re: Isomorphism

Quote:
 Originally Posted by Siron If you can prove your homomorphism is a surjection and a bijection that will implicate it's a bijection...
Indeed, (P ^ Q) => Q.

(I think he meant "injection")

July 11th, 2011, 06:00 AM   #7
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Re: Isomorphism

Quote:
Originally Posted by The Chaz
Quote:
 Originally Posted by Siron If you can prove your homomorphism is a surjection and a bijection that will implicate it's a bijection...
Indeed, (P ^ Q) => Q.

(I think he meant "injection")
Yes, indeed I meant 'injection' .

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