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 April 18th, 2012, 10:14 PM #1 Newbie   Joined: Apr 2012 Posts: 16 Thanks: 0 isomorphism Let G be an abelian group of order $n$. Define $\phi : G \longrightarrow G$ by $\phi (a)= a^m$, where $a\in G$. Prove if gcd(m,n) = 1 then $\phi$ is an isomorphism.
 April 22nd, 2012, 02:08 AM #2 Senior Member   Joined: Jul 2011 Posts: 227 Thanks: 0 Re: isomorphism First, prove that $\phi$ is an homomorphism. Afterwards you need to prove $\phi$ is bijective thus on-to-one and onto (surjective). To prove if $\phi$ is one-to-one (or injective) you have to show: $\forall a,b \in G: \phi(a)=\phi(b) \Rightarrow (a)=(b)$ $\phi(a)=\phi(b) \Rightarrow a^m=b^m \Rightarrow (a^m)(b^{-1})^m=(ab^{-1})^m=e$ We know that $gcd(m,n)=1$ thus you can write $m=kn+r$ with $0< r < n$ (note that $r \neq 0$ because of $gcd(m,n)=1$) Thus $(ab^{-1})^m=(ab^{-1})^{kn+r}=[(ab^{-1})^n]^k(ab^{-1})^r=e$ There's a theorem which says that if $G$ is a finite group then the order (in this case $n$) is a multiple of the order of every element $x \in G$ that means $[(ab^{-1})^n]^k=e$ and thus $(ab^{-1})^r=e$ but because $0< r, $r$ has to be minimal and thus $r=1$ therefore $a=b$ and $\phi$ is on-to-one. This is just an attempt so I'm not absolutely sure it's entirely correct, but maybe you can do something with my proof.
 July 2nd, 2012, 02:17 AM #3 Newbie   Joined: Jul 2012 Posts: 4 Thanks: 0 Re: isomorphism Note that: (a) for each $a \in G$ $a^n=1$; (b) gcd(m,n)=1 is equivalent to the statement, there exist integers k and l such that km+ln=1; (c) to show that $\phi$ is a bijection is to show its kernel is trivial and it is surjective. Since for each $a \in G$ $a^{1}=a^{km+ln}=a^{km}a^{ln}=a^{km}$ it follows that: 1. $\phi(a^k)=a$ and so $\phi$ is surjective; 2. If $\phi(a)=1$ then $a^m=1$ and $a=a^{mk}=1$ and so the kernel of $\phi$ is trivial ($\phi$ is injective)

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