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I need to show the following: If G is abelian group of order n, then $f(x)=x^m$, where (n,m) are co-prime, is an automorphism of G

I know it needs Lagrange theorem, but would appreciate some pointers.

To show injective, I take $x^m=y^m$ and then try to prove that this implies $x=y$. But can't seem to do that.

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Use the fact that $am+bn=1$ for some $a,b\in Z$. –  Boris Novikov Mar 21 '13 at 16:14

2 Answers 2

up vote 4 down vote accepted

We don't need that $G$ is abelian in order to show that $f$ is bijective.

Choose $u,v \in \mathbb{Z}$ with $un+vm=1$ and verify that $g(x):=x^v$ is inverse to $f$.

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It does have to do with Lagrange's Theorem because you need $x^n=1$ for all $x\in G$. (This is result is called the Little Lagrange Theorem by @PeteL.Clark.) –  lhf Mar 21 '13 at 16:18

Boris's comment and Martin's answer give you what you need for establishing that $f$ is bijective.

Then, we do need that $G$ is abelian to establish that the homomorphism property holds. That is, we need $\color{blue}{\bf\text{commutativity}}$ to conclude that for all $x, y \in G$

$f(xy) = (xy)^m = \color{blue}{\bf{\underbrace{(xy)(xy)\cdots (xy)}_{\large m\;times} = x^my^m}} = f(x)f(y)\quad\quad\quad\quad\quad\quad\tag{because $G$ is abelian}$

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Thanks, I get it. –  user66306 Mar 21 '13 at 16:28

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