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Let $G$ be abelian. Let $n \in \mathbb{Z}^+$ such that the order of $G$ and $n$ are relatively prime. Show that the function $\varphi : G \rightarrow G$ defined by $\varphi (a) = a^n $ is an automorphism for $a \in G$.

I'm stuck in proving 1-1-ness of the function. Let $a,b$ in $G$ such that $\varphi (x) = \varphi (y)$ then $a^n = b^n$. How can I show that $\varphi$ is 1-1 from here on out?

Alex M.
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llawliet_78
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  • What you want to show is essentially the same as http://math.stackexchange.com/questions/1064817/let-g-to-be-finite-abelian-group-of-order-og-let-n-to-be-prime-number-and/ where I have given an answer not using Bezout's identity (indeed turning the answer into a proof of said identity). – Tobias Kildetoft Nov 30 '15 at 19:49

4 Answers4

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I assume you have shown that $\phi$ is a homomorphism.

Showing that a homomorphism is injective is equivalent to showing that its kernel is trivial.

Let $a \in \ker \phi$. We want to show that $a=1$. Let $g$ be the order of $G$ and let $k$ be the order of $a$. By Lagrange's theorem, $k \mid g$. Since $a \in \ker \phi$, we know that $a^n = 1$. Hence $k \mid n$. Since $\gcd(n,g) = 1$, there exist $x,y \in \mathbb{Z}$ such that $nx + gy = 1$. Now $k$ divides both $g$ and $n$, so it divides (a multiple of) their sum $nx + gy$. So $k \mid nx + gy = 1$. Therefore $k=1$ (orders of elements are $>0$). So $a$ has order $1$. It follows that $a$ is the neutral element.

$\ker \phi$ is trivial, so $\phi$ is injective.

rwols
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    No need to use Bezout: $k \mid g, k \mid n \implies k \mid \gcd(g,n)=1$. – lhf Nov 30 '15 at 12:31
  • I agree with @lhf that Bezout is not needed here (and in fact, writing this out properly gives a group theoretic proof of Bezout's identity). – Tobias Kildetoft Nov 30 '15 at 12:46
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Fun fact to know that it works also the other way around: let $G$ be abelian, if $\varphi(a)=a^n$ is an automorphism of $G$, then gcd$(n,|G|)=1$.

Proof: assume that gcd$(n,|G|) \neq 1$ and let $p$ be a prime dividing both $n$ and $|G|$. By Cauchy's Theorem, there is an $a \neq 1$, such that $a^p = 1$. But then $\varphi(a)=a^n=(a^p)^{\frac{n}{p}}=1^{\frac{n}{p}}=1=1^n=\varphi(1)$. Since $\varphi$ is injective, it follows that $a=1$, a contradiction.

Nicky Hekster
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Hint: use the fact that $ \gcd(|G|, n ) = 1 $, hence, by Euclid's algorithm (or from the mentioned Bézout's identity), there exist integer $ x,y $ such that $$ x|G| + yn = 1 $$ So

$$ a = a ^{x|G|}a^{yn} = a^{yn} = b^{yn} = b $$

Jytug
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You can either show that $\phi(x)=\phi(y)$ implies $x=y$ or if $x\neq y$ then $\phi(x)\neq\phi(y)$.

Let's use the latter, if $x\neq y$ then we have that $xy^{-1}\neq e$. This gives us that $$\phi(xy^{-1})=\phi(x)\phi(y^{-1})=\phi(x)\phi(y)^{-1}=x^n (y^{n})^{-1}$$ for $x^n (y^{n})^{-1}=e$ we must have that $x^n=y^{-n}$ which is not possible as $x\neq y$

Zelos Malum
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