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I am having trouble with showing that the function in this problem is injective. I've been trying it for a while already. Surjectivity wasn't hard, and neither was proving that it was a homomorphism.

Let $G$ be a finite abelian group and let $n$ be a positive integer relatively prime to $|G|$.

  1. Show that the mapping $\varphi(x)=x^n$ is an automorphism.
  2. Show that every $x\in G$ has an $nth$ root, i.e., for every $x$ there exists some $y \in G$ such that $y^n=x$.
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That's a strangely worded problem, given that part 1 implies part 2 via the surjectivity... – Arturo Magidin Apr 13 '12 at 3:18
up vote 4 down vote accepted

In fact, the result holds if we assume that $G$ is abelian of exponent $k$, where $\gcd(k,n)=1$ (an group is of "exponent $k$" if $x^k=1$ for all $x\in G$). This gives your result as a special case, since when $G$ is finite, by Lagrange's Theorem it is of exponent $|G|$.

To show $\varphi(x)=x^n$ is injective, we just need to show that the kernel is trivial

If $\varphi(x) = 1$, then $x^n=1$. Therefore, the order of $x$ divides $n$. But by assumption, the order of $x$ also divides $k$. Hence, it divides $\gcd(k,n)$. Therefore...

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Hint: In a finite set a function is surjective iff is injective iff is bijective.

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