Finite groups and one-to-one functions on them.

Question

I am having trouble with this problem:

Assume that $(\mathbb{G}, *)$ is a finite group and there exists a positive integer $n$ such that gcd($n, |\mathbb{G}|)=1$.

Prove that the function $F_n: \mathbb{G} \rightarrow \mathbb{G} $ defined $(\forall x)(F_n(x)=x^n)$ is one-to-one.

I know that since $n$ and $|\mathbb{G}|$ are relatively prime $\exists x,y \in \mathbb{Z}$ such that $nx + |\mathbb{G}|y=1$. I'm just not sure how to use that to prove that $F_n$ is one-to-one.

At first I thought that using the above equation would allow me to solve for $x, y$ and then allow me to somehow show that $x=y \Rightarrow x^n = y^n \Rightarrow F_n(x)=F_n(y)$, therefore $F_n$ is one-to-one. I don't think that will work, though.

Help? Can someone point me in the right direction?

Answer 1

Suppose we had two elements $g, h$ such that $F_n(g) = F_n(h)$; that is, $g^n = h^n$. It follows that \begin{align*} g^{nx} &= g^{1 - |G|y} \\ &= g \cdot \left(g^{|G|}\right)^{-y} \\ &= g \end{align*} after using the fact that $g^{|G|} = e$ for every $g \in G$. Make the same computation for $h$ to conclude that

$$g = g^{nx} = h^{nx} = h$$

Answer 2

Let $|G|=g$ so we have $(g,n)=1$.

This means we can find integers $p, q$ with $pg+nq=1$

Note also that $x^g=1$ for all elements of $G$.

Can you put that together to show that $F_q$ is an inverse for $F_n$?

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Since $x^g=1$ we have also $x^{pg}=1$

Note that $F_q(F_n(x))=F_q(x^n)=x^{nq}=x^{nq}\cdot 1=x^{nq}x^{pg}=x^{nq+pg}=x^1=x$

and similarly $F_n(F_q(x))=x^{nq}=x$

Since $F_n$ has an inverse, it must be a bijection and hence one-to-one.