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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?

L.J.
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2 Answers2

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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 $F_n$?

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.

Mark Bennet
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  • I can mostly see the logic of your answer now. I think you assumed that the operation in question was multiplication, where I only meant for it to be an arbitrary operation. Thus, I was a little confused by $x^g=1$ at first, but now I believe that you meant $1$ is the identity element for multiplication. I'm not exactly sure what you mean by "show that $F_q$ is the inverse for $F_n$", though. I managed to solve the problem using T. Bongers' answer. Did you have something different in mind? – L.J. Mar 19 '14 at 17:24
  • @JoePirus T. Bongers answer is exactly equivalent to mine. Common notations for the identity in a group are $e$ and $1$. I'll add a section to the end of my answer for the rest. – Mark Bennet Mar 19 '14 at 18:41
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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

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

  • Okay, since gcd(n, |G|)=1 we obtain the fact that nx+ – L.J. Mar 19 '14 at 17:05
  • Oops. Given that $gcd(n, |G|)=1$, it is true that $nx + |G|y = 1$ where $x, y \in Z$. This is equivalent to $nx = 1 - |G|y$. Thus it is true that $g^{nx} = g^{1 - |G|y}$. From here we end up using Cauchy's theorem regarding finite groups and obtain the fact that $g^{|G|}=id_G$. and ultimately $g^{nx}=g$ and similarly for $h^{nx}$. Thus we show that $F_n(g)=F_n(h)$ implies that $g=h$, so $F_n$ is 1:1. – L.J. Mar 19 '14 at 17:18