# The order of group $G$ is odd. Prove the mapping $f:G\to G$ by $f(x) = x^2$ is injective.

The order of group $G$ is odd. Prove the mapping $f:G\to G$ by $f(x) = x^2$ is injective

For what it is worth this is what I have tried.

Assume $x,y \in G$, $f(x) = f(y)$. We want to show $x = y$.

• Case (i) Either $x$ or $y$ is $e$ (the identity element of $G$). Then w.l.o.g. let $x = e$. So $x^2 = y^2$. $e^2 = e = y^2$. So $y^{-1} = y$. So $y = e = x$ and we are done or the order of $y$ is $2$ which is a contradiction since $|G|$ is odd.

• Case (ii) Neither $x$ nor $y$ is $e$. $x^2 = y^2$.

Now I am out of gas. I stared at some Cayley tables of $Z_n$ under modular addition for odd $n$. Sure enough the elements on the diagonal are the group elements. I cannot even think of any other examples of groups of odd order.

Let the order of the group be $2n - 1$. $$x^2 = y^2 \implies (x^2)^n = (y^2)^n \implies x^{2n-1}x = y^{2n-1}y$$ and by Lagrange's theorem, $$x^{2n-1} = e$$ so $$x^{2n-1}x = y^{2n-1}y \implies x = y$$
In general: let $|G|=n$ and let $m$ a positive integer with gcd$(m,n)=1$. Then the map $f: G \rightarrow G$, defined by $f(x)=x^m$ is bijective.
Proof By Bézout's Theorem we can find $a,b \in \mathbb{Z}$, such that $am+bn=1$. Note that Lagrange's Theorem tells us that $x^n=y^n=e$.
Assume that $x^m=y^m$. Then $x=x^{am+bn}=(x^{m})^a \cdot (x^{n})^b=(y^{m})^a \cdot e^b=(y^{m})^a \cdot (y^{n})^b=y$. So $f$ is injective and hence bijective, since $|G|$ is finite (Pigeonhole Principle).