The function $\Psi:G \to G$ with $g \mapsto g^m$ is an automorphism.

Question

Let $G$ be an Abelian group of order $n$ and for a positive integer $m$ we have that $\text{gcd}(m,n)=1$.

Then the function $\Psi:G \to G$ with $g \mapsto g^m$ is an automorphism.

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My try:

First I show that the function is homomorphism: Let $g_1,g_2 \in G$ and consider $\Psi(g_1g_2)=(g_1g_2)^m$ and this is true since $G$ is a group,besides it's Abelian and so $\Psi(g_1g_2)=g_1^{m}g_2^{m}=\Psi(g_1)\Psi(g_1)$.

The function is clearly surjective.

To show that it's injective we need to show $\text{ker}(\Psi)=\left\{e_G\right\}$, if we assume there exists $e_G\ne g \in\text{ker}(\Psi)$ then $\Psi(g)=e_G=g^m$,but I don't know how to continue.

Any help is appreciated.

Answer 1

Any $f\colon X\to X$ is bijective if it is surjective/injective when $X$ is finite. In this case, you can also use the following. Since $\gcd(m, n) = 1$ there are integers $r, s$ such that $rm+sn = 1$, so if $g^m = 1$, then $g^{rm+sn} = g^{sn} = 1$ because $|G| = n$, but at the same time $g^{rm+sn} = g^1 = g$, hence $\Psi$ is injective.

To obtain surjectivity, observe that by $rm+sn=1$, we have for any $g\in G, (g^r)^m=g$.

Answer 2

A (purely) group-theoretical argument here:

Suppose $g^m=e_G$, then $\text{ord}(g)|m$. By Lagrange's theorem we have $\text{ord}(g)|n$. Therefore $\text{ord}(g)$ is a common divisor of two relatively prime numbers, so $\text{ord(g)}=1$ and $g=e_G$.

Answer 3

To show injectivity, suppose that $g^m = g_{1}^{m}$.

Then $g^mg_{1}^{-m} = 1 \implies g^mg_{-1}^m = 1 \implies (gg_{1}^{-1})^{m}=1$. Since $(m,n)=1, mx+ny=1$ and so $1=(gg_{1}^{-1})^{m} = (gg_{1}^{-1})^{1-ny} = gg_{1}^{-1} \implies g = g_{1}$ because $|G| = n$.