Fix $n\in\Bbb N$. Let $G$ be a group where $|g|\mid n,\forall g\in G$. If $\gcd(n,m)=1$, prove $f: G \to G,g\to g^m$, is injective, without Lagrange.

abstract-algebragroup-theory

Fix a natural number $n$. Let $G$ be a group with the property $|g| \mid n, \forall g \in G$. If $\gcd(n,m)=1$, prove that $f: G \to G$ via $g \to g^m$ is injective, without Lagrange's theorem

I noticed some post like the one below uses Lagrange's Theorem (and some other theorems). But we have not learned those yet.
Let $G$ be a finite group of order $d$ and $n$ be an integer with $\gcd(n, d)=1$. Prove the mapping $f:G\to G$, $f(x)=x^n$ is bijective.

So far, I know two ways to show injectivity:

  • Assume $f(x) = f(y)$ and show $x = y$. So this means $x^m = y^m$. Some ideas I had was to somehow use $\gcd(n,m) = 1 \implies an+bm = 1$ for some integer $a,b$. But I had no idea where to go.
  • Show $\ker f = \{e\}$, the identity. But I had no idea where to go.

Best Answer

Because $\gcd(m,n)=1$, there exist integers $r,s$ such that $rn+sm=1$. Also $x^n=e$ for all $x\in G$.

So suppose that $x^m=y^m$. Then $x^{sm} = (x^m)^s = (y^m)^s = y^{sm}$.

So $x^{1-rn}=x^{sm}=y^{sm}=y^{1-rn}$.

And $x^{1-rn} = x(x^n)^{-r}$...

Can you take it from there?

I don't think Lagrange's Theorem is relevant here. Note that $G$ is not assumed to be finite.

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