if $k$ is a positive integer and $G$ a finite group such that $G=\{x^k:x\in G\}$ , then is it true that g.c.d.$(|G|,k)=1$ ?

Question

If $G$ is a finite group of order $n$ and $k$ is a positive integer such that g.c.d.$(n,k)=1$ , then I know that $G=\{x^k:x\in G\}$ ; is the converse true ? that is if $k$ is a positive integer and $G$ is a finite

group such that $G=\{x^k:x\in G\}$ , then is it true that g.c.d.$(|G|,k)=1$ ?

I have a feeling that the converse is true . Say g.c..d$(|G|,k)=d>1$ , then there is a prime $p|d$ , then there is an element $g$ of $G$ of order $p$ , but I don't know what else to do . Please help.

Answer 1

The map $x \to x^k$ is surjective iff the map $x \to x^k$ is injective (because $G$ is finite).

In particular, this property implies that $x^k=1 \Rightarrow x=1$. If $gcd(|G|,k) \neq 1$, there is a prime $p$ dividing $k$ and an element $x$ of order $p$. But then $x^k=1 \Rightarrow x=1$, a contradiction.

Answer 2

1. Let's denote $G^k:=\{x^k:x\in G\}$. There is a $p$ such that $(|G|, k) = d$ and $p | d$. Using Cauchy theorem there is an element $g \in G$ of order $p$.

2. $|G| = n = p^sm$ and $(m, p) = 1$. Then $H = G^m \neq \{1\}$ because $g^m \neq 1$.

3. $\{1\} = G^n = G^{m \cdot p^s} = G^{m \cdot d^s} = ((G^{d})^{d^{s-1}})^m = (G^{d^{s-1}})^m = \dots=G^m \neq \{1\}$.

Answer 3

Yes it is true.

Let $f:G\to G$ by $x\mapsto x^k$ by our assumption $f$ is one to one and onto. Let $d=Gcd(|G|,k)$ and assume $d>1$.

we have a prime $p$ dividing $d$, so there is an element $a$ of order $p$ in $G$. But $f(a)=a^k$ and since $p|k$ we have $a^k=e$. Beside that $e^k=e$ so $f$ is not one one contradiction.