TITLE: Suppose that $G$ is a finite group with $\gcd(|G|,3)=1$. Show that $G=\{x^3:x\in G\}$
QUESTION [6 upvotes]: Suppose that $G$ is a finite group with $\gcd(|G|,3)=1$. If $\theta:G\longrightarrow G$, $x\longmapsto x^3$ is a homomorphism of groups, we want to prove that $$G=\{x^3:x\in G\}.$$
To this end, we claim that $\ker \theta =1$. Indeed, if $a\in \ker \theta$, then $g^3=1_G$. So $|g| \mid 3$. On the other hand, $|g|\mid |G|$. So, $|g|\mid \gcd(|G|,3)$ and therefore $g=1_G$.
Hence $\theta$ is injective which maps the finite group $G$ into itself. Therefore, $\theta$ is surjective (and hence $\theta\in \mathrm{Aut}(G)$). Then,
$$G=\mathrm{im}\theta=\{x^3:x\in G\}$$
and the result follows.

My question is whether we could work as follows: Take an $g\in G$. Then, using Bezout's identity,
$$g=g^1=g^{|G|a+3b}=(g^b)^3\in \mathrm{im}(\theta)$$
where $a,b$ are integers and $g^b\in G$. So, could we omit using that way the whole premise about the homomorphism $\theta$?

Thanks.
Update: Many apologies, I forgot to define $\theta$!

REPLY [2 votes]: The map $\theta_i:x\mapsto x^i$ for $i>1$ is not in general a homomorphism for finite groups $G$. It is a homomorphism whenever a group $G$ is abelian, though, as is easy to see.
Thus we cannot use the fact that $\theta$ is a homomorphism, since it isn't!
It is a homomorphism on the subgroup $X=\langle x\rangle$ though. And if $|X|$ is prime to $i$, then $\theta_i$ must be injective, since no element of $X$ can be mapped to $1$ (other than $1$) as $|X|$ can contain no elements of order $i$ by Langrage's theorem. Injective maps from a finite set to itself are bijective, and so there are (unique) $i$th roots of any given element.
Since every element has an $i$th root if $\gcd(|G|,i)=1$, the map (not homomorphism) $\theta_i:G\to G$ is bijective.