Let $G$ a finite group without order elements $3$. Suppose
$$(ab)^3=a^3b^3$$
for all $a,b \in G$
Let $a \in G$. Demonstrate that there is $x \in G$ such as $x^3=a$
My attempt:
Since the group has no elements of order $3$, then
$$x^3=e \Longrightarrow x=e$$
Let $a,b \in G$ such as $a^3=b^3$
$$a^3=b^3 \Rightarrow a^3b^{-3}=e \Rightarrow (ab^{-1})^3=e$$
Therefore $ab^{-1}=e \Rightarrow a=b$
This means that no two cubes of different elements are equal. So if we cube all $n$ elements in the group we get $n$ different results. Hence every element of the group must be a cube.
Let $a \in G$, every element of the group must be a cube, then for $x \in G$
$$x^3=a$$
Is my attempt correct?
Other attempt:
Let $a \in G$ and $m=\text{ord}(a)$ Since the group has no elements of order $3$, then $m$ is not divisible by $3$, thus $\text{gdc}(m,3)=1$ by Bezout's Theorem exist $s,p \in \mathbb{Z}$ such as $ms+3p=1$,
$$a=a^1=a^{ms+3p}=a^{ms}a^{3p}=(a^m)^s(a^p)^3=e^s(a^p)^3=(a^p)^3$$
If $x=a^p$ then $x^3=a$, $x \in G$ because $a^p \in G$
Is my other attempt correct?
