If in a finite group such that $\forall x\exists y:x=y^3$ then its order can't be divided by $3$ Let $G$ be a finite group. If for every $x \in G$ there is a $y \in G$ such $y^3=x$, then the order of $G$ is not divisible by $3$.
I thought using Burnside's lemma, But I cannot see how.
It seems to be the opposite to this:
Finite group that has order not divisible by 3
$$\forall x\in G.\exists y\in G.[y^{3}=x]$$
 A: Say the order of $G$ is divisibly by $3$. Then by Cauchy's theorem, there exists an element $g_1$ of order $3$. By assumption, there exists $g_2$ such that $g_2^3=g_1$. Note that the order of $g_2$ is necessarily $9$ (it cannot be $3$, since otherwise $g_1=1$). Then there is a $g_3$ such that $g_3^3=g_2$. The order of $g_3$ is $27$ (if it were $9$, then $g_2$ has order $3$). Continuing in this fashion, there must be an element of order $3^k$ for each $k$. As $G$ is a finite group, this is a contradiction.
A: What you want to prove is that if $.^3$ is a surjective map, then $3 \nmid |G|$. In general:

If $G$ is finite group of order $n$ and $k$ an integer then the map $f: a \mapsto a^k$ is a bijection if and only if $\gcd(k,n)=1$. 

Proof Assume first $\gcd(k,n)\neq 1$. Then we can find a prime $p$ with $p \mid k$ and $p \mid n$. By Cauchy's Theorem there is an $a \in G$ with ${\rm order}(a)=p$. Then $a^k=a^{p \cdot \frac{k}{p}}=1^\frac{k}{p}=1$. Since $f$ is injective this yields $a=1$, a contradiction.  To prove the other direction: let $\gcd(k,n)=1$. We only need to show injectivity of the map $f$, since $G$ is finite and hence $f$ would be automatically surjective. Since $\gcd(k,n)=1$, we can find $a,b \in \mathbb{Z}$ with $ak+bn=1$. So if $g^k=h^k$, then
$$\begin{align}
g&=g^{ ak+bn}\\
&=(g^k)^a \cdot (g^n)^b\\
&=(h^k)^a \cdot 1^b\\
&=(h^k)^a \cdot (h^n)^b \\
&= h^{ ak+bn}\\
&=h.
\end{align}$$
A: Consider $\varphi : G \rightarrow G$ defined for every $x \in G$ by
$$\varphi(x)=x^3$$
By hypothesis, $\varphi$ is surjective. But because $G$ is finite, this implies that $\varphi$ is injective.
Now, if $G$ has order dividible by $3$, it must have an element of order $3$, which contradicts the fact that $\varphi$ is injective.
