Let $G$ be a finite group whose order is not divisible by $3$. Show that for every $g∈G$ there exists an $h∈G$ such that $g=h^3$.
How can I solve this problem? Can anyone help me please?
Let $G$ be a finite group whose order is not divisible by $3$. Show that for every $g∈G$ there exists an $h∈G$ such that $g=h^3$.
How can I solve this problem? Can anyone help me please?
Hint: Since $3\nmid |G|,$ $gcd(|G|,3)=1$ and it follows from Bézout's identity that we can find a and b, such that $3a+|G|b=1.$
Since $3$ does not divide $|G|$, we have one of the following situations:
Case I: $|G| \equiv 2 \pmod 3$. Hence $|G|+1 \equiv 0 \pmod 3$ so $(\underbrace{g^{(|G|+1)/3}}_{\text{let this be } h})^3=g^{|G|+1}=g$ by Lagrange's Theorem.
Case II: $|G| \equiv 1 \pmod 3$. Hence $2|G|+1 \equiv 0 \pmod 3$ so $(\underbrace{g^{(2|G|+1)/3}}_{\text{let this be } h})^3=g^{2|G|+1}=g$ by Lagrange's Theorem.
For all $\,x\in G\,$ , define
$f_x:G\to G\;\;,\;\;f_x(g):=x^{-3}g$
Now,
$f_x(g)=f_x(h)\Longleftrightarrow x^{-3}g=x^{-3}h\Longleftrightarrow g=h\Longrightarrow \;\;f_x\,\,\,\text{is}\,\,1-1$
End the argument now (where do we use that $\,3\nmid |G|\,$?)