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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?

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    Consider the cyclic subgroup generated by $g$...2012-12-31

3 Answers 3

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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 $3a+|G|b=1.$

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    More generally you can show (with the same proof) that if $(|G|, n) = 1$, then the map $g \mapsto g^n$ is$a$bijection $G \rightarrow G$.2012-12-31
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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.

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    And $\frac13(|G|+1)(2|G|+1)$ works in all cases.2012-12-31
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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 $\,3\nmid |G|\,$?)

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    I still don't follow. It sounds suspiciously like you are showing that for every non-trivial $x \in G$ there exists a $g \ne 1$ such that $g = x^3$, which is kinda backwards.2012-12-31