Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Consider the cyclic subgroup generated by $g$... – Henning Makholm Dec 31 '12 at 2:42

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

share|cite|improve this answer I still don't see it clearly from what you wrote. – DonAntonio Dec 31 '12 at 2:50
@DonAntonio Given $g\in G,$ consider $h=g^a.$ $g=g^{3a+|G|b}=(g^a)^3.$ – ՃՃՃ Dec 31 '12 at 2:53
+1 Very nice! Didn't see that one...though I think directly is slightly clearer: $$g\in G\Longrightarrow g=g^1=g^{3a}g^{b|G|}=(g^a)^3$$ – DonAntonio Dec 31 '12 at 2:57
@DonAntonio You're right, that makes it clearer :-) – ՃՃՃ Dec 31 '12 at 2:59
Clearly one can also use $|g|$ in place of $|G|$, but using the latter allows us to find cube roots for all elements of $G$ with just one exponent $a$. – peoplepower Dec 31 '12 at 3:03

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.

share|cite|improve this answer
And $\frac13(|G|+1)(2|G|+1)$ works in all cases. – peoplepower Dec 31 '12 at 3:15

For all $\,x\in G\,$ , define

$$f_x:G\to G\;\;,\;\;f_x(g):=x^{-3}g$$


$$f_x(g)=f_x(h)\Longleftrightarrow x^{-3}g=x^{-3}h\Longleftrightarrow g=h\Longrightarrow \;\;f_x\,\,\,\text{is}\,\,1-1$$

End the argument now (where do we use that $\,3\nmid |G|\,$?)

share|cite|improve this answer
I don't see how this helps? $f_x$ is trivially injective regardless of $G$. – Erick Wong Dec 31 '12 at 3:02
Yes, but if $\,3\mid |G|\,$ then there exists $\,x\in G\,\,\,s.t.\,\,x^3=1\Longrightarrow f_x=Id_G\,$ , so that in that case $\,f_x(g)=1\Longleftrightarrow g=1\,$ and we don't get $\,\exists\,1\neq g\in G\,\,\,s.t.\,\,x^{-3}g=1\Longrightarrow g=x^3\,$ . Of course, we also need finitiness to deduce injective iff surjective. – DonAntonio Dec 31 '12 at 3:08
Anyway, I'd go with use's answer: simplicity and elegancy. – DonAntonio Dec 31 '12 at 3:08
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. – Erick Wong Dec 31 '12 at 3:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.