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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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3  
Consider the cyclic subgroup generated by $g$... –  Henning Makholm Dec 31 '12 at 2:42

3 Answers 3

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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Yes...so? 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
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+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.

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

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

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

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 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

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