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Let $G$ be a finite group whose cardinality is not divisible by $3$. How can one show that for every $g$ belonging to $G$ there exists an $h \in G$ such that $g=h^3$?

quid
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user304330
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2 Answers2

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For every $g$, the map $h \mapsto h^3$ on $\langle g \rangle$ is injective (otherwise there'd be an element of order $3$, contradicting the assumption), whence this map is surjective and the claim follows.

quid
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    Why would it have an element of order three ($G$ is not abelian)? – WimC Jan 11 '16 at 17:26
  • Right. I changed the argument a bit. – quid Jan 11 '16 at 17:47
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    @quid : actually, your original intention can be fixed, but the argument is not much different from other ones: choose integers $x,y$ with $3x + |G|y = 1$. If $g,h \in G$ with $g^{3} = h^{3}$, then we also clearly have $g^{|G|} = h^{|G|}$,so that $g = g^{3x + |G|y} = (g^{3})^{x}(g^{|G|})^{y} = (h^{3})^{x}(h^{|G|})^{y} = h^{3x + |G|y } = h$. – Geoff Robinson Jan 11 '16 at 18:23
  • @GeoffRobinson I would not have though of doing it that way; thanks for the information. – quid Jan 11 '16 at 18:26
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An alternative strategy is to choose $g \in G$, and note that $|\langle g \rangle|$ divides $|G|$ by Lagrange's theorem, so that the order of $g$, say $n$, is not divisible by $3$. Then there are integers $a,b$ with $na+3b = 1.$Then we have $g = g^{1} = g^{na+3b} = g^{na}g^{3b} = 1_{G}g^{3b} = (g^{b})^{3}$.

Geoff Robinson
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