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Let $G$ be a group and $g\in G$.

Prove or Disprove: If $o(g^3)=2$ then $o(g)=6$.

I tried to disprove with $g=(14)(32)$ then $g^3=(14)(32)$ so that $o(g^3)=2$ but $o(g)=2\neq6$.

Is it right?

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  • $\begingroup$ Did you mean $g$ instead of $a$? $\endgroup$
    – jjagmath
    Feb 6, 2022 at 19:11
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    $\begingroup$ @Jay To disprove something, a particular counterexample is enough. $\endgroup$
    – jjagmath
    Feb 6, 2022 at 19:13

2 Answers 2

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Yes, that's correct. The order of $g=(14)(32)=g^3$ is indeed two. Any transposition will do in fact; even better, pick any element of order two in any group.

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  • $\begingroup$ There is maybe a simpler counterexample? $\endgroup$
    – Xavi
    Feb 6, 2022 at 19:16
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    $\begingroup$ Yes, @Xavi; like I said: any transposition will do. $\endgroup$
    – Shaun
    Feb 6, 2022 at 19:17
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    $\begingroup$ Or really any element $g$ in any group with $o(g)=2$ $\endgroup$ Feb 6, 2022 at 20:23
  • $\begingroup$ Indeed, yes, @chickenNinja123. Thank you. $\endgroup$
    – Shaun
    Feb 6, 2022 at 20:24
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(I). Example. Let $G=\{1,g\}$ where $g^2=1\ne g.$ Then $o(g)=2.$ And $g^3=g(g^2)=g\cdot 1=g\ne 1$ but $(g^3)^2=(g^2)^3=1^3=1,$ so $o(g^3)=2.$

(II). If $G$ is any group with identity $1$ and if $o(g^3)=2$ then $[o(g)=6\iff g^3\ne g].$

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