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Let $(G,*)$ be a group. And let $ x$ be a element of odd order of $G$ , then prove or disprove that , there is a element $y$ in $G$ such that , $y^2 = x$

Please provide some hint, i am not able to show any contradicting examples nor able to prove it.

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  • $\begingroup$ What examples of groups with elements of odd order have you tried? $\endgroup$ Jun 30, 2019 at 12:37
  • $\begingroup$ @MatthewLeingang $ Z_n$ $\endgroup$
    – Rkb
    Jun 30, 2019 at 12:40
  • $\begingroup$ Good. If you can prove the statement for $Z_n$, then you can prove it in general, because if $x$ has odd order, it generates a subgroup isomorphic to some $Z_n$ (where $n$ is odd). $\endgroup$ Jun 30, 2019 at 12:41
  • $\begingroup$ @MatthewLeingang i didn't get it please could you elaborate $\endgroup$
    – Rkb
    Jun 30, 2019 at 12:45
  • $\begingroup$ Yeah no it's your homework problem $\endgroup$ Jun 30, 2019 at 13:06

3 Answers 3

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Suppose $x$ has order $2n+1$. Then $x^{n+1}$ can serve as $y$, since $y^2=x^{2n+2}=x^{2n+1}x=x$. Therefore the statement is true.

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Let $x^{2k-1} = e$ for some $k\ge 1$, then $x^{2k} = x$, then let $y=x^k$.

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  • $\begingroup$ This is essentially the same as Parcly Taxel's answer with $k=n+1$ $\endgroup$ Jun 30, 2019 at 13:56
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hint: If $y^2=x$ what is the relation between the order of $y$ and that of $x$?

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    $\begingroup$ Both have same order..i had tried this...but not able to find any disproving example $\endgroup$
    – Rkb
    Jun 30, 2019 at 12:41

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