# Prove or disprove that there is a element $y$ in $G$ such that , $y^2 = x$

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.

• What examples of groups with elements of odd order have you tried? Jun 30, 2019 at 12:37
• @MatthewLeingang $Z_n$
– Rkb
Jun 30, 2019 at 12:40
• 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). Jun 30, 2019 at 12:41
• @MatthewLeingang i didn't get it please could you elaborate
– Rkb
Jun 30, 2019 at 12:45
• Yeah no it's your homework problem Jun 30, 2019 at 13:06

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.
Let $$x^{2k-1} = e$$ for some $$k\ge 1$$, then $$x^{2k} = x$$, then let $$y=x^k$$.
• This is essentially the same as Parcly Taxel's answer with $k=n+1$ Jun 30, 2019 at 13:56
hint: If $$y^2=x$$ what is the relation between the order of $$y$$ and that of $$x$$?