# prove of disprove :'For every $x\in G$ there exists some $y\in G$ such that $x=y^2$, where $G$ is a group."

I am working on a question in the book: A Book of Abstract Algebra by Pinter. The question asks to prove or disprove the following statement:

For every $x\in G$ there exists some $y\in G$ such that $x=y^2$, where $G$ is a group.

Now I am quite stumped by this. I tried the following: $$x=yy$$ thus $$x^{-1} = y^{-1}y^{-1}$$ Now since $y \in G$ we need to find an element $z=y^{-1}$ such that $x z^2=e$. I am not sure if this is the way to go to show the final result. If someone could help me out along the way that would be greatly appreciated. Thanks in advance!

P.S. Another approach might be $$y^{-1}x = xy^{-1} =y$$ Now because a group is not always commutative this could imply the statement is untrue.... I am not sure.. Thanks!

• Look for a counterexample, such as the group of residues $\pmod{2}$. Nov 2, 2013 at 22:21
• Ah thanks! I guess specifically choosing $x={1}$ we find that no other element satisfies $x=y^2$ in the group of residues $\mod 2$. Thanks for your quick response! Nov 2, 2013 at 22:24
• No problem - you are welcome. Nov 2, 2013 at 22:25
• Note that if $G$ is a finite group of odd order, this is actually true. Nov 2, 2013 at 22:31
• This is true in some groups and false in others. Aug 22, 2019 at 8:00

Another very obvious counter example: In $\mathbb{R} \backslash \{0\}$ under mulitplication, there is no element such that $x^2 = -1$.

The statement is however true if $G$ is a finite group of odd order. To see this, let $|G| = 2n+1$. Then for any $x\in G$,

$$(x^{n+1})^2 = x^{2n+2} = x$$ since $x^{2n+1} = 1$ by Lagrange. So $y=x^{n+1}$ would do the trick.

If you consider the group $\mathbb{Z}/2\mathbb{Z}$ i.e. the group of residues $\pmod{2}$ under addition, in which every element squared gives the identity, there can be no element whose "square" is equal to the non-identity element 1.

Consider the group $$S_3$$ .

There exists no $$x$$ such that $$x^2=(12)$$

• Thanks for the answer :) Aug 22, 2019 at 16:44
• @Slugger;my pleasure,if you feel that this answer is simple and good ,do consider accepting it Aug 23, 2019 at 2:01
• I have already accepted an answer I'm afraid Aug 23, 2019 at 17:00
• @Slugger;no worries, just suggested as its my first answer here Aug 24, 2019 at 2:44