# Prove or disprove by counterexample that in every group, every element has a square root

For every $x \in G$, there is some $y \in G$ such that $x=y^2$. (This is the same as saying that every element of G has a square root)

Now, I'm not sure but I've been trying to think of counter-examples and I thought of the group of integers under multiplication.

Because if $x=5$ there is no integer $y$ where $y^2=5$

Is this right?

You can take the group $\mathbb{Z}/2\mathbb{Z}$ and note that $\bar{0}+\bar{0} = \bar{1}+\bar{1} = \bar{0}$, so $\bar{1}$ is not a square.

Consider $G = \Bbb Q^*$ under multilplication. There is no $x \in \Bbb Q^*$ for which $2 = x^2$.

• wouldn't you have to exclude 0 for it to be a group? for the rational example.
– mika
Commented Feb 8, 2015 at 23:41
• That's the job of "*". Though, I will have to exclude zero in the first one, so I'll just remove that.
– user207710
Commented Feb 8, 2015 at 23:45
• Oh i see, thanks!
– mika
Commented Feb 8, 2015 at 23:50

For a familiar counterexample, just take $(\mathbb{R}^*, *)$, the nonzero real numbers under multiplication. Since negative numbers don't have square roots, this is a counterexample to the claim.

(Z,+) also works. (Odd numbers.)

Your answer isn't quite right. Is the set of integers under multiplication a group? Try thinking about the integers under a different operation which makes it a group. Isn't it funny that $1/2$ is not an integer...

• oh yes! I just realize the mistake.
– mika
Commented Feb 8, 2015 at 22:31