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 $2 = x^2$.
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.
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...
