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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?

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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.

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

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

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

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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...

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

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