Element of a cyclic group of even order has two square roots?

Question

The problem Im struggling with is "Let $G$ be a cyclic group of order $d$. If $d$ is even, then each element of $G$ has either two or zero square roots in $G$"

Does this mean I need to find a cyclic group with an odd number of elements where each element $e$ can be expressed as $e^2$ .

Also, if $d$ is odd, then is it also true that if $d$ is odd then every element of $G$ has exactly one square root.

Answer 1

Cyclic group is generated by some element $g.$ Since the group is order $d,$ our elements are $g, g^2,\dots, g^{d}=e.$

Since $d$ is even, $g^{2k}$ has square roots $g^{k}$ and $g^{k+d/2},$ which you can check are distinct.

The other case is an element of the form $g^{2k+1}.$ For any element of the cyclic group, $g^m,$ its square is $g^{2m},$ which cannot be $g^{2k+1}$. Again this is because $d$ is even: $2m$ is still even modulo $d.$ If $d$ was odd, this wouldn't be the case.

Answer 2

Since $G$ is cyclic it is abelian, so $\{ x^2 | x\in G\} = G^2$ is a (normal) subgroup of G. This is also valid for when $d$ is odd.

See if you can study the quotient $G/G^2$ and the map $x \mapsto x^2$ for $x\in G$.