Odd Order Group Element is Square

Theorem
Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $x \in G$.

Then:
 * $\exists y \in G: y^2 = x$


 * the order $\order x$ is odd
 * the order $\order x$ is odd

Proof
Let $\order x$ be odd.

Then:
 * $\exists n \in \Z_{> 0}: x^{2 n - 1} = e$

from the definition of the order of an element.

Conversely, suppose that
 * $\exists n \in \Z_{> 0}: x^{2 n - 1} = e$

Then $\order x$ is a divisor of $2 n - 1$ from Element to Power of Multiple of Order is Identity.

Hence $\order x$ is odd.

So $\order x$ is odd $\exists n \in \Z_{> 0}: x^{2 n - 1} = e$.

Then:

Hence the result.