# 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

### Corollary

Let $\struct {G, \circ}$ be a finite group.

Then:

$\forall x \in G: \exists y \in G: y^2 = x$

if and only if $\order G$ 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 if and only if $\exists n \in \Z_{> 0}: x^{2 n - 1} = e$.

Then:

 $\displaystyle \order x$ $=$ $\displaystyle 2 n - 1$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle x^{2 n - 1}$ $=$ $\displaystyle e$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle x^{2 n}$ $=$ $\displaystyle x$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle \paren {x^n}^2$ $=$ $\displaystyle x$

Hence the result.

$\blacksquare$