Odd Order Group Element is Square

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $x \in G$ such that $\left\vert{x}\right\vert$ is odd.

Then:
 * $\exists y \in G: y^2 = x$ iff $\left\vert{x}\right\vert$ is odd

where $\left\vert{x}\right\vert$ denotes the order of $x$.

Corollary
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

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

iff $\left\vert{G}\right\vert$ is odd.

Proof
Let $\left\vert{x}\right\vert$ be odd.

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

from the definition of the order of an element.

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

Then $\left\vert{x}\right\vert$ is a divisor of $2 n - 1$ from Element to the Power of Multiple of Order.

Hence $\left\vert{x}\right\vert$ is odd.

So $\left\vert{x}\right\vert$ is odd iff $\exists n \in \Z_+^*: x^{2 n - 1} = e$.

Then:

Hence the result.

Proof of Corollary
Suppose $\left\vert{G}\right\vert$ is odd.

Then from Order of Element Divides Order of Finite Group, all elements of $G$ are of odd order.

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

from the main result.

Now suppose that:
 * $\forall x \in G: \exists y \in G: y^2 = x$

From the main result it follows that all elements of $G$ are of odd order.

Suppose $\left\vert{G}\right\vert$ were even.

Then from Cauchy's Group Theorem it follows that there must exist elements in $G$ of even order.

From that contradiction we conclude that $\left\vert{G}\right\vert$ is odd.