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 $$\left|{x}\right|$$ is odd.

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

where $$\left|{x}\right|$$ 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 $$\left|{G}\right|$$ is odd.

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

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

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

Then:

$$ $$ $$ $$

Hence the result.

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