Odd Order Group Element is Square/Corollary

Corollary to Odd Order Group Element is Square
Let $\struct {G, \circ}$ be a finite group.

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

$\order G$ is odd.

Proof
Suppose $\order G$ 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 Odd Order Group Element is Square.

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

From Odd Order Group Element is Square it follows that all elements of $G$ are of odd order.

$\order G$ were even.

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

From that contradiction we conclude that $\order G$ is odd.