Odd Order Group Element is Square/Corollary

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Corollary to Odd Order Group Element is Square

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

Then:

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

if and only if $\order G$ 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.

$\Box$


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.

Aiming for a contradiction, suppose $\order G$ 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.

$\blacksquare$


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