Odd Order Group Element is Square/Corollary

Corollary to Odd Order Group Element is Square

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

Then:

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

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 $\order G$ is even.

From Cauchy's Lemma (Group Theory) it follows that there exist elements in $G$ of even order.

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

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.12 \ \text{(b)}$