Odd Order Group Element is a 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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert{x}\right\vert\)	\(=\)	\(\displaystyle 2 n - 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle x^{2 n - 1}\)	\(=\)	\(\displaystyle e\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle x^{2 n}\)	\(=\)	\(\displaystyle x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle \left({x^n}\right)^2\)	\(=\)	\(\displaystyle x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Hence the result.

$\blacksquare$

Sources

• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $6.15$