Odd Order Group Element is Square

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Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $x \in G$.


Then:

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

if and only if:

the order $\order x$ is odd


Corollary

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

Let $\order x$ be odd.

Then:

$\exists n \in \Z_{> 0}: x^{2 n - 1} = e$

from the definition of the order of an element.

Conversely, suppose that

$\exists n \in \Z_{> 0}: x^{2 n - 1} = e$

Then $\order x$ is a divisor of $2 n - 1$ from Element to Power of Multiple of Order is Identity.

Hence $\order x$ is odd.


So $\order x$ is odd if and only if $\exists n \in \Z_{> 0}: x^{2 n - 1} = e$.


Then:

\(\displaystyle \order x\) \(=\) \(\displaystyle 2 n - 1\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x^{2 n - 1}\) \(=\) \(\displaystyle e\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x^{2 n}\) \(=\) \(\displaystyle x\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \paren {x^n}^2\) \(=\) \(\displaystyle x\)

Hence the result.

$\blacksquare$


Sources