Odd Order Group Element is Square

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Theorem

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

Let $x \in G$.

Let the order $\order x$ be odd.


Then:

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


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:

\(\ds \order x\) \(=\) \(\ds 2 n - 1\)
\(\ds \leadsto \ \ \) \(\ds x^{2 n - 1}\) \(=\) \(\ds e\)
\(\ds \leadsto \ \ \) \(\ds x^{2 n}\) \(=\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds \paren {x^n}^2\) \(=\) \(\ds x\)
\(\ds \leadsto \ \ \) \(\ds \exists y \in G: \, \) \(\ds y^2\) \(=\) \(\ds x\) where $x^n$ is precisely that $y$

Hence the result.

$\blacksquare$


Warning

It is completely false to say:

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

if and only if:

the order $\order x$ is odd

An order $2$ element in $C_4$ refutes the converse.

This mistake can arise by supposing that this:

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

implies:

$\exists n \in \N: \paren {x^n}^2 = x$

The second step can only be used if every $x$ can be expressed in the terms of $y^2$.


Sources