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:
- $\exists y \in G: y^2 = x$
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
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: An Introduction to Groups: Exercise $15$