# Even Order Group has Order 2 Element

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## Theorem

Let $G$ be a group whose identity is $e$.

Let $G$ be of even order.

Then:

- $\exists x \in G: \order x = 2$

That is:

- $\exists x \in G: x \ne e: x^2 = e$

## Proof 1

In any group $G$, the identity element $e$ is self-inverse with Identity is Only Group Element of Order 1, and is the only such.

That leaves an odd number of elements.

Each element in $x \in G: \order x > 2$ can be paired off with its inverse, as $\order {x^{-1} } = \order x > 2$ from Order of Group Element equals Order of Inverse.

Hence there must be at least one element which has not been paired off with any of the others which is therefore self-inverse.

The result follows from Group Element is Self-Inverse iff Order 2.

$\blacksquare$

## Proof 2

This is a direct corollary of the stronger result Even Order Group has Odd Number of Order 2 Elements.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.12 \ \text{(a)}$ - 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.9$: Exercise $5.7$ - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 38 \delta$ - 1974: Thomas W. Hungerford:
*Algebra*... (previous) ... (next): $\S 1.1$: Exercise $14$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $6$: An Introduction to Groups: Exercise $10$ - 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): $\S 3.4$: Cyclic groups: Exercise $8$