Order of Boolean Group is Power of 2

Theorem

Let $G$ be a Boolean group.

Let $\order G$ denote the order of $G$.

Then:

$\order G = 2^n$

where $n \in \Z_{\ge 0}$ is a positive integer.

Proof

The case where $n = 0$ is clear:

$\order {\set e} = 1$

and $e^2 = e$.

Aiming for a contradiction, suppose $\order G = m \times 2^k$ for some odd integer $m$.

Then $m$ itself has an odd prime $p$ as a integer (which may of course equal $m$ if $m$ is itself prime).

Then by Cauchy's Lemma (Group Theory) there exists $g \in G$ such that $\order g = p$.

Hence it is not the case that $g^2 = e$.

Hence $\order G$ has no prime factor which is odd.

The result follows.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $17$