Index of Trivial Subgroup is Cardinality of Group

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

Let $\set e$ be the trivial subgroup of $G$.

Then:
 * $\index G {\set e} = \order G$

where:
 * $\index G {\set e}$ denotes the index of $\set e$ in $G$
 * $\order G$ denotes the cardinality of $G$.

Proof
By definition of cardinality and the trivial subgroup:
 * $\order {\set e} = 1$

From Lagrange's Theorem:
 * $\index G {\set e} = \dfrac {\order G} {\order {\set e} } = \dfrac {\order G} 1 = \order G$