Index of Trivial Subgroup is Cardinality of Group

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

Let $\left\{{e}\right\}$ be the trivial subgroup of $G$.

Then:
 * $\left[{G : \left\{{e}\right\} }\right] = \left\vert{G}\right\vert$

where:
 * $\left[{G : \left\{{e}\right\} }\right]$ denotes the index of $\left\{{e}\right\}$ in $G$
 * $\left\vert{G}\right\vert$ denotes the cardinality of $G$.

Proof
By definition of cardinality and the trivial subgroup:
 * $\left\vert{\left\{{e}\right\} }\right\vert = 1$

From Lagrange's Theorem:
 * $\left[{G : \left\{{e}\right\} }\right] = \dfrac {\left\vert{G}\right\vert} {\left\vert{\left\{{e}\right\} }\right\vert} = \dfrac {\left\vert{G}\right\vert} 1 = \left\vert{G}\right\vert$