# 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:

$\index G {\set e} = \dfrac {\order G} {\order {\set e} } = \dfrac {\order G} 1 = \order G$

$\blacksquare$