Index of Trivial Subgroup is Cardinality of Group
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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:
- $\index G {\set e} = \dfrac {\order G} {\order {\set e} } = \dfrac {\order G} 1 = \order G$
$\blacksquare$
Sources
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.9$