# Statement(s) > [!NOTE] Theorem (Langrage) > Let $G$ be a [[Finite Group|finite group]] and $H$ a [[Subgroup|subgroup]]. Then $|G|=[G:H]\cdot |H|\tag{1}$where $|\;|$ denotes [[Cardinality|cardinality]] and $[G:H]$ denotes the [[Index of Subgroup|index]] of $H$ in $G.$ > **Note**: If $H$ is a normal subgroup, then we may write $\lvert G \rvert=\lvert G/H \rvert\cdot |H|$ instead of $(1)$ since $[G:H]=|G/H|=|G/H^l|=|G/H^r|$ denotes the cardinality of the [[Coset space|coset space]] $H$ in $G.$ # Proof(s) **Proof 1:** Let $g_{1}H,g_{2}H,\dots g_{m}H$ be the distinct left cosets of $H.$ Since [[Coset Space Partitions Subgroup]], by [[Cardinality of Union of Disjoint Sets]], $|G|=|g_{1}H|+|g_{2}H|+\dots+|g_{m}H|.$ Now by [[Subgroup and its Left Coset, Right Coset have Same Cardinality]], $|g_{1}H|=|g_{2}H|=\dots=|g_{m}H|$Hence $|G|=m \cdot |H|$ $\blacksquare$ # Application(s) **Consequences**: [[Order of Element of Finite Group Divides Order of The Group]]. **Examples**: # Bibliography