Group theory/Lagrange's theorem/Section

Joseph-Louis Lagrange (1736 Turin - 1813 Paris)

Theorem

Let ${}G$ be a finite group, and let ${}H\subseteq G$ denote a subgroup

of ${}G$. Then the cardinality ${}{\#\left(H\right)}$ divides the cardinality ${}{\#\left(G\right)}$.

Proof

We consider the left cosets ${}gH:={\left\{gh\mid h\in H\right\}}$ for all ${}g\in G$. The mapping

$H\longrightarrow gH,h\longmapsto gh,$

is a bijection between ${}H$ and ${}gH$, so that all cosets have the same number of elements (namely ${}{\#\left(H\right)}$). The cosets form (as they are the equivalence classes) a partition of ${}G$. Hence, ${}{\#\left(G\right)}$ is a multiple of ${}{\#\left(H\right)}$.

$\Box$

Corollary

Let ${}G$ be a finite group, and let ${}g\in G$ denote an element. Then the order of ${}g$ divides the

group order.

Proof

Let ${}H$ be the subgroup generated by ${}g$. Due to fact, we have

${}\operatorname {ord} \,(g)=\operatorname {ord} \,(H)\,.$

Therefore, due to fact, this number divides the group order of ${}G$.

$\Box$

Definition

For a subgroup ${}H\subseteq G$, the cardinality of the (left- or right--)cosets is called the index of ${}H$ in ${}G$, denoted as

$\operatorname {ind} _{G}H.$

In the preceding definition, the number is in general to be understood as the cardinality of a set. However, the index is mainly used if it is finite, that is, if there are only finitely many cosets. This is, for finite ${}G$, always the case but can also hold for infinite ${}G$, as already the example ${}\mathbb {Z} n\subseteq \mathbb {Z}$, ${}n\geq 1$, shows. If ${}G$ is a finite group, and ${}H\subseteq G$ is a subgroup, then Lagrange's theorem yields the simple index formula

${}{\#\left(G\right)}={\#\left(H\right)}\cdot \operatorname {ind} _{G}H\,.$