ラグランジュの定理 (群論)
ラグランジュの定理の系(2) ― 有限群 G の任意の元 g の位数は群 G の位数 |G| を割り切る。
| ⟨ g ⟩ | | | G | ⟺ g | G | = e {\displaystyle |\langle g\rangle |{\bigg |}|G|\Longleftrightarrow g^{|G|}=e} .
{\displaystyle {\begin{aligned}a_{1}H&=\{a_{1}h_{1},a_{1}h_{2},a_{1}h_{3},\cdots ,a_{1}h_{m}\},\\a_{2}H&=\{a_{2}h_{1},a_{2}h_{2},a_{2}h_{3},\cdots ,a_{2}h_{m}\},\\a_{3}H&=\{a_{3}h_{1},a_{3}h_{2},a_{3}h_{3},\cdots ,a_{3}h_{m}\},\\\cdots \\a_{k}H&=\{a_{k}h_{1},a_{k}h_{2},a_{k}h_{3},\cdots ,a_{k}h_{m}\},\\G/H&=\{a_{1}H,a_{2}H,a_{3}H,\cdots ,a_{k}H\},\\G&=a_{1}H\cup a_{2}H\cup a_{3}H\cup \cdots \cup a_{k}H\ \ (i\neq j\Rightarrow a_{i}H\cap a_{j}H=\emptyset ).\end{aligned}}}
このとき、|H| = |a1H| = |a2H| = |a3H| = ... = |akH| = m となるので、|G| = km となる。k = |G/H| = [G : H], m = |H| となるので、
| G | = [ G : H ] ⋅ | H | . {\displaystyle |G|=[G:H]\cdot |H|.}
Q.E.D.