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    Lagrange's theorem: $|G|/|H|=[G:H]$ for finite $G$

    E-NNTQB

    Area
    Algebra
    Topics
    Cosets and Lagrange, Subgroups
    Status
    draft

    Show that if \(G\) is finite and \(H\le G\), then \[ [G:H]=\frac{|G|}{|H|}. \]

    Solution

    The left cosets of \(H\) partition \(G\). For each \(g\in G\), multiplication by \(g\) gives a bijection \[ H\longrightarrow gH,\qquad h\mapsto gh, \] so every coset has \(|H|\) elements. Since there are \([G:H]\) cosets, \[ |G|=[G:H]\,|H|. \] Therefore \[ [G:H]=\frac{|G|}{|H|}. \]

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