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    $|HK|=|H||K|/|H\cap K|$

    E-AMD-BTVLG2S2

    Area
    Algebra
    Topics
    Cosets and Lagrange, Subgroups
    Status
    draft

    Let \(H, K \leq G\) a finite group, and without using the normalizers of \(H\) or \(K\), show that \(\abs{HK} = \abs{H} \abs{K}/\abs{H\intersect K}\).

    Solution

    Consider the surjection \[ \mu:H\times K\to HK,\qquad (h,k)\mapsto hk. \] Fix \(x=h_0k_0\in HK\). Then \[ hk=x=h_0k_0 \] if and only if \[ d:=h_0^{-1}h=k_0k^{-1}\in H\cap K. \] Equivalently, \[ (h,k)=(h_0d,d^{-1}k_0) \] for a unique \(d\in H\cap K\). Hence every fiber of \(\mu\) has cardinality \(|H\cap K|\).

    Therefore \[ |H||K|=|H\times K|=|HK|\,|H\cap K|, \] so \[ \boxed{|HK|=\frac{|H||K|}{|H\cap K|}}. \]

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