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

    P-EN5JE

    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 \[ \phi:H\times K\longrightarrow HK, \qquad \phi(h,k)=hk. \] Fix \(hk\in HK\). If \(x\in H\cap K\), then \[ \phi(hx,x^{-1}k)=hk. \] Conversely, if \(h_1k_1=hk\), then \[ h^{-1}h_1=kk_1^{-1}\in H\cap K. \] Writing this element as \(x\) gives \[ (h_1,k_1)=(hx,x^{-1}k). \] Hence every fiber of \(\phi\) has exactly \(|H\cap K|\) elements. Counting \(H\times K\) by fibers yields \[ |H||K|=|HK|\,|H\cap K|, \] so \[ |HK|=\frac{|H||K|}{|H\cap K|}. \]

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