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    The centralizer of a subgroup is a subgroup

    P-5IV34

    Area
    Algebra
    Topics
    Centralizers and Normalizers, Subgroups
    Status
    draft
    • Show that the centralizer \(C_G(H)\) of a subgroup is again a subgroup.
    Solution

    Recall \[ C_G(H)=\{g\in G:gh=hg\text{ for every }h\in H\}. \]

    1. The identity lies in \(C_G(H)\).

    For every \(h\in H\), one has \(eh=he=h\).

    2. If \(x,y\in C_G(H)\), then \(xy^{-1}\in C_G(H)\).

    Fix \(h\in H\). Since \(y\) commutes with \(h\), so does \(y^{-1}\). Therefore \[ xy^{-1}h=xhy^{-1}=hxy^{-1}, \] using first that \(y^{-1}h=hy^{-1}\) and then that \(xh=hx\). Hence \(xy^{-1}\) commutes with every element of \(H\).

    3. Therefore \(C_G(H)\le G\).

    By 1 the centralizer is nonempty, and 2 verifies the one-step subgroup criterion.

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