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    $C_H(x)=H\cap C_G(x)$ for $H\leq G$

    E-AMD-UDJALSZY

    Area
    Algebra
    Topics
    Centralizers and Normalizers, Subgroups
    Status
    draft

    Show that for \(H\leq G\), \(C_H(x) = H \intersect C_G(x)\).

    Solution

    1. An element \(h\) lies in \(C_H(x)\) if and only if it lies in \(H\cap C_G(x)\).

    By definition, \[ h\in C_H(x) \quad\Longleftrightarrow\quad h\in H\text{ and }hx=xh. \] Also, \[ h\in C_G(x)\quad\Longleftrightarrow\quad h\in G\text{ and }hx=xh. \] Since \(H\le G\), for \(h\in H\) the condition \(h\in C_G(x)\) is exactly \(hx=xh\). Therefore \[ h\in C_H(x) \quad\Longleftrightarrow\quad h\in H\cap C_G(x). \]

    2. Hence \(C_H(x)=H\cap C_G(x)\).

    The two subsets have the same elements by 1.

    Source Collections