Search the corpus

    $H\operatorname{char} K\operatorname{char} G$ implies $H\operatorname{char} G$

    E-AMD-4M7MSCEI

    Area
    Algebra
    Topics
    Automorphisms, Normal Subgroups, Subgroups
    Status
    draft

    Show that \(H \operatorname{char} K \operatorname{char} G \implies H \operatorname{char} G\).

    So “characteristic” is a transitive relation for subgroups.

    Solution

    Let \(\phi\in\Aut(G)\). Since \(K\operatorname{char}G\), \[ \phi(K)=K, \] so the restriction \[ \phi|_K:K\longrightarrow K \] is an automorphism of \(K\). Since \(H\operatorname{char}K\), every automorphism of \(K\) preserves \(H\); hence \[ \phi(H)=(\phi|_K)(H)=H. \] As this holds for every \(\phi\in\Aut(G)\), we have \[ H\operatorname{char}G. \] Thus characteristicity is transitive.

    Source Collections