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    A normal $p$-subgroup is contained in every Sylow $p$-subgroup

    E-AMD-4ZKQKRLH

    Area
    Algebra
    Topics
    Normal Subgroups, Sylow Theory, p-Groups
    Status
    draft

    Show that any normal \(p\)-subgroup of a finite group \(G\) is contained in every Sylow \(p\)-subgroup of \(G\).

    Solution

    Let \(N\trianglelefteq G\) be a normal \(p\)-subgroup and let \(P\in\Syl_p(G)\). Since \(N\trianglelefteq G\), the product \(NP\) is a subgroup of \(G\). Moreover, \[ |NP|=\frac{|N||P|}{|N\cap P|} \] is a power of \(p\), so \(NP\) is a \(p\)-subgroup containing \(P\).

    By maximality of the Sylow \(p\)-subgroup \(P\), we must have \(NP=P\). Therefore \(N\le P\). Since \(P\) was arbitrary, \(N\) is contained in every Sylow \(p\)-subgroup of \(G\).

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