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

    P-F6RAE

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

    Let \(G\) be a finite group, and let \(N \trianglelefteq G\) be a normal \(p\)-subgroup of \(G\) (for a prime \(p\)). Prove that \(N\) is contained in every Sylow \(p\)-subgroup of \(G\).

    Solution

    Let \(P\) be any Sylow \(p\)-subgroup of \(G\). Since \(N\trianglelefteq G\), the product \(NP\) is a subgroup of \(G\). Moreover, \[ |NP|=\frac{|N||P|}{|N\cap P|}, \] which is a power of \(p\). Thus \(NP\) is a \(p\)-subgroup of \(G\) containing the Sylow \(p\)-subgroup \(P\).

    By maximality of \(P\) among \(p\)-subgroups, \[ NP=P. \] Therefore \[ N\subseteq NP=P. \] Since \(P\) was arbitrary, \(N\) is contained in every Sylow \(p\)-subgroup of \(G\).

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