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    Sylow's first theorem: existence of prime-power subgroups

    T-WRMBM

    Every power of $p$ dividing $\abs G$ is the order of a subgroup, and every $p$-subgroup lies in a Sylow $p$-subgroup.Suggest a slogan

    Let \(G\) be a finite group and \(p\) a prime, and write \(\abs G=p^a m\) with \(p\nmid m\). Then \(G\) has a subgroup of order \(p^n\) for every \(0\leq n\leq a\); in particular, \(G\) has a Sylow $p$-subgroup, of order \(p^a\). Moreover, every \(p\)-subgroup of \(G\) is contained in a Sylow \(p\)-subgroup [@DF04].

    For \(a\geq1\) this strengthens Cauchy's theorem, which gives a subgroup of order \(p\).

    Guide Appearances

    Dependencies

    • T-XDMK2Cauchy's theoremuses
    • L-DJKXLFixed-point congruence for $p$-groupsuses
    • D-7TQ2MSylow $p\dash$subgroupuses

    Backlinks

    • T-EF2MZSylow's second theorem: Sylow subgroups are conjugateuses

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