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    Groups of prime order are cyclic

    E-AMD-CM4ZVMMU

    Area
    Algebra
    Topics
    Classification, Cosets and Lagrange, Cyclic Groups
    Status
    draft

    Show that every group of order \(p\) is cyclic.

    Solution

    Let \(|G|=p\) with \(p\) prime, and choose \(g\in G\) with \(g\ne e\). By Lagrange’s theorem, the order of \(g\) divides \(p\). Since \(g\) is not the identity, its order is not \(1\), so \[ |g|=p. \] Hence the cyclic subgroup \(\langle g\rangle\) has \(p\) elements. Since \(|G|=p\) as well, \[ \langle g\rangle=G. \] Thus every group of prime order is cyclic (and therefore isomorphic to \(\ZZ/p\ZZ\)).

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