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    Subgroup proof

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    If |G|=p^n, where p is a prime, show that G has subgroups G_0, G_1, . . ., G_n with 1= G_0 <= G_1 <= . . . <= G_n =G such that [G_i : G_{i-1}] = p, 1 <= i <= n.

    I need a rigorous, detailed proof of this to study please.

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    Solution Preview

    We know that if |G| is divisible by p, then G has a subgroup of order p.
    Now |G| = p^n. Let G_n = G.
    Since |G| ...

    Solution Summary

    This provides an example of proving given subgroups.