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Math Help - Commutator subgroup and a center

  1. #1
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    Commutator subgroup and a center

    Hi, all

    C_{i}(G) denotes a center in central series and [G, G] denotes a commutator subgroup of G.

    My textbook says [C_{i}(G), C_{i}(G)] < C_{i-1}(G).

    Since C_{i-1}(G) is a normal subgroup of C_{i}(G) and C_{i}(G) / C_{i-1}(G) is abelian, it is plausible that C_{i-1}(G) contains [C_{i}(G), C_{i}(G)].

    What bothering me is that why [C_{i}(G), C_{i}(G)] is a proper subgroup of C_{i-1}(G) rather than being just a subgroup?

    I will appreciate if someone gives me an intuitive enlightment of the above one.
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  2. #2
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    Quote Originally Posted by aliceinwonderland View Post
    Hi, all

    C_{i}(G) denotes a center in central series and [G, G] denotes a commutator subgroup of G.

    My textbook says [C_{i}(G), C_{i}(G)] < C_{i-1}(G).

    Since C_{i-1}(G) is a normal subgroup of C_{i}(G) and C_{i}(G) / C_{i-1}(G) is abelian, it is plausible that C_{i-1}(G) contains [C_{i}(G), C_{i}(G)].

    What bothering me is that why [C_{i}(G), C_{i}(G)] is a proper subgroup of C_{i-1}(G) rather than being just a subgroup?

    I will appreciate if someone gives me an intuitive enlightment of the above one.
    I think your book might have a mistake and that it should be \leq rather than <.
    Consider, G=S_3 since Z(G) = \{ (1)\}.
    Then central series just consists of trivial subgroups.
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