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Algebra Proof Question

  1. Dec 6, 2007 #1


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    I have questions about this proof. First question is the italic part.

    A finite group G with no characteristic subgroups other than G and 1 is either simple or a direct product of isomorphic simple groups.

    Proof: Choose a minimal normal subgroup H of G whose order is minimal among all nontrivial normal subgroups. Write H = H_1, and consider all subgroups of G of the form [itex]H_1 x ... x H_n[/itex], where n >= 1, H_i is normal in G and H_i is isomorphic to H. Let M be such a subgroup of largest possible order. We show that M = G by showing that M char G; to see this, it suffices to show that [itex]\alpha(H_i)[/itex] is contained in M for every i and every automorphism [itex]\alpha[/itex] of G. Of course, [itex]\alpha(H_i)= H = H_1[/itex]. We show that [itex]\alpha(H_i)[/itex] is normal in G. If a is in G, then [itex]a = \alpha(b)[/itex] for some b in G, and [itex]a\alpha(H_i)a^{-1}[/itex] = [itex]\alpha(bH_{i}b^{-1})[/itex] is contained H_i since it is normal in G. If [itex]\alpha(H_i)[/itex] is not contained in M, then [itex]\alpha(H_i) n M[/itex] (n is intersect) is not contained in [itex]\alpha(H_i)[/itex] and [itex]|\alpha(H_i) n M| < |\alpha(H_i)| = H[/itex]. But [itex]\alpha(H_i) n M[/itex] is normal in G, so this implies that the order of that group is 1 because of the minimality of H. The subgroup <M,[itex]\alpha(H_i)[/itex]>=M x [itex]\alpha(H_i)[/itex] is a subgroup of the same type as M but of larger order, a contradiction. We conclude that M char G, and so M = G. Finally, H = H_1 must be simple: if N is a nontrivial normal subgroup of H, then N is a normal subgroup of M = H_1 x ... H_n = G, and this contradicts the minimal choice of H.


    Nevermind, I got it.

    You can read it for pleasure I guess. :smile:
    Last edited: Dec 6, 2007
  2. jcsd
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