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Math Help - Characteristic groups

  1. #1
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    Characteristic groups

    I got stuck with two questions dealing with characteristic groups

    the first - G is a finite group characteristically simple and N is a minimal normal sub group in G (contains no proper normal subgroups of G). if H=N_1 \times N_2 \times \cdots \times N_k is a direct sum of isomorphic copies of N=N_1 where k is maximal, then it is normal in G.

    I can show that N is characteristically simple, and also all its isomorphic copies. I tried to show that if gN_i g^{-1} \cap N_j \neq \{ e \} then gN_i g^-1 \cap N_j = N_j, and show that either there is another copy of N that I can join to the sum, or H is normal.

    The problem is I can't see how I can use the fact that G is characteristically simple. I know that Z(G) and the commutator subgroups are characteristic in G but couldn't find what to do with it.


    The second question is to show that if N is normal in G then \Phi(N)\subseteq\Phi(G) where \Phi(A) is the intersection of all maximal groups in A. I thougt to use \Phi(G)\cap N and show that there is an intersection of some maximal groups in N that gives it. another thing is if H is maximal in G, and \Phi(N) is not contained in H then maybe H\Phi(G) will give a contradictory to the maximality of H (which is well defined bacuase \Phi(G) char N normal in G).

    any insights?
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  2. #2
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    Quote Originally Posted by Prometheus View Post

    the first - G is a finite group characteristically simple and N is a minimal normal sub group in G (contains no proper normal subgroups of G). if H=N_1 \times N_2 \times \cdots \times N_k is a direct sum of isomorphic copies of N=N_1 where k is maximal, then it is normal in G.
    this is a strange problem! the reason is that i think H=G if G is characteristically simple. that's because \alpha(N_1) is a copy of N_1 for any \alpha \in \text{Aut}(G). why? thus H_0=\prod_{\alpha \in \text{Aut}(G)} \alpha(N_1) \leq H.

    but H_0 is a non-trivial characterstic subgroup of G. thus H_0=G, because G is characteristically simple.



    The second question is to show that if N is normal in G then \Phi(N)\subseteq\Phi(G) where \Phi(A) is the intersection of all maximal groups in A.
    do you know this fact that the Frattini subgroup, \Phi(G), is the set of all the non-generators of G? if you know this, then your problem is not that hard. first note that if G has no maximal subgroup

    (this might happen if G is not finite), then by definition \Phi(G)=G, and there's nothing to prove. otherwise, if \Phi(N) \subseteq M, for any maximal subgroup M of G, then we're done. so suppose there

    exists a maximal subgroup M of G such that \Phi(N) \nsubseteq M. now since \Phi(N) is characterstic in N and N is normal in G, \Phi(N) is normal in G. thus \Phi(N)M is a subgroup of G properly containing M.

    thus \Phi(N)M=G, by maximality of M. therefore (M \cap N)\Phi(N)=N. why? that means <\{\Phi(N), M \cap N \} >=N and hence M \cap N=N, because the elements of \Phi(N) are non-generators.

    so we've proved that N =M \cap N \subseteq M, which gives us \Phi(N) \subseteq N \subseteq M. contradiction!
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  3. #3
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    in the first question, I meant that k is maximal in finding k-1 isomorphic copies of N=N_1 so that H=N_1 \times N_2 \times \cdots \times N_k is a subgroup of G. not all the copies of N are in there. (or maybe they are, but this needs a proof).

    any way, thanks for the second question
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  4. #4
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    Quote Originally Posted by Prometheus View Post
    in the first question, I meant that k is maximal in finding k-1 isomorphic copies of N=N_1 so that H=N_1 \times N_2 \times \cdots \times N_k is a subgroup of G. not all the copies of N are in there. (or maybe they are, but this needs a proof).

    any way, thanks for the second question
    well, the product of finitely many normal subgroups is obviously normal. for that we don't need so many conditions that are givem in your problem! that's why i think something is missing here!
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  5. #5
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    the isomorphism is between N and N_i. it is not an automorphism \alpha of G, and then taking \alpha(N) = N_i.
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