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Question about normal subgroups/Lattice Isomorphism Theorem

  1. Dec 23, 2013 #1
    I was just brushing up on some Algebra for the past couple of days. I realize that the lattice isomorphism theorem deals with the collection of subgroups of a group containing a normal subgroup of G. Now, in general, if N is a normal subgroup of G, all of the subgroups of larger order than N do not necessarily contain N. But, I was wondering what dictates whether or not this is the case, or whether or not subgroups of larger order exist that don't contain the normal subgroup of smaller order besides Lagrange's Theorem and the various theorems which extend it, i.e. Cauchy's, Sylow's, Hall's?

    In other words, if N is normal in G, clearly, by Lagrange's Theorem, a subgroup H of G does not contain N if the order of N does not divide the order of H, which can be extended to if the respective orders of N and G are relatively prime, and so on, provided that such a group exists. One example of existence of such a group follows from Cauchy: if p is a prime dividing the order of G then there exists a subgroup H of order p, and since the order of N is less than p, it follows that the order of H and the order of N are relatively prime, so that their intersection is trivial. Similar cases arise from Sylow's, and Hall's theorems, and I understand that. But what I mainly want to know what structure-related or just basic group-specific properties dictate the existence of such groups, rather than order related properties.

    I mean, it is obvious in the case that I mentioned before that the product HN is a subgroup of G and the order of HN is the maximal case, the product of the orders of H and N, and moreover, HN contains N, so that an inclusive chain of subgroups exists from 1 to G through HN and N, and from the lattice isomorphism theorem, it is apparent that such chains determine the structure of G since every subgroup must contain a normal subgroup. So, it would seem that existence of such subgroups may depend on the structure. Maybe the isomorphism type of G?

    For instance, the example that I have been considering seems to suggest that if G is cyclic, the normal subgroups which are contained in all larger subgroups is the subgroup that is isomorphic to the direct product of cyclic groups of relatively prime orders corresponding to each factor in a "relatively prime factorization" of the order of G, while the normal subgroups which are not contained in all larger subgroups are the ones whose isomorphism type is a direct product of cyclic groups which does not have a cyclic group factor for each factor in some "relatively prime factorization" of the order of G. And also in this sense, the collection of cyclic groups as determined from a "relatively prime factorization" of G form a non-unique "basis" for G.

    But I'm asking how this extends to groups with other structure.

    Thanks in advance
  2. jcsd
  3. Jan 10, 2014 #2


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    I don't understand what your question is but is is easy to construct groups that contain subgroups that are disjoint from a normal subgroup. For instance take the Cartesian product of Z with Z. Each factor is normal and disjoint from the other. Is that your question?
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