Question about normal subgroups/Lattice Isomorphism Theorem

In summary, the lattice isomorphism theorem deals with the collection of subgroups of a group containing a normal subgroup of G. The existence of subgroups of larger order that do not contain the normal subgroup of smaller order depends on the structure of the group and its isomorphism type. For example, in cyclic groups, the normal subgroups that are not contained in larger subgroups have an isomorphism type that does not have a cyclic group factor for each factor in a "relatively prime factorization" of the group's order. However, for groups with other structures, it is possible to construct subgroups that are disjoint from the normal subgroup.
  • #1
epr1990
26
0
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
 
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  • #2
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?
 

FAQ: Question about normal subgroups/Lattice Isomorphism Theorem

What is a normal subgroup?

A normal subgroup is a subgroup of a group that is invariant under conjugation by elements of the larger group. In other words, if every element of the larger group is mapped to a different element by conjugation with an element of the normal subgroup, then the subgroup is considered normal.

How is a normal subgroup related to the Lattice Isomorphism Theorem?

The Lattice Isomorphism Theorem states that there is a one-to-one correspondence between the normal subgroups of a group G and the subgroups of the quotient group G/N, where N is a normal subgroup of G. This means that the normal subgroups of G can be visualized as "shadows" of the subgroups of G/N.

Can a group have more than one normal subgroup?

Yes, a group can have multiple normal subgroups. In fact, every group has at least two normal subgroups: the trivial subgroup (containing only the identity element) and the group itself. Other normal subgroups may exist depending on the structure of the group.

How can the Lattice Isomorphism Theorem be applied in practice?

The Lattice Isomorphism Theorem is useful in many areas of mathematics, including group theory, ring theory, and algebraic geometry. It can be used to simplify the study of group properties, classify groups, and prove other theorems.

Is the Lattice Isomorphism Theorem a generalization of other theorems?

Yes, the Lattice Isomorphism Theorem is a generalization of other theorems, including the Third Isomorphism Theorem and the Fundamental Theorem of Finite Abelian Groups. It also has connections to other mathematical concepts, such as Galois theory and algebraic topology.

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