Normal Subgroups and Factor Groups

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Introduction

In abstract algebra, particularly in group theory, the concept of a group being the product of its subgroup and their factor group is a fundamental idea. This concept is closely related to the notion of a normal subgroup and the construction of the factor group. In this article, we will explore the conditions under which a group G is isomorphic to the direct product of a subgroup H and the factor group G/H.

Normal Subgroups and Factor Groups

A normal subgroup H of a group G is a subgroup that is invariant under conjugation by members of G. In other words, for any g in G and h in H, the element g^(-1)hg is also in H. The factor group G/H is the set of all left cosets of H in G, with the group operation defined as (gH)(g'H) = (gg')H. The factor group G/H is a group under this operation, and it is denoted by G/H.

The Direct Product of a Subgroup and a Factor Group

The direct product of a subgroup H and a factor group G/H is a group that consists of all ordered pairs (h, gH), where h is an element of H and gH is an element of G/H. The group operation is defined as (h1, g1H)(h2, g2H) = (h1h2, g1g2H). The direct product H × (G/H) is a group under this operation.

Conditions for Isomorphism

To determine when a group G is isomorphic to the direct product of a subgroup H and the factor group G/H, we need to find the conditions under which the following isomorphism holds:

H \times (G/H) \cong G

This isomorphism implies that there exists a bijective homomorphism between the two groups. In other words, there exists a one-to-one correspondence between the elements of the two groups that preserves the group operation.

Theorem 1: Isomorphism between H × (G/H) and G

Let G be a group and H a normal subgroup of G. Then, the following isomorphism holds:

H \times (G/H) \cong G

if and only if H is a subgroup of the center of G, denoted by Z(G).

Proof

Suppose that H is a subgroup of Z(G). Let φ: H × (G/H) → G be the map defined by φ(h, gH) = gh. We need to show that φ is a bijective homomorphism.

First, we show that φ is a homomorphism. Let (h1, g1H) and (h2, g2H) be elements of H × (G/H). Then,

φ((h1, g1H)(h2, g2H)) = φ((h1h2, g1g2H)) = g1g2h1h2

= g1h1g2h2 = φ(h1, g1H)φ(h2, g2H)

Therefore, φ is a homomorphism.

Next, we show that φ is bijective. Let g be an element of G. Then, g = gh for some h in H. Therefore, φ(h, gH) = gh = g. This shows that φ is surjective.

To show that φ is injective, suppose that φ(h1, g1H) = φ(h2, g2H). Then, gh1 = gh2 for some h in H. Since H is a subgroup of Z(G), we have h1 = h2. Therefore, (h1, g1H) = (h2, g2H), which shows that φ is injective.

Conversely, suppose that H × (G/H) is isomorphic to G. Then, there exists a bijective homomorphism φ: H × (G/H) → G. Let h be an element of H. Then, φ(h, eH) = h, where e is the identity element of G. Since φ is a homomorphism, we have φ(h, gH) = gh for some g in G. Therefore, h is in the center of G, which implies that H is a subgroup of Z(G).

Theorem 2: Isomorphism between H × (G/H) and G

Let G be a group and H a normal subgroup of G. Then, the following isomorphism holds:

H \times (G/H) \cong G

if and only if G is a direct product of H and a subgroup K of G/H.

Proof

Suppose that G is a direct product of H and a subgroup K of G/H. Then, G = HK, where H ∩ K = {e}. Let φ: H × (G/H) → G be the map defined by φ(h, gH) = hg. We need to show that φ is a bijective homomorphism.

First, we show that φ is a homomorphism. Let (h1, g1H) and (h2, g2H) be elements of H × (G/H). Then,

φ((h1, g1H)(h2, g2H)) = φ((h1h2, g1g2H)) = h1g1h2g2

= (h1g1)(h2g2) = φ(h1, g1H)φ(h2, g2H)

Therefore, φ is a homomorphism.

Next, we show that φ is bijective. Let g be an element of G. Then, g = hg for some h in H and gH in G/H. Therefore, φ(h, gH) = hg = g. This shows that φ is surjective.

To show that φ is injective, suppose that φ(h1, g1H) = φ(h2, g2H). Then, h1g1 = h2g2 for some h in H. Since H ∩ K = {e}, we have h1 = h2. Therefore, (h1, g1H) = (h2, g2H), which shows that φ is injective.

Conversely, suppose that H × (G/H) is isomorphic to G. Then, there exists a bijective homomorphism φ: H × (G/H) → G. Let h be an element of H. Then, φ(h, eH) = h, where e is the identity element of G. Since φ is a homomorphism, we have φ(h, gH) = hg for some g in G. Therefore, G = HK, where H ∩ K = {e}.

Conclusion

In conclusion, a group G is isomorphic to the direct product of a subgroup H and the factor group G/H if and only if H is a subgroup of the center of G or G is a direct product of H and a subgroup K of G/H. This result provides a characterization of the conditions under which a group can be expressed as the product of a subgroup and a factor group.

References

[1] Artin, E. (1954). Galois Theory. University of Notre Dame Press.
[2] Dummit, D. S., & Foote, R. M. (2004). Abstract Algebra. John Wiley & Sons.
[3] Lang, S. (2002). Algebra. Springer-Verlag.
[4] Rotman, J. J. (1995). An Introduction to the Theory of Groups. Springer-Verlag.
Q&A: When is a Group the Product of its Subgroup and their Factor Group? ====================================================================

Q: What is the significance of a group being the product of its subgroup and their factor group?

A: A group being the product of its subgroup and their factor group is a fundamental concept in abstract algebra, particularly in group theory. It provides a way to decompose a group into smaller subgroups and factor groups, which can be used to study the properties of the group.

Q: What are the conditions under which a group G is isomorphic to the direct product of a subgroup H and the factor group G/H?

A: A group G is isomorphic to the direct product of a subgroup H and the factor group G/H if and only if H is a subgroup of the center of G or G is a direct product of H and a subgroup K of G/H.

Q: What is the center of a group?

A: The center of a group G, denoted by Z(G), is the set of all elements in G that commute with every element in G. In other words, an element g is in the center of G if and only if gh = hg for all h in G.

Q: What is a direct product of two subgroups?

A: A direct product of two subgroups H and K of a group G is a subgroup HK of G, where H ∩ K = {e}. In other words, the direct product of two subgroups is a subgroup that consists of all products of elements from the two subgroups.

Q: How can we determine if a group G is a direct product of a subgroup H and a subgroup K of G/H?

A: To determine if a group G is a direct product of a subgroup H and a subgroup K of G/H, we need to check if G = HK, where H ∩ K = {e}. This can be done by showing that every element of G can be written as a product of an element from H and an element from K.

Q: What are some examples of groups that are isomorphic to the direct product of a subgroup and their factor group?

A: Some examples of groups that are isomorphic to the direct product of a subgroup and their factor group include:

The group of integers under addition, which is isomorphic to the direct product of the subgroup of even integers and the factor group of odd integers.
The group of 2x2 invertible matrices over the real numbers, which is isomorphic to the direct product of the subgroup of diagonal matrices and the factor group of upper triangular matrices.

Q: What are some applications of the concept of a group being the product of its subgroup and their factor group?

A: The concept of a group being the product of its subgroup and their factor group has many applications in abstract algebra, particularly in group theory. Some examples include:

The study of group representations, which involves the decomposition of a group into smaller subgroups and factor groups.
The study of group actions, which involves the decomposition of a group into smaller subgroups and factor groups to study the action of the group on a set.
The study of group homomorphisms, which involves the decomposition of a group into smaller subgroups and factor groups to study the homomorphisms between groups.

Q: What are some common mistakes to avoid when working with groups that are isomorphic to the direct product of a subgroup and their factor group?

A: Some common mistakes to avoid when working with groups that are isomorphic to the direct product of a subgroup and their factor group include:

Assuming that a group is a direct product of a subgroup and their factor group without checking the conditions.
Failing to check if the subgroup and the factor group are indeed subgroups of the group.
Failing to check if the subgroup and the factor group are indeed isomorphic to the direct product of the subgroup and their factor group.

Q: What are some resources for learning more about groups that are isomorphic to the direct product of a subgroup and their factor group?

A: Some resources for learning more about groups that are isomorphic to the direct product of a subgroup and their factor group include:

Textbooks on abstract algebra, such as "Abstract Algebra" by David S. Dummit and Richard M. Foote.
Online resources, such as the Wikipedia article on group theory.
Research papers on group theory, such as "Group Theory" by Steven H. Weintraub.