Problem 18 - Algebra

Question

Let \(\mathbf{H}=\mathbf{H}_{G}\) be the special cohomology functor for a finite group \(G\). Show that: $$ \begin{array}{l} \mathbf{H}^{0}\left(I_{G}\right)=0 ; \mathbf{H}^{0}(Z) \sim \mathbf{H}^{1}(I)=\mathbf{Z} / n \mathbf{Z} \text { where } n=\\#(G) ; \\ \mathbf{H}^{\circ}(Q / Z)=\mathbf{H}^{1}(Z)=\mathbf{H}^{2}(I)=0 \\ \mathbf{H}^{\prime}(Q / Z)=\mathbf{H}^{2}(Z)=\mathbf{H}^{3}(I) \approx G^{\wedge}=\operatorname{Hom}(G, \mathbf{Q} / \mathbf{Z}) \text { by definition. } \end{array} $$

Step-by-Step Solution

Answer

We derived the following results for the special cohomology functor \(\mathbf{H}\) of a finite group \(G\): 1. \(\mathbf{H}^0(I_G)=0\) 2. \(\mathbf{H}^0(Z) \cong \mathbf{H}^1(I) = \mathbf{Z}/n\mathbf{Z}\) 3. \(\mathbf{H}^\circ(Q/Z) = \mathbf{H}^1(Z) = \mathbf{H}^2(I) = 0\) 4. \(\mathbf{H}'(Q/Z) = \mathbf{H}^2(Z) = \mathbf{H}^3(I) \cong G^\wedge = \operatorname{Hom}(G, \mathbf{Q}/\mathbf{Z})\) This is done by showing the properties of augmentation ideal, cohomology groups, center, and quotient group and applying the definitions of the dual group and Hom-set.

Step 1: Show \(\mathbf{H}^0(I_G)=0\)

The \(0\)-th cohomology group \(\mathbf{H}^0(I_G)\) is given by \(I_G / [I_G, G]\). Since the augmentation ideal \(I_G\) is generated by elements \(\{g - 1\mid g \in G\}\), we observe that \([I_G, G] = I_G\). Therefore, \(\mathbf{H}^0(I_G) = I_G/ I_G = 0\). This completes the first part of the problem.

Step 2: Show \(\mathbf{H}^0(Z) \cong \mathbf{H}^1(I) = \mathbf{Z}/n\mathbf{Z}\)

By definition, \(\mathbf{H}^0(Z)= Z/[Z,G]\) and \(\mathbf{H}^1(I)=I/NI\), with \(N\) being the norm ideal related to \(I_G\). Now, we observe that \([Z,G]=1\) because the center commutes with all elements in \(G\). As a result, we have \(\mathbf{H}^0(Z)=Z/1=Z\). Also, since \(G\) is finite, the norm acts trivially on \(I\), and so \(NI=I\). Therefore, \(\mathbf{H}^1(I)=I/NI=I/I=Z\). Hence, \(\mathbf{H}^0(Z)\cong \mathbf{H}^1(I)\). Since \(|Z|=n\), it follows that \(\mathbf{H}^1(I)=\mathbf{Z}/n\mathbf{Z}\).

Step 3: Show \(\mathbf{H}^\circ(Q/Z) = \mathbf{H}^1(Z) = \mathbf{H}^2(I) = 0\)

It can be shown that for any index \(i\), we have \(\mathbf{H}^i(Q / Z) = 0\) if \(i\) is not a multiple of \(|G|\), and \(\mathbf{H}^i(Q / Z) = Z\) if \(i\) is a multiple of \(|G|\). Since none of the indices in this step are multiples of \(|G|\), it follows that \(\mathbf{H}^\circ(Q/Z) = \mathbf{H}^1(Z) = \mathbf{H}^2(I) = 0\).

Step 4: Show \(\mathbf{H}'(Q/Z) = \mathbf{H}^2(Z) = \mathbf{H}^3(I) \cong G^\wedge = \operatorname{Hom}(G, \mathbf{Q}/\mathbf{Z})\)

In this step, we need to show the isomorphism for \(\mathbf{H}^3(I)\). Since \(3\) is a multiple of \(|G|\), we have \(\mathbf{H}'(Q / Z) = \mathbf{H}^2(Z) = \mathbf{H}^3(I) = Z\). Now we will consider the isomorphism of this cohomology group with the dual group of \(G\). By definition, \(G^\wedge = \operatorname{Hom}(G, \mathbf{Q}/\mathbf{Z})\). It is a well-known fact that the dual of a finite abelian group is isomorphic to the group itself. Thus, we have \(\mathbf{H}^3(I) \cong G^{\wedge} = \operatorname{Hom}(G, \mathbf{Q}/\mathbf{Z})\). By completing all four steps, we have shown all the required equalities and isomorphisms, and the exercise is complete.

Key Concepts

Cohomology Functor

The cohomology functor plays a pivotal role in understanding the properties of algebraic structures, particularly finite groups, in the realm of abstract algebra. It is essentially a sequence of functors \( \mathbf{H}^n \) from the category of groups to the category of abelian groups anchored on the integer \( n \). For a finite group \( G \), the \( n \)-th cohomology group \( \mathbf{H}^n(G) \) captures information about the group's structure and can provide insights into their extensions, representations, and action on other structures.

Specifically, in the exercise, the \( \mathbf{H}^0 \) functor reflects a fundamental property that relates to the augmentation ideal, demonstrating a foundational aspect of group cohomology: trivial cohomology at the zero level when applied to certain ideals like the augmentation ideal \( I_G \) of a group. Understanding how \( \mathbf{H}^n \) behaves for different \( n \) and various subgroup structures within \( G \) is key to unraveling the deeper algebraic properties it encapsulates.

Augmentation Ideal

The augmentation ideal \( I_G \) is a profound concept within the study of group rings and group cohomology. It is formed by considering the kernel of the augmentation map, which is the group homomorphism from the group ring \( \mathbb{Z}[G] \) to the integers \( \mathbb{Z} \), mapping each element of the group to \( 1 \). The elements of \( I_G \) are thus formal differences \( g - 1 \) where \( g \) ranges over the group elements.

The importance of the augmentation ideal is highlighted in the exercise, where we see that \( \mathbf{H}^0(I_G) \) vanishes. This is intuitive as the ideal itself is generated by elements that define the boundaries of cycles, revealing a cancellation in the zeroth cohomology group and underscoring the connection between algebraic topology and group theory.

Group Homomorphisms

Group homomorphisms are the cement that holds much of group theory together, serving as the maps between groups that preserve group structure and operations. Formally, a function \( f: G \rightarrow H \) between two groups is a homomorphism if for all elements \( a \) and \( b \) in \( G \) it holds that \( f(ab) = f(a)f(b) \).

In our exercise, the augmentation map, which sends each group element to \( 1 \), is a primary example of a group homomorphism. Interestingly, cohomology groups themselves can be thought of as arising from homomorphisms in the context of derived functors, where they measure the failure of exactness in sequences of abelian groups. In the case of our exercise, the norm map related to \( I_G \) and the map from \( G \) to \( \mathbb{Q}/\mathbb{Z} \) are critical homomorphisms that shape the cohomology groups in question.

Abelian Groups

In the universe of group theory, abelian groups, named after the mathematician Niels Henrik Abel, are the groups where the binary operation is commutative, meaning that \( a * b = b * a \) for all elements \( a \) and \( b \). This simple yet profound property has profound implications, allowing for more in-depth analysis and simpler computations within the group. In the context of cohomology, abelian groups often serve as the target groups for cohomology functors.

In the particular case of our exercise, the relevant abelian group is \( \mathbf{Z}/n\mathbf{Z} \) which emerges from considering the cohomology of the center \( Z \) of \( G \) and from the augmentation ideal \( I \) under the first cohomology functor. Additionally, the dual of a finite abelian group, a concept inextricably linked to the cohomology of finite groups, is used to express the isomorphism in the final step of our problem, showcasing their central role in group theory and homological algebra.