Question

If \(\mathcal{A}\) and \(\mathcal{B}\) are two groups then their direct product, \(A \times \mathcal{B}\), is defined to be the set of ordered pairs \((X, Y)\), with \(X\) an element of \(\mathcal{A}, Y\) an element of \(\mathcal{B}\) and multiplication given by \((X, Y)\left(X^{\prime}, Y^{\prime}\right)=\left(X X^{\prime}, Y Y^{\prime}\right)\). Prove that \(\mathcal{A} \times \mathcal{B}\) is a group. Denote the cyclic group of order \(n\) by \(\mathcal{C}_{n}\) and the symmetry group of a regular \(n\)-sided figure (an \(n\)-gon) by \(\mathcal{D}_{n}-\) thus \(\mathcal{D}_{3}\) is the symmetry group of an equilateral triangle, as discussed in the text. (a) By considering the orders of each of their elements, show (i) that \(\mathcal{C}_{2} \times \mathcal{C}_{3}\) is isomorphic to \(\mathcal{C}_{6}\), and (ii) that \(\mathcal{C}_{2} \times \mathcal{D}_{3}\) is isomorphic to \(\mathcal{D}_{6}\). (b) Are any of \(\mathcal{D}_{4}, \mathcal{C}_{8}, \mathcal{C}_{2} \times \mathcal{C}_{4}, \mathcal{C}_{2} \times \mathcal{C}_{2} \times \mathcal{C}_{2}\) isomorphic?

Short answer

Proved \(\mathcal{A} \times \mathcal{B}\) is a group. \(\mathcal{C}_2 \times \mathcal{C}_3\) is isomorphic to \(\mathcal{C}_6\) and \(\mathcal{C}_2 \times \mathcal{D}_3\) to \(\mathcal{D}_6\). None of \(\mathcal{D}_4, \mathcal{C}_8, \mathcal{C}_2 \times \mathcal{C}_4, \mathcal{C}_2 \times \mathcal{C}_2 \times \mathcal{C}_2\) are isomorphic.

Step-by-step solution

- Define Group Properties

To prove that \(\mathcal{A} \times \mathcal{B}\) is a group, we need to show that it satisfies the group properties: closure, associativity, identity, and inverses.

- Closure

Take any two elements \((X, Y)\) and \((X', Y')\) in \(\mathcal{A} \times \mathcal{B}\). The product is \((X, Y)(X', Y') = (XX', YY')\). Since \(\mathcal{A}\) and \(\mathcal{B}\) are groups, \((XX', YY')\) is also in \(\mathcal{A} \times \mathcal{B}\). Thus, the set is closed under the group operation.

- Associativity

For any \((X, Y), (X', Y'), (X'', Y'') \in \mathcal{A} \times \mathcal{B}\), we need to show \((X, Y) \left((X', Y') (X'', Y'')\right) = \left((X, Y)(X', Y')\right)(X'', Y'')\). Because \(\mathcal{A}\) and \(\mathcal{B}\) are associative, \((X(X'X''), Y(Y'Y'')) = ((XX')X'', (YY')Y'')\). Therefore, \(\mathcal{A} \times \mathcal{B}\) is associative.

- Identity Element

An identity in \(\mathcal{A} \times \mathcal{B}\) is \((e_A, e_B)\), where \(\mathcal{A}\) has identity element \(\mathcal{e_A}\) and \(\mathcal{B}\) has identity element \(\mathcal{e_B}\). For any \((X, Y)\), \((X, Y)(e_A, e_B) = (Xe_A, Ye_B) = (X, Y)\). So, \((e_A, e_B)\) is the identity for \(\mathcal{A} \times \mathcal{B}\).

- Inverse Element

In \(\mathcal{A} \times \mathcal{B}\), the inverse of \((X, Y)\) is \((X^{-1}, Y^{-1})\). Since \(\mathcal{A}\) and \(\mathcal{B}\) are groups, \((X^{-1}, Y^{-1})\) is an element of \(\mathcal{A} \times \mathcal{B}\). Therefore, \((XX^{-1}, YY^{-1}) = (e_A, e_B)\), confirming it is an inverse.

- Isomorphism \(\mathcal{C}_2 \times \mathcal{C}_3 \to \mathcal{C}_6\)

The order of \(\mathcal{C}_2\) is 2 and the order of \(\mathcal{C}_3\) is 3. Their product group \(\mathcal{C}_2 \times \mathcal{C}_3\) has elements of order LCM of 2 and 3, which is 6. Thus, \(\mathcal{C}_2 \times \mathcal{C}_3\) is isomorphic to \(\mathcal{C}_6\).

- Isomorphism \(\mathcal{C}_2 \times \mathcal{D}_3 \to \mathcal{D}_6\)

The group \(\mathcal{D}_3\) has order 6 and is the symmetry group of an equilateral triangle. Considering \(\mathcal{C}_2 \times \mathcal{D}_3\), the resulting group represents the symmetries of a regular hexagon (6 sides), \(\mathcal{D}_6\). Therefore, \(\mathcal{C}_2 \times \mathcal{D}_3\) is isomorphic to \(\mathcal{D}_6\).

- Checking for Isomorphisms Among Given Groups

The orders of \(\mathcal{D}_4\), \(\mathcal{C}_8\), \(\mathcal{C}_2 \times \mathcal{C}_4\) and \(\mathcal{C}_2 \times \mathcal{C}_2 \times \mathcal{C}_2\) are calculated. Each has a distinct number of elements and does not match in structure. Thus, \(\mathcal{D}_4, \mathcal{C}_8, \mathcal{C}_2 \times \mathcal{C}_4\) and \(\mathcal{C}_2 \times \mathcal{C}_2 \times \mathcal{C}_2\) are not isomorphic.

Key concepts

direct product of groups

The direct product of two groups, say \(\text{\mathcal{A}}\) and \(\text{\mathcal{B}}\), is formed by taking ordered pairs of their elements. An element in this product, written as \(\(X, Y\)\), includes an element \(\text{X}\) from \(\text{\mathcal{A}}\) and an element \(\text{Y}\) from \(\text{\mathcal{B}}\). The operation is performed component-wise: if you have two elements \(\((X, Y)\)\) and \(\((X', Y')\)\), their product is \(\((XX', YY')\)\). This means the product in \(\text{\mathcal{A}}\) is combined with the product in \(\text{\mathcal{B}}\). To check if \(\text{\mathcal{A}} \times \mathcal{B}\) is a group, we confirm:

• Closure: The operation keeps results within the set.
• Associativity: Grouping of operations doesn't affect the result.
• Identity Element: An element exists that leaves others unchanged when used in the operation.
• Inverse Element: For each element, there's another that brings it back to the identity.

cyclic groups

A cyclic group is one where all elements can be generated by repeatedly applying a group operation to a single element, called a generator. For instance, if \(\text{\mathcal{C}_{n}}\) is a cyclic group of order \(n\), this means applying the group's operation to the generator \('g'\) will produce all elements in the group: \(\text{e, g, g}^2, ..., g^{n-1}\). Here \(e\) is the identity element.
This aspect makes cyclic groups very structured and predictable. For example, the cyclic group \(\text{\mathcal{C}_6}\) contains six elements and can be completely generated by its smallest element (other than the identity). A notable property is that the direct product of cyclic groups could itself be cyclic. Specifically, \(\text{\mathcal{C}_2} \times \text{\mathcal{C}_3}\) is isomorphic to \(\text{\mathcal{C}_6}\) because their least common multiple (LCM) is 6.

isomorphism

In group theory, an isomorphism is a way to show two groups are structurally the same, even if their elements or operations look different. Formally, an isomorphism is a bijective function that maps one group to another and preserves their group operations. If groups \(\mathcal{G}\) and \(\mathcal{H}\) are isomorphic, denoted as \(\mathcal{G} \cong \mathcal{H}\), they have the same structure.
The exercise provides two isomorphism examples:

• Cyclic Groups: \(\mathcal{C}_2 \times \mathcal{C}_3 \cong \mathcal{C}_6\). The product of orders 2 and 3 results in a group with order 6.
• Symmetry Groups: \(\mathcal{C}_2 \times \mathcal{D}_3 \cong \mathcal{D}_6\). Here, the symmetry operations of \(\mathcal{D}_6\) include those of \(\mathcal{D}_3\) plus an additional symmetry from \(\mathcal{C}_2\).

These show how different looking groups share intrinsic properties.

symmetry groups

Symmetry groups, denoted as \(\mathcal{D}_n\), represent the symmetries of an \(n\)-sided polygon, including rotations and reflections. The symmetry group \(\mathcal{D}_n\) is therefore crucial in geometry and group theory.
For example, \(\mathcal{D}_3\) represents the symmetries of an equilateral triangle, while \(\mathcal{D}_6\) represents those of a hexagon. These groups combine operations such as rotations and mirror reflections that map the polygon onto itself.

• Order of \(\mathcal{D}_n\): It is \(2n\), due to \(n\) rotations and \(n\) reflections.
• Direct Product in Symmetry: Combining reflection groups like \(\mathcal{C}_2\) with \(\mathcal{D}_3\) results in \(\mathcal{D}_6\), showing shared properties of larger symmetry groups.

Understanding symmetry groups helps in many areas, from crystallography to solving puzzles like Rubik's cube.