Question

Consider the set of numbers 1,3,5,7 with multiplication (mod 8 ) as the law of combination. Write the multiplication table to show that this is a group. [To multiply two numbers (mod 8 ), you multiply them and then take the remainder after dividing by \(8 . \text { For example, } 5 \times 7=35 \equiv 3(\bmod 8) .]\) Is this group isomorphic to the cyclic group of order 4 or to the 4 's group?

Short answer

The set {1, 3, 5, 7} with multiplication (mod 8) forms a group. It is isomorphic to the 4's group (Klein four-group).

Step-by-step solution

- Define the Set and Operation

Consider the set of numbers \(\text{1, 3, 5, 7}\) with multiplication (mod 8) as the operation. This means to find the result of multiplying two numbers, multiply them and then take the remainder when the product is divided by 8.

- Create the Multiplication Table

Create the multiplication table for the set {1, 3, 5, 7} using multiplication (mod 8).

- Calculate Entries

Calculate each entry of the table: - \(1 \times 1 \equiv 1 (\bmod 8)\) - \(1 \times 3 \equiv 3 (\bmod 8)\) - \(1 \times 5 \equiv 5 (\bmod 8)\) - \(1 \times 7 \equiv 7 (\bmod 8)\) - \(3 \times 1 \equiv 3 (\bmod 8)\) - \(3 \times 3 \equiv 9 \rightarrow 9 \div 8 = 1 R 1 \rightarrow 9 \equiv 1 (\bmod 8)\) - \(3 \times 5 \equiv 15 \rightarrow 15 \div 8 = 1 R 7 \rightarrow 15 \equiv 7 (\bmod 8)\) - \(3 \times 7 \equiv 21 \rightarrow 21 \div 8 = 2 R 5 \rightarrow 21 \equiv 5 (\bmod 8)\) - \(5 \times 1 \equiv 5 (\bmod 8)\) - \(5 \times 3 \equiv 15 \rightarrow 15 \equiv 7 (\bmod 8)\) - \(5 \times 5 \equiv 25 \rightarrow 25 \div 8 = 3 R 1 \rightarrow 25 \equiv 1 (\bmod 8)\) - \(5 \times 7 \equiv 35 \equiv 3 (\bmod 8)\) - \(7 \times 1 \equiv 7 (\bmod 8)\) - \(7 \times 3 \equiv 21 \equiv 5 (\bmod 8)\) - \(7 \times 5 \equiv 35 \equiv 3 (\bmod 8)\) - \(7 \times 7 \equiv 49 \rightarrow 49 \div 8 = 6 R 1 \rightarrow 49 \equiv 1 (\bmod 8)\)

- Compile the Table

Use the results from Step 3 to compile the full multiplication table: | * | 1 | 3 | 5 | 7 ||---|---|---|---|---|| 1 | 1 | 3 | 5 | 7 || 3 | 3 | 1 | 7 | 5 || 5 | 5 | 7 | 1 | 3 || 7 | 7 | 5 | 3 | 1 |

- Check Group Properties

Verify if the set and operation form a group: 1. **Closure**: The table shows that the product of any two elements in the set is still within the set {1,3,5,7}. 2. **Associativity**: Multiplication is inherently associative. 3. **Identity**: The number 1 acts as the identity element since multiplying any number by 1 gives that number. 4. **Inverses**: Each element has an inverse in the set that results in the identity element (1) when multiplied. From the table, the inverses are: - \(1^{-1} = 1\) - \(3^{-1} = 3\) - \(5^{-1} = 5\) - \(7^{-1} = 7\)

- Identify Group Type

Compare the group's structure to known groups: The table corresponds neatly to the Klein four-group (also known as the 4's group), characterized by each element being self-inverse and the group containing four elements. The cyclic group of order 4, on the other hand, would have a different structure where there is an element that generates all other elements in the group.

Key concepts

modular arithmetic

Modular arithmetic is a system of arithmetic for integers where numbers wrap around after reaching a certain value, known as the modulus. In simpler terms, it deals with the remainder when one integer is divided by another. For example, in this problem, we use modulus 8, written as 'mod 8'. This means that when we multiply two numbers, we then take the remainder when the product is divided by 8. For instance, in our example, when we multiply 5 by 7, we get 35. Dividing 35 by 8 gives us a remainder of 3, so 5 times 7 mod 8 is 3. This is calculated as:\[ 35 \rightarrow 35 \ 8 = 4 \text {R} 3 \rightarrow 35 \ 8 \ 3 \ \quad \text; {mod } 8 \].Modular arithmetic is essential for understanding how operations such as addition, subtraction, and multiplication work in contexts where you want to limit the results to a range of numbers. This concept has wide applications from clock arithmetic to coding and cryptography.

multiplication table

The multiplication table is a simple tool to illustrate the results of multiplying any pair of elements in a set under a particular operation—in this case, under modular arithmetic with mod 8. Here, we are constructing a multiplication table for the set {1, 3, 5, 7}.To do this, you multiply each pair of elements and then reduce the result modulo 8. For instance:

• 1 × 3 = 3 (mod 8)
• 5 × 7 = 35 ≡ 3 (mod 8)

Using these calculations, we can form a table where the rows and columns are the elements of the set, and the cells contain the results of the modular multiplication. This makes it easy to see all the product pairs at a glance and helps to verify the group's properties, such as closure and finding inverses easily. It visually demonstrates how the group's operation works.

Klein four-group

The Klein four-group, also known as the 4's group or V4, is a well-known group in abstract algebra and group theory. It consists of four elements where each element is its own inverse, meaning that multiplying an element by itself yields the identity element. In the Klein four-group, the identity element is typically 1.To illustrate this with our set {1, 3, 5, 7} and multiplication mod 8, we observe the following properties:

• 1 × 1 ≡ 1 (mod 8)
• 3 × 3 ≡ 1 (mod 8)
• 5 × 5 ≡ 1 (mod 8)
• 7 × 7 ≡ 1 (mod 8)

This shows that each element is indeed its own inverse. Additionally, there is no single element that can generate all other elements of the group by repeated application of the operation. Instead, the combination of any two different elements results in one of the other elements in the set.

cyclic group

A cyclic group is a group that can be generated by a single element. This means every element of the group can be expressed as some power (or multiple, in the case of additive groups) of a particular element, known as a generator.In contrast to the Klein four-group, the cyclic group of order 4 has a very different structure. An example of a cyclic group of order 4 is the set {1, 2, 3, 4} under addition mod 4. In this group, 1 is the generator because:

• 1 + 1 = 2 (mod 4)
• 1 + 1 + 1 = 3 (mod 4)
• 1 + 1 + 1 + 1 = 4 ≡ 0 (mod 4)

All elements can be obtained from repeated addition of 1. For a group to be cyclic, there must be at least one element that can generate all the other elements through the group's operation. In our problem, it's clear from the multiplication table that none of the elements {1, 3, 5, 7} generates the entire set through repeated multiplication mod 8, hence our group is not cyclic.