Problem 3 - Abstract Algebra: Theory and Applications

Question

Find all the subgroups of the quaternion group, \(Q_{8}\). Which subgroups are normal? What are all the factor groups of \(Q_{8}\) up to isomorphism?

Step-by-Step Solution

Answer

What are the factor groups of \(Q_8\) up to isomorphism? The distinct subgroups of the quaternion group, \(Q_8\), are: 1. \(\langle 1 \rangle = \{1\}\) 2. \(\langle -1 \rangle = \{-1, 1\}\) 3. \(\langle i \rangle = \{i, -i, 1, -1\}\) 4. \(\langle j \rangle = \{j, -j, 1, -1\}\) 5. \(\langle k \rangle = \{k, -k, 1, -1\}\) The normal subgroups among these are: 1. \(\langle 1 \rangle = \{1\}\) 2. \(\langle -1 \rangle = \{-1, 1\}\) The factor groups of \(Q_8\) up to isomorphism are \(Q_{8}\) itself and the Klein four-group.

Step 1: Write down all the elements of the quaternion group, \(Q_{8}\).

The quaternion group, \(Q_{8}\), has the following elements: \(Q_{8} = \{1, -1, i, -i, j, -j, k, -k\}\), where the multiplication rules are \(i^2 = j^2 = k^2 = ijk = -1\) and \(ij = k, jk = i, ki = j\).

Step 2: List the cyclic subgroups generated by each element of \(Q_{8}\).

We can list down the cyclic subgroups generated by each element of \(Q_8\) as follows: \begin{align*} \langle 1 \rangle &= \{1\} \\ \langle -1 \rangle &= \{-1, 1\} \\ \langle i \rangle &= \{i, -i, 1, -1\} \\ \langle j \rangle &= \{j, -j, 1, -1\} \\ \langle k \rangle &= \{k, -k, 1, -1\} \\ \langle -i \rangle &= \{-i, i, 1, -1\} \\ \langle -j \rangle &= \{-j, j, 1, -1\} \\ \langle -k \rangle &= \{-k, k, 1, -1\} \end{align*}

Step 3: Remove duplicate subgroups.

After removing duplicates, we are left with the following distinct subgroups: \begin{align*} \langle 1 \rangle &= \{1\} \\ \langle -1 \rangle &= \{-1, 1\} \\ \langle i \rangle &= \{i, -i, 1, -1\} \\ \langle j \rangle &= \{j, -j, 1, -1\} \\ \langle k \rangle &= \{k, -k, 1, -1\} \end{align*}

Step 4: Check for normal subgroups.

A subgroup N of a group G is normal if \(gNg^{-1} = N\) for all elements g in G. Among the subgroups listed above, the following subgroups are normal in \(Q_{8}\): \begin{align*} \langle 1 \rangle &= \{1\} \\ \langle -1 \rangle &= \{-1, 1\} \\ \end{align*}

Step 5: Find the factor groups up to isomorphism.

Since there are only two normal subgroups in \(Q_{8}\), namely \(\{1\}\) and \(\{-1, 1\}\), the factor groups of \(Q_{8}\) up to isomorphism are \(Q_{8} / \{1\}\) and \(Q_{8}/\{-1, 1\}\). For \(Q_{8} / \{1\}\), the factor group is isomorphic to \(Q_{8}\) itself. For \(Q_{8}/\{-1, 1\}\), the factor group is defined by the cosets of the normal subgroup \(\{1, -1\}\). These cosets are: \(\{1,-1\}\), \(\{i,-i\}\), \(\{j,-j\}\), and \(\{k,-k\}\) which is isomorphic to the Klein four-group (also denoted as V or \(Z_2 \times Z_2 \)), since it has 4 elements and every non-identity element has order 2. Thus, the factor groups of \(Q_{8}\) up to isomorphism are \(Q_{8}\) itself and the Klein four-group.

Key Concepts

Subgroups

A subgroup is essentially a smaller group within a larger group. For the quaternion group, denoted as \(Q_8\), we begin by identifying the elements of the group. \(Q_8\) consists of the elements \(\{1, -1, i, -i, j, -j, k, -k\}\). Each of these elements can generate a cyclic subgroup, which is a minimal subgroup containing that element.

When listing cyclic subgroups, it's essential to recognize that some elements, such as \(-i\), \(-j\), and \(-k\), generate subgroups that are identical to those generated by \(i\), \(j\), and \(k\) respectively. So, to avoid repetition, we only keep distinct subgroups. After this elimination, the significant subgroups are:

• \(\langle 1 \rangle = \{1\}\)
• \(\langle -1 \rangle = \{-1, 1\}\)
• \(\langle i \rangle = \{i, -i, 1, -1\}\)
• \(\langle j \rangle = \{j, -j, 1, -1\}\)
• \(\langle k \rangle = \{k, -k, 1, -1\}\)

These subgroups show how different subsets of \(Q_8\) can still retain the structure of a group.

Normal Subgroups

A normal subgroup is a special type of subgroup that remains invariant under conjugation by any element of the group. In simpler terms, if \(N\) is a normal subgroup of a group \(G\), then every element of \(G\) can "shuffle" the elements of \(N\) without changing the subgroup itself. Mathematically, this is expressed as \(gNg^{-1} = N\) for every \(g \in G\).

In the quaternion group \(Q_8\), we determine which of its subgroups are normal. Here, \(\langle 1 \rangle = \{1\}\) is trivially normal because it only contains the identity element, which does not change under group operations. Similarly, \(\langle -1 \rangle = \{-1, 1\}\) is also normal since conjugation by any element of \(Q_8\) does not result in elements outside of \{-1, 1\}.

• \(\langle 1 \rangle = \{1\}\)
• \(\langle -1 \rangle = \{-1, 1\}\)

These are the only normal subgroups of \(Q_8\), influencing possible factor groups derived from \(Q_8\).

Factor Groups

Factor groups, also known as quotient groups, emerge when a normal subgroup is used to partition the original group into distinct sets called cosets. Each coset serves as a single element in the new group. Factor groups essentially provide a way of simplifying the group's structure.

For \(Q_8\), the factor groups are determined through its normal subgroups. The first factor group \(Q_8 / \{1\}\) remains isomorphic to \(Q_8\) since the only subgroup is trivial. The second factor group, \(Q_8 / \{-1, 1\}\), results in cosets that significantly simplify the group structure, reflecting the Klein four-group structure. This is because:

• The cosets are \(\{1,-1\}, \{i,-i\}, \{j,-j\}, \{k,-k\}\).
• Each coset, just like each non-identity element in the Klein four-group, has an order of 2.

Consequently, the factor groups of \(Q_8\), up to isomorphism, are the quaternion group itself and the Klein four-group. This offers insight into the fundamental structure and properties of \(Q_8\).