Problem 12
Question
Let \(G\) be the group of upper triangular matrices with non-zero diagonal elements. Let \(H\) be the subgroup consisting of those matrices whose diagonal element is 1 . (Actually prove that \(H\) is a subgroup). How would you describe the factor group \(G / H ?\)
Step-by-Step Solution
Verified Answer
The subgroup H consists of upper triangular matrices with diagonal elements equal to 1 and can be represented as:
\[H = \left\{ \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} \mid b \in \mathbb{R} \right\}\]
H is indeed a subgroup of G, as it satisfies the criteria for identity element, inverses, and closure. The factor group G/H consists of cosets of H in G and can be described as pairs of non-zero real numbers (a, c), representing the diagonal elements of the upper triangular matrices in G:
\[G/H \cong \mathbb{R}^* \times \mathbb{R}^*\]
1Step 1: Define the group G and subgroup H
The group G consists of upper triangular matrices with non-zero diagonal elements. An upper triangular matrix looks like this:
\[G = \left\{ \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \mid a,b,c \in \mathbb{R}, a,c \neq 0 \right\}\]
The subgroup H consists of those matrices whose diagonal elements are 1:
\[H = \left\{ \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} \mid b \in \mathbb{R} \right\}\]
2Step 2: Prove that H is a subgroup
To show that H is a subgroup of G, we need to check the following criteria:
1. Identity element: There exists an identity element e in H such that for any element h in H, e*h = h*e = h;
2. Inverses: For any element h in H, there exists an inverse h^(-1) also in H such that h*h^(-1) = h^(-1)*h = e;
3. Closure: For any two elements h1 and h2 in H, the product h1 * h2 is also in H.
Let's check these one by one:
1. Identity element: The identity element for matrix multiplication is the identity matrix:
\[\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\]
This matrix is in H since its diagonal elements are 1, so the identity element exists in H.
2. Inverses: Let's consider an element h in H:
\[h = \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}\]
To find its inverse, we want to find a matrix such that their product is an identity matrix:
\[h^{-1} = \begin{pmatrix} 1 & -b \\ 0 & 1 \end{pmatrix}\]
Multiplying h and h^(-1):
\[h * h^{-1} = \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} * \begin{pmatrix} 1 & -b \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\]
Hence, h^(-1) is also in H, and this holds for any element in H.
3. Closure: Let h1 and h2 be any two elements in H:
\[h1 = \begin{pmatrix} 1 & b1 \\ 0 & 1 \end{pmatrix}, h2 = \begin{pmatrix} 1 & b2 \\ 0 & 1 \end{pmatrix}\]
Now let's compute the product of h1 and h2:
\[h1 * h2 = \begin{pmatrix} 1 & b1 \\ 0 & 1 \end{pmatrix} * \begin{pmatrix} 1 & b2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & b1+b2 \\ 0 & 1 \end{pmatrix}\]
Since b1 and b2 are real numbers, their sum is also a real number, so (b1+b2) ∈ ℝ, ensuring that the resulting matrix is in H.
Hence, H is a subgroup of G since it meets all the criteria.
3Step 3: Describe the factor group G/H
The factor group G/H consists of the cosets of H in G. Let's consider a matrix g in G:
\[g = \begin{pmatrix} a & b \\ 0 & c \end{pmatrix}\]
Then the coset gH contains all elements obtained by multiplying g by elements of H:
\[gH = \left\{ g * h \mid h \in H \right\} = \left\{ \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} * \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix} \mid x \in \mathbb{R} \right\} = \left\{ \begin{pmatrix} a & b+ax \\ 0 & c \end{pmatrix} \mid x \in \mathbb{R} \right\}\]
The factor group G/H consists of these cosets:
\[G/H = \left\{ gH \mid g \in G \right\}\]
Since the coset space is defined by the values of a and c, we can describe the factor group G/H as a set of pairs of non-zero real numbers (a, c):
\[G/H \cong \mathbb{R}^* \times \mathbb{R}^*\]
Here, \(\mathbb{R}^*\) denotes the non-zero real numbers. In conclusion, the factor group G/H can be described as pairs of non-zero real numbers, representing the diagonal elements of the upper triangular matrices in G.
Key Concepts
Upper Triangular MatrixSubgroupCosetsReal Numbers
Upper Triangular Matrix
An upper triangular matrix is a type of square matrix where all the elements below the main diagonal are zero. This means the matrix has a form where non-zero elements only appear on or above the diagonal. For example:
This structure simplifies many calculations, such as matrix multiplication and finding determinants. Determinants, for example, can be found simply by multiplying the diagonal elements in an upper triangular matrix. This simplicity and structured constraints make upper triangular matrices valuable in various mathematical applications.
- Given a 2x2 upper triangular matrix: \[ \begin{pmatrix} a & b \ 0 & c \end{pmatrix} \]
This structure simplifies many calculations, such as matrix multiplication and finding determinants. Determinants, for example, can be found simply by multiplying the diagonal elements in an upper triangular matrix. This simplicity and structured constraints make upper triangular matrices valuable in various mathematical applications.
Subgroup
A subgroup is essentially a group within a group. It shares the properties of a group and fits completely within a larger group, maintaining the operations and identity of the overall larger group. To be a subgroup, a subset must fulfill certain conditions:
This containment allows us to form new concepts within the larger group such as quotients or factor groups.
- Identity element: The subgroup must contain the group's identity element.
- Closure: The operation defined for the group still applies within the subgroup. If you take two elements from the subgroup and perform a group operation (like multiplication), the result must also be in the subgroup.
- Inverses: Every element in the subgroup must have an inverse also within the subgroup.
This containment allows us to form new concepts within the larger group such as quotients or factor groups.
Cosets
Cosets are key players when we study the structure of groups, especially in forming factor groups. In simple terms, a coset is formed by taking an element from a group and multiplying it with all elements of a subgroup. This leads to two types of cosets if we use a group \(G\) and a subgroup \(H\):
- Left coset: Generated by a fixed element \(g\) in \(G\) as \(gH\), which means multiply \(g\) with every element in \(H\).
- Right coset: Created similarly but as \(Hg\), meaning we compute every element in \(H\) multiplied by \(g\).
Real Numbers
Real numbers are the broad class of numbers used in mathematics that can fill the entire number line. They include all the rational numbers (such as integers like -1, 0, 1, fractions like 1/2) and the irrational numbers (numbers that cannot be represented as simple fractions, like \(\pi\) and \(\sqrt{2}\)).
- Non-zero real numbers: These are real numbers excluding zero. This subset is crucial when we describe the factor group \(G/H\) because diagonal elements in upper triangular matrices must be non-zero to ensure invertibility.
Other exercises in this chapter
Problem 10
If \(N\) is a nilpotent \(n \times n\) matrix, show that \(I+N\) is invertible.
View solution Problem 11
Let \(R\) be the set of all upper triangular \(n \times n\) matrices \(\left(a_{i j}\right)\) with \(a_{i j}\) in some field \(k\), so \(a_{i j}=0\) if \(i>j .\
View solution Problem 13
Let \(R\) be the ring of \(n \times n\) matrices over a field \(k .\) Let \(L\) be the subset of matrices which are 0 except on the first column. (a) Show that
View solution Problem 14
Let \(F\) be any field. Let \(D\) be the subgroup of diagonal matrices in \(G L_{n}(F)\). Let \(N\) be the normalizer of \(D\) in \(G L_{n}(F)\). Show that \(N
View solution