Consider the following group under matrix multiplication:
\[ H = \left\{ \begin{bmatrix} 1 & p & q \\ 0 & 1 & r \\ 0 & 0 & 1 \end{bmatrix} : p, q, r \in \mathbb{R} \right\}. \]
Then the center of the group is isomorphic to
Consider the following group under matrix multiplication:
\[ H = \left\{ \begin{bmatrix} 1 & p & q \\ 0 & 1 & r \\ 0 & 0 & 1 \end{bmatrix} : p, q, r \in \mathbb{R} \right\}. \]
Then the center of the group is isomorphic to
Show Hint
- \( (\mathbb{R} \setminus \{0\}, \times) \)
- \( (\mathbb{R}, +) \)
- \( (\mathbb{R}^2, +) \)
- \( (\mathbb{R}, +) \times (\mathbb{R} \setminus \{0\}, \times) \)
The Correct Option is B
Solution and Explanation
Step 1: Understand the structure of the group.
The given group \( H \) consists of all upper triangular \( 3 \times 3 \) real matrices with 1s on the diagonal. Each element can be written as \\ \[ A(p, q, r) = \begin{bmatrix} 1 & p & q \\ 0 & 1 & r \\ 0 & 0 & 1 \end{bmatrix}, \] where \( p, q, r \in \mathbb{R}. \)
Step 2: Compute the product of two general elements.
Let \[ A(p, q, r) \text{ and } A(p', q', r') \in H. \] Then, under matrix multiplication, \[ A(p, q, r)A(p', q', r') = \begin{bmatrix} 1 & p + p' & q + q' + pr' \\ 0 & 1 & r + r' \\ 0 & 0 & 1 \end{bmatrix}. \]
Step 3: Find the center of the group.
The center \( Z(H) \) of a group consists of all elements that commute with every other element of the group. So, we need \[ A(p, q, r)A(p', q', r') = A(p', q', r')A(p, q, r) \forall \, p', q', r'. \] Step 4: Compute the commutation condition.
Using the multiplication formula: \[ A(p, q, r)A(p', q', r') = \begin{bmatrix} 1 & p + p' & q + q' + pr' \\ 0 & 1 & r + r' \\ 0 & 0 & 1 \end{bmatrix}, \] and \[ A(p', q', r')A(p, q, r) = \begin{bmatrix} 1 & p + p' & q + q' + p'r \\ 0 & 1 & r + r' \\ 0 & 0 & 1 \end{bmatrix}. \] For these to be equal, the (1,3) entries must be the same: \[ q + q' + pr' = q + q' + p'r \Rightarrow pr' = p'r. \] Step 5: Simplify the condition.
Since this must hold for all \( p', r' \in \mathbb{R} \), the only way is when \( p = 0 \) and \( r = 0 \). So, elements in the center have the form \[ A(0, q, 0) = \begin{bmatrix} 1 & 0 & q \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}. \] Step 6: Identify the structure of the center.
These elements depend only on \( q \in \mathbb{R} \), and multiplication of such matrices gives \[ A(0, q, 0)A(0, q', 0) = A(0, q + q', 0), \] showing the operation is addition on real numbers. Hence, \( Z(H) \cong (\mathbb{R}, +). \)
Final Answer: The center of the group is isomorphic to \( (\mathbb{R}, +). \)
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