← 2014 Paper 2

UPSC 2014 Maths Optional Paper 2 Q1a — Step-by-Step Solution

10 marks · Section A

Normal subgroups; quotient groups · Algebra · asked 2× in 13 yrs · Read the full method →

Question

Let $G$ be the set of all real $2\times 2$ matrices $\begin{bmatrix}x & y\\ 0 & z\end{bmatrix}$, where $xz\ne 0$. Show that $G$ is a group under matrix multiplication. Let $N$ denote the subset $\left\{\begin{bmatrix}1 & a\\ 0 & 1\end{bmatrix}:a\in\mathbb R\right\}$. Is $N$ a normal subgroup of $G$? Justify your answer.

Technique

Direct verification of group axioms; explicit conjugation calculation to check normality.

Solution

Part 1 — $G$ is a group

Closure. For $A=\begin{pmatrix}x_1&y_1\\0&z_1\end{pmatrix}$, $B=\begin{pmatrix}x_2&y_2\\0&z_2\end{pmatrix}$:

The product is upper-triangular with $(1,1)$ entry $x_1 x_2$ and $(2,2)$ entry $z_1 z_2$. Their product $x_1 x_2 z_1 z_2=(x_1 z_1)(x_2 z_2)\ne 0$. So $AB\in G$ ✓.

Associativity. Inherited from matrix multiplication (associative for all matrices).

Identity. $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}\in G$ (with $x=z=1,\,y=0$; $xz=1\ne 0$). And $IA=AI=A$ for all $A\in G$ ✓.

Inverse. For $A=\begin{pmatrix}x&y\\0&z\end{pmatrix}$ with $xz\ne 0$:

This is upper-triangular with $(1,1)\cdot(2,2)=1/(xz)\ne 0$, so $A^{-1}\in G$ ✓.

Hence $G$ is a group under matrix multiplication.

Part 2 — Is $N$ normal in $G$?

$N=\{n_a:a\in\mathbb R\}$ where $n_a=\begin{pmatrix}1&a\\0&1\end{pmatrix}$.

To check normality, compute $gn_a g^{-1}$ for general $g=\begin{pmatrix}x&y\\0&z\end{pmatrix}\in G$.

This is $n_{xa/z}\in N$. So $gNg^{-1}\subseteq N$ for every $g\in G$.

Answer