June, 2012
ELECTIVE COURSE : MATHEMATICS
MTE-6 : ABSTRACT ALGEBRA
1. (a) Let \(S=\{(x, y) \mid x, y \in \mathbf{R}\}\). Show that \(S\) is a 4 ring with identity with the operations defined by :
\((x, y)+(u, v)=(x+u, y+v)\)
\((x, y) \cdot(\mathrm{u}, \mathrm{v})=(x \mathrm{u}-y \mathrm{v}, x \mathrm{v}+y \mathrm{u})\).
Answer:
To show that \(S\) is a ring with identity under the given operations, we need to verify the following properties:
1. Closure under addition
2. Associativity of addition
3. Commutativity of addition
4. Existence of an additive identity
5. Existence of additive inverses
6. Closure under multiplication
7. Associativity of multiplication
8. Distributive laws
9. Existence of a multiplicative identity
Let’s go through them step by step:
1. Closure under addition:
For any \((x, y),(u, v) \in S,(x, y)+(u, v)=(x+u, y+v)\), and since \(x+u, y+v \in\) \(\mathbf{R}\), the result is also in \(S\).
2. Associativity of addition:
For any \((x, y),(u, v),(a, b) \in S\),
\[
\begin{aligned}
& ((x, y)+(u, v))+(a, b) \\
& =(x+u, y+v)+(a, b) \\
& =((x+u)+a,(y+v)+b) \\
& =(x+(u+a), y+(v+b)) \\
& =(x, y)+(u+a, v+b) \\
& =(x, y)+((u, v)+(a, b)) .
\end{aligned}
\]
3. Commutativity of addition:
For any \((x, y),(u, v) \in S\),
\[
(x, y)+(u, v)=(x+u, y+v)=(u+x, v+y)=(u, v)+(x, y) .
\]
4. Existence of an additive identity:
The element \((0,0) \in S\) acts as the additive identity, since for any \((x, y) \in S\),
\[
(x, y)+(0,0)=(x+0, y+0)=(x, y)
\]
5. Existence of additive inverses:
For any \((x, y) \in S\), the element \((-x,-y) \in S\) is the additive inverse, since
\[
(x, y)+(-x,-y)=(x-x, y-y)=(0,0) .
\]
6. Closure under multiplication:
For any \((x, y),(u, v) \in S,(x, y) \cdot(u, v)=(x u-y v, x v+y u)\), and since \(x u-\) \(y v, x v+y u \in \mathbf{R}\), the result is also in \(S\).
7. Associativity of multiplication:
For any \((x, y),(u, v),(a, b) \in S\),
\[
\begin{aligned}
& ((x, y) \cdot(u, v)) \cdot(a, b) \\
& =(x u-y v, x v+y u) \cdot(a, b) \\
& =((x u-y v) a-(x v+y u) b,(x u-y v) b+(x v+y u) a) \\
& =(x(u a-v b)-y(v b+u a), x(u b+v a)+y(u a-v b)) .
\end{aligned}
\]
Now consider the product \((x, y) \cdot((u, v) \cdot(a, b))\) :
\[
\begin{aligned}
& (x, y) \cdot(u, v) \cdot(a, b) \\
& =(x, y) \cdot(u a-v b, u b+v a) \\
& =(x(u a-v b)-y(u b+v a), x(u b+v a)+y(u a-v b)) .
\end{aligned}
\]
Comparing the two products, we find that they are equal:
\[
((x, y) \cdot(u, v)) \cdot(a, b)=(x, y) \cdot((u, v) \cdot(a, b)) .
\]
8. Distributive laws:
For any \((x, y),(u, v),(a, b) \in S\),
\[
\begin{aligned}
& (x, y) \cdot((u, v)+(a, b)) \\
& =(x, y) \cdot(u+a, v+b)
\end{aligned}
\]
\[
\begin{aligned}
& =(x(u+a)-y(v+b), x(v+b)+y(u+a)) \\
& =(x u+x a-y v-y b, x v+x b+y u+y a) \\
& =(x u-y v, x v+y u)+(x a-y b, x b+y a) \\
& =(x, y) \cdot(u, v)+(x, y) \cdot(a, b) .
\end{aligned}
\]
Similarly,
\[
\begin{aligned}
& ((u, v)+(a, b)) \cdot(x, y) \\
& =(u+a, v+b) \cdot(x, y) \\
& =((u+a) x-(v+b) y,(u+a) y+(v+b) x) \\
& =(u x+a x-v y-b y, u y+a y+v x+b x) \\
& =(u x-v y, u y+v x)+(a x-b y, a y+b x) \\
& =(u, v) \cdot(x, y)+(a, b) \cdot(x, y) .
\end{aligned}
\]
Thus, the distributive laws hold.
9. Existence of a multiplicative identity:
The element \((1,0) \in S\) acts as the multiplicative identity, since for any \((x, y) \in S\),
\[
(x, y) \cdot(1,0)=(x \cdot 1-y \cdot 0, x \cdot 0+y \cdot 1)=(x, y) .
\]
Since all the required properties are satisfied, we can conclude that S is a ring with identity under the given operations.
Thus, \(S=\{(x, y) \mid x, y \in \mathbf{R}\}\) forms a ring with identity under the given operations.
(b) Define a relation \(S\) on \(\mathbf{R} \times \mathbf{R}\) by \(\left(x_{1}, y_{1}\right) S\) \(\left(x_{2}, y_{2}\right)\) iff \(3 x_{1}+y_{1}=3 x_{2}+y_{2}\). Show that \(S\) is an equivalence relation. What does each class represent geometrically?
Answer:
To show that \(S\) is an equivalence relation on \(\mathbf{R} \times \mathbf{R}\), we need to prove that it satisfies the following three properties:
1. Reflexivity: \(\forall\left(x_1, y_1\right) \in \mathbf{R} \times \mathbf{R},\left(x_1, y_1\right) S\left(x_1, y_1\right)\)
2. Symmetry: \(\forall\left(x_1, y_1\right),\left(x_2, y_2\right) \in \mathbf{R} \times \mathbf{R}\), if \(\left(x_1, y_1\right) S\left(x_2, y_2\right)\) then \(\left(x_2, y_2\right) S\left(x_1, y_1\right)\)
3. Transitivity: \(\forall\left(x_1, y_1\right),\left(x_2, y_2\right),\left(x_3, y_3\right) \in \mathbf{R} \times \mathbf{R}\), if \(\left(x_1, y_1\right) S\left(x_2, y_2\right)\) and \(\left(x_2, y_2\right) S\left(x_3, y_3\right)\), then \(\left(x_1, y_1\right) S\left(x_3, y_3\right)\)
Let’s go through them step by step:
1. Reflexivity:
For any \(\left(x_1, y_1\right) \in \mathbf{R} \times \mathbf{R}\), we have:
\[
3 x_1+y_1=3 x_1+y_1
\]
Thus, \(\left(x_1, y_1\right) S\left(x_1, y_1\right)\), so reflexivity holds.
2. Symmetry:
Suppose \(\left(x_1, y_1\right) S\left(x_2, y_2\right)\). Then:
\[
3 x_1+y_1=3 x_2+y_2
\]
By symmetry, this implies:
\[
3 x_2+y_2=3 x_1+y_1
\]
Thus, \(\left(x_2, y_2\right) S\left(x_1, y_1\right)\), so symmetry holds.
3. Transitivity:
Suppose \(\left(x_1, y_1\right) S\left(x_2, y_2\right)\) and \(\left(x_2, y_2\right) S\left(x_3, y_3\right)\). Then:
\(3 x_1+y_1=3 x_2+y_2\) and \(3 x_2+y_2=3 x_3+y_3\)
Adding both equations, we get:
\[
\left(3 x_1+y_1\right)+\left(3 x_2+y_2\right)=\left(3 x_2+y_2\right)+\left(3 x_3+y_3\right)
\]
Simplifying, we obtain:
\[
3 x_1+y_1=3 x_3+y_3
\]
Thus, \(\left(x_1, y_1\right) S\left(x_3, y_3\right)\), so transitivity holds.
Since reflexivity, symmetry, and transitivity are satisfied, \(S\) is an equivalence relation on \(\mathbf{R} \times \mathbf{R}\)
Let’s complete the geometric interpretation of the equivalence classes.
Each equivalence class can be represented by the equation:
\(3 x+y=k\), where \(k \in \mathbf{R}\) is a constant.
This equation represents a straight line with a slope of -3 and a \(y\)-intercept of \(k\). Each equivalence class corresponds to a line in the \(\mathbf{R}^2\) plane with a slope of -3 and a different \(y\) intercept.
So, the equivalence classes under the relation \(S\) represent a family of parallel lines with a slope of -3 in the \(\mathbf{R}^2\) plane. Each equivalence class contains all the points on one these parallel lines.
(c) Let \(G=\left\{\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \mid a, b, c, d, \in \mathbf{R}, a d \neq 0\right\}\).
Let \(H=\left\{\left[\begin{array}{ll}1 & b \\ 0 & 1\end{array}\right] \mid b \in R, b>0\right\}\). Is \(H a\) subgroup of G. Give reasons for your answer.
Answer:
To determine if \(H\) is a subgroup of \(G\), we need to verify that \(H\) satisfies the following three conditions:
1. The identity element of \(G\) is in \(H\).
2. If \(h_1, h_2 \in H\), then \(h_1 h_2 \in H\).
3. If \(h \in H\), then \(h^{-1} \in H\).
Let’s examine each of these conditions:
1. Identity element of \(G\) :
The identity element of \(G\) is the \(2 \times 2\) identity matrix:
\[
I=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]
\]
This matrix is in \(H\), since it can be obtained by setting \(b=0\). So, the identity element of \(G\) is in \(H\).
2. Closure under matrix multiplication:
Let \(h_1, h_2 \in H\), where
\[
h_1=\left[\begin{array}{cc}
1 & b_1 \\
0 & 1
\end{array}\right], \quad h_2=\left[\begin{array}{cc}
1 & b_2 \\
0 & 1
\end{array}\right]
\]
with \(b_1, b_2>0\). Then, their product \(h_1 h_2\) is given by:
\[
h_1 h_2=\left[\begin{array}{cc}
1 & b_1 \\
0 & 1
\end{array}\right]\left[\begin{array}{cc}
1 & b_2 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & b_1+b_2 \\
0 & 1
\end{array}\right]
\]
Since \(b_1, b_2>0\), their sum \(b_1+b_2>0\) as well. Therefore, \(h_1 h_2 \in H\), and closure under matrix multiplication holds.
3. Existence of inverses:
Let \(h \in H\), where
\[
h=\left[\begin{array}{ll}
1 & b \\
0 & 1
\end{array}\right]
\]
with \(b>0\). The inverse of \(h\) is given by:
\[
h^{-1}=\left[\begin{array}{cc}
1 & -b \\
0 & 1
\end{array}\right]
\]
However, this inverse does not belong to \(H\), since the condition \(b>0\) is not satisfied.
Therefore, the existence of inverses within \(H\) does not hold.
\[
\text { Since the third condition is not satisfied, we can conclude that } H \text { is not a subgroup of } G \text {. }
\]