ABSTRACT ALGEBRA
1. (a) Write two distinct equivalence relations on 5 the set \(S=\{a, b, c\}\), Justifying your answer.
Answer:
Two distinct equivalence relations on the set \(S=\{a, b, c\}\) :
Equivalence Relation 1:
\[
R_1=\{(a, a),(b, b),(c, c)\}
\]
1. Reflexivity: Every element is related to itself. \((a, a),(b, b)\), and \((c, c)\) are in \(R_1\).
2. Symmetry: If \((x, y) \in R_1\), then \((y, x) \in R_1\). Since there are no other pairs in \(R_1\) besides the reflexive ones, symmetry holds.
3. Transitivity: If \((x, y) \in R_1\) and \((y, z) \in R_1\), then \((x, z) \in R_1\). Since there are no other pairs in \(R_1\) besides the reflexive ones, transitivity holds.
Equivalence Relation 2:
\[
R_2=\{(a, a),(b, b),(c, c),(a, b),(b, a),(a, c),(c, a),(b, c),(c, b)\}
\]
1. Reflexivity: Every element is related to itself. \((a, a),(b, b)\), and \((c, c)\) are in \(R_2\).
2. Symmetry: If \((x, y) \in R_2\), then \((y, x) \in R_2\). We have both \((a, b)\) and \((b, a)\), as well as \((a, c)\) and \((c, a)\), and \((b, c)\) and \((c, b)\) in \(R_2\).
3. Transitivity: If \((x, y) \in R_2\) and \((y, z) \in R_2\), then \((x, z) \in R_2\). Since all possible pairs are in \(R_2\), transitivity holds.
These two equivalence relations, \(R_1\) and \(R_2\), are distinct and satisfy reflexivity, symmetry, and transitivity on the set \(S=\{a, b, c\}\).
(b) Let \(G\) be a group and \(\mathrm{I}, \mathrm{K}\) be normal 2 subgroups of \(G\). Let \(H \cap K=\{e\}\). Then show that \(h k=k h\) for each \(h \in H, k \in K\).
Answer:
Let \(G\) be a group and \(H, K\) be normal subgroups of \(G\) such that \(H \cap K=\{e\}\). We need to show that \(h k=k h\) for each \(h \in H, k \in K\).
Since \(H\) and \(K\) are normal subgroups, they satisfy the following conditions:
1. \(g h g^{-1} \in H\) for all \(g \in G, h \in H\).
2. \(g k g^{-1} \in K\) for all \(g \in G, k \in K\).
Now, let’s take an arbitrary \(h \in H\) and \(k \in K\). We want to show that \(h k=k h\).
Consider the element \(g=h\), and apply condition \(1:\)
\[
h k h^{-1} \in K
\]
Now multiply on the right by \(h\) :
\[
\left(h k h^{-1}\right) h=h k\left(h^{-1} h\right)=h k \in K
\]
So, we have shown that \(h k \in K\).
Next, consider the element \(g=k\), and apply condition 2:
\[
k h k^{-1} \in H \text {. }
\]
Now multiply on the right by \(k\) :
\[
\left(k h k^{-1}\right) k=k h\left(k^{-1} k\right)=k h \in H
\]
So, we have shown that \(k h \in H\).
Since \(h k \in K\) and \(k h \in H\), the product \(h k\) and \(k h\) must both belong to the intersection of \(H\) and \(K\), i.e., \(H \cap K\).
As given, \(H \cap K=\{e\}\), so \(h k=k h=e\), which implies that \(h k=k h\) for each \(h \in\) \(H, k \in K\).
(c) Give an example, with justification, of a ring \(R\) and its ideals \(I\) and \(J\) such that \(IJ\neq I\cap J\) .
Answer:
Let’s consider the ring \(R=\mathbb{Z}[x]\), which is the set of polynomials with integer coefficients. We will find two ideals \(I\) and \(J\) such that \(I J \neq I \cap J\).
Let \(I=2 \mathbb{Z}[x]\), which is the set of all polynomials with even integer coefficients. This is an ideal of \(\mathbb{Z}[x]\) because:
1. \(0 \in I\) (the additive identity is in \(I\) ).
2. If \(f(x), g(x) \in I\), then \(f(x)-g(x) \in I\) (closed under subtraction).
3. If \(f(x) \in I\) and \(g(x) \in \mathbb{Z}[x]\), then \(f(x) g(x) \in I\) (closed under multiplication by any element of the ring).
Now, let’s define another ideal \(J=(x)\), which is the set of all polynomials with integer coefficients that are multiples of \(x\). This is also an ideal of \(\mathbb{Z}[x]\) because:
1. \(0 \in J\) (the additive identity is in \(J\) ).
2. If \(f(x), g(x) \in J\), then \(f(x)-g(x) \in J\) (closed under subtraction).
3. If \(f(x) \in J\) and \(g(x) \in \mathbb{Z}[x]\), then \(f(x) g(x) \in J\) (closed under multiplication by any element of the ring).
Now, let’s compute the product of the two ideals: \(I J\). If \(f(x) \in I\) and \(g(x) \in J\), then \(f(x) g(x) \in I J\) is a polynomial with even integer coefficients and is a multiple of \(x\). Thus, \(I J \subseteq 2 x \mathbb{Z}[x]\)
Next, let’s compute the intersection of the two ideals: \(I \cap J\). Since all elements of \(I\) have even coefficients and all elements of \(J\) are multiples of \(x\), their intersection contains polynomials that are both multiples of \(x\) and have even coefficients. Thus, \(I \cap J \subseteq 2 x \mathbb{Z}[x]\).
However, these two sets are not equal. In particular, the polynomial \(2 x \in I J\) because \(2 x=(2)(x)\), where \(2 \in I\) and \(x \in J\). On the other hand, \(x \in J\) but \(x \notin I\), so \(2 x \notin I \cap\) \(J\). Therefore, \(I J \neq I \cap J\).