Elementary Group Theory Solution Exercise: 3.1

Elementary Group Theory Solution 

                                                           Exercise: 3.1

 1.

a.

Soln:

(i) m = 3, n = 5

Given, m * n = m + n

So, 3 * 5 = 3 + 5 = 8

(ii) m = 2, n = - 5

So, 2 * (-5) = 2 + (-5) = 2 – 5 = - 3.

b.

(i) m = 3, n = 5

Given, m * n = m – n

So, 2 * (-5) = 2 + (-5) = 2 – 5 = - 3.

(ii) m = 2, n = - 5

Given, m + n = m – n.

So, 2 * (-5) = 2 – (-5) = 2 + 5 = 7.

 

c.

(i) m = 3, n = 5

Given, m * n = nm + m + n

So, 3 * 5 = 3 * 5 + 3 + 5 = 23.

(ii) m = 2, n = - 5

Given, m * n = mn + m + n

So, 2 * (-5) = 2 * (-5) + 2 + (-5) = - 10 + 2 – 5 = - 13.

 

2.

Soln:

a.

S = {-1,0,1}

Consider 1,1 ԑ S.

Here, a * b = a + b

So, 1 * 1 = 1 + 1 = 2 ∉  S.

Thus, a * b = a + b Is not a binary operation on S = {-1,0,1}

 

b.

S = {1,2,4}

Consider 2,4 ԑ S.

Here, a * b = a.b

So, 2 * 4 = 2 * 4 = 8 ∉  S.

Thus, a * b = abIs not a binary operation on S = {1,2,4}

 

c.

S = {2,4,6,8,10,…}

Here, a * b = a + b

We know that the addition of two even numbers is always an even number which belong to the set S.

So, a * b = a + b is a binary operation on the set S = {2,4,6,8,10,….}

 

d.

S = Set of integers.

Given, a * b = a – b

Since, the difference of two integers always yields an integer.

Thus, for all,a,b ԑ S, a * b = a – b ԑ S.

So, a * b = a – b is a binary operation on the set ‘S’ of integers.

 

3.

Soln:

Cayley’s table.

X	-1	1
-1	1	-1
1	-1	1

 

From the above table,

(-1) * (-1) = 1 ԑ S, (-1) * 1 = - 1 ԑ S.

1 * (-1) = - 1 ԑ S, 1 * 1 = 1 ԑ S.

So, multiplication is a binary operatio on set a S.

 

4.

Soln:

The set of positive integers is denoted by Z

a.

Any m, n ԑ Z+${\mathrm{Z}}^{+}$à m + n ԑ Z+${\mathrm{Z}}^{+}$.

So, Z+${\mathrm{Z}}^{+}$ is closed under addition.

 

b.

Any m,n ԑ Z+${\mathrm{Z}}^{+}$àmn = nm

So, Z+${\mathrm{Z}}^{+}$ is commutative under multiplication.

 

c.

Consider, 2,3,4 ԑ Z+${\mathrm{Z}}^{+}$.

2 – (3 – 4) = 2 – (-1) = 2 + 1 = 3.

And (2 – 3) – 4 = - 1 – 4 = - 5

Thus, 2 – (3 – 4) ≠ (2 – 3) – 4.

SO, Z+${\mathrm{Z}}^{+}$ is not associative under subtraction.

 

5.

Soln:

a.

For closure,

Consider 1,2 ԑ Z.

Then , m * n = 12$\frac{1}{2}$(m – n)

So, 1 * 2 = 12$\frac{1}{2}$(1 – 2) = −12$-\frac{1}{2}$∉ Z.

So, Z is not closed.

For Associative,

Consider 2,4,8 ԑ Z.

= 2 * (4 * 8) = 2 * {12(4−8)}$\left\{\frac{1}{2}\left(4-8\right)\right\}$

= 2 * (-2) = 12$\frac{1}{2}$ {2 – (-2)} = 2

And (2 * 4) * 8 = {12(2−4)}$\left\{\frac{1}{2}\left(2-4\right)\right\}$ * 8

= (-1) * 8 = 12$\frac{1}{2}$ {(-1) – 8} = 12$\frac{1}{2}$ (-9) = −92$-\frac{9}{2}$∉ Z.

Hence, the operation * defined on Z is now associative,

For Commutative,

For, m , n ԑ Z, m * n = 12$\frac{1}{2}$(m – n)

= −12$-\frac{1}{2}$(n – m) = - (n * m).

So, the operation * is not commutative on Z.

 

b.

For closure,

For, m,n ԑ Z, m * n = n ԑ Z.

So, the operation ‘*’ is closed on Z.

For associative,

Since, for m,n and p ԑ Z.

(m * n) * p = n * p = p ԑ Z.

And m * (n * p) = m * p = p ԑ Z.

So, (m * n) * p = m * (n * p)

So, the operation ‘*’ is associative.

For commutative.

Since, for m,n ԑ Z, m * n = n ԑ Z.

And n * m = m ԑ Z.

But m ≠ n; so the operation * is not commutative on Z/

 

c.

For closure

Since, for m,n ԑ Z, m * n = m + n + 1 ԑ Z.

So, * is closed on Z,

For associative

Since, for n,m,p ԑ Z.

(m * n) * p = (m + n + 1) * p

= (m + n + 1) + p + 1 = m + n + p + 2.

And m * (n * p) = m * (m + p + 1)

= m + n (n + p + 1) + 1 = m + n + p + 2.

So, (m * n) * p = m * (n * p), for all m,n,p ԑ Z.

So, * is associative on Z.

For commutative.

For m,n ԑ Z.

m * n = m + n + 1.

And n * m = n + m + 1 = m + n + 1

So, m * n = n * m, for all m,n ԑ Z.

So, * is commutative on Z.