SC421: Introduction to Modern Algebra Assignment 1 Jan 15, 2025
DA-IICT, B.Tech, winter 2024
1. In the following determine whether the systems described are groups. If they are not,
point out which of the Group axioms fail to hold.
(a) G = set of all integers, a ∗ b = a − b.
soln:
This is not a group. The given operation is not associative.
a − (b − c) = (a − b) + c ̸= (a − b) − c. Also there is no identity element in Z w.r.t
the given operation.
a − e = a ⇒ e = 0, and e − a = a ⇒ e = 2a.
(b) G = set of all positive integers, a ∗ b = ab the usual product of integers.
soln:
This is not a group because the multiplicative inverse of every positive integers
doesn’t exist.
(c) G = set of all rational numbers with odd denominators, a ∗ b = a + b the usual
addition.
soln:
This is a group. The non-trivial property to verify is closure. Others are trivial.
let a = pq , b = rs ∈ G. Then a + b = ps+qr
qs
. qs is odd since q and s are odd and it
cannot have an even factor.
(d) Let S be a set. Which of the following is a group?
(i) ⟨P(S), ∪⟩, (ii) ⟨P(S), ∩⟩, (iii) ⟨P(S), ∆⟩, where A∆B = A ∪ B\A ∩ B
soln:
(i) For A, B ∈ P(S), A ∪ B ∈ P(S). Hence P(S) is closed under ∪.
Set union is an associative operation.
ϕ ∪ A = A ∪ ϕ = A, ∀A ∈ P(S). So ϕ acts as the identity element with respect to
∪ in P(S).
However for any A ∈ P(S) we can’t find a subset of S whose union with A will
give the empty set.
Hence ⟨P(S), ∪⟩ is not a group.
(ii) P(S) is closed under ∩.
∩ is associative.
The set S acts as the identity with respect to ∩ in P(S).
However for any A ∈ P(S) we can’t find a B ∈ P(S) such that A ∩ B = S.
Hence ⟨P(S), ∩⟩ is not a group.
(iii) A∆B is some subset of S. By definition P(S) is a collection of all possible
subsets of S. So A∆B ∈ P(S), ∀ A, B ∈ P(S). So P(S) is closed under ∆.
The operation ∆ is associative. This is easily seen using a venn diagram for three
subsets A, B and C. B∆C is represented by the shaded region of set B and set
1
C. Finally A∆(B∆C) is shown as the shaded region of the three sets. It is clear
that only those elements remain which occur exactly in one of the three sets or
in all the three sets.
It is clear from the figure we arrive at the same final set if we construct (A∆B)∆C.
A A
B C
B C
B C
The empty set ϕ acts as the identity element with respect to ∆.
A∆ϕ = A ∪ ϕ\A ∩ ϕ = A\ϕ = A
Every subset is the inverse of itself with respect to ∆.
A∆A = A ∪ A\A ∩ A = A\A = ϕ
which as mentioned above acts as the identity. So the set P(S) satisfies all the
group axioms under ∆. Hence it is a group under ∆.
x 0
2. Let G = { |x, y ∈ R}. Let the binary operation on G be the matrix multipli-
y 0
cation.
(a) Is G closed under multiplcation ?
soln:
x1 0 x2 0
Consider and in G.
y1 0 y2 0
x1 0 x2 0 x1 x2 0
= ∈G
y1 0 y2 0 y1 y2 0
So G is closed under multiplication.
2
E∈G
(b) Find a matrix such that AE = A, ∀A ∈ G}.
1 0
The matrix is a possibility for E. We may take b = 0.
b 0
(c) Is EA = A, ∀A ∈ G?
soln:
1 0 x 0 x 0
EA = =
b 0 y 0 bx 0
∴ EA ̸= A, ∀ A ∈ G
(d) Is G a group?
soln:
G is not a group since no element in G acts as the identity according to the Group
Axioms.
(e) Show that in a group G, a ∗ e = a, ∀a ∈ G =⇒ e ∗ a = a
soln:
Here we will assume all the group axioms to be valid except the axiom for the
identity element.
a∗e=a ∀ a∈G
∴ a−1 ∗ e = a−1
∴ a−1 ∗ e ∗ a = a−1 ∗ a = e
∴ a ∗ a−1 ∗ e ∗ a = a ∗ e = a
∴ e∗e∗a=e∗a=a
So a ∗ e = a =⇒ e ∗ a = a when G is a group.
3. (a) A permutation is a one-one mapping of a set of n natural numbers onto itself.
Justify that the set of all permutations on n objects denoted as Sn forms a group
under the operation of composition of permutations.
soln:
Consider the set of integers N = {1, 2, 3, ..., n}.
Let f and g be two permutations on N . Then f and g are one one and onto on
N.
So h = f ◦ g is also one one and onto. Hence Sn is closed under composition of
permutations.
The composition of functions is associative since
h ◦ (f ◦ g)(a) = h(f (g(a))) and (h ◦ f ) ◦ g(a) = h(f (g(a))) for any a ∈ N .
The identity permutation given by the function e(a) = a ∀a ∈ N is the identity
element w.r.t composition. If f is any permutation then f (e(a)) = f (a) and
e(f (a)) = f (a). So
f ◦e=e◦f =f
Every one-one and onto function has an inverse function. The inverse function
acts as the inverse permutation of a given permutation. Hence the collection of
all permutations Sn forms a group under the operation of composition.
1 2 3
(b) Consider the set {1, 2, 3}. Consider the two permutations σ = and
3 1 2
1 2 3
τ= of S3 the permutation group on three elements. Write down all
2 1 3
3
the elements of this group using composites of σ and τ . Work out the group table
.
soln
1 2 3 1 2 3 2 1 2 3
σ= ,τ= ,σ = ,
3 1 2 2 1 3 2 3 1
3 1 2 3 2 1 2 3 2 1 2 3
σ = = e = τ , στ = ,σ τ=
1 2 3 1 3 2 3 2 1
2 2
Therefore the six elements of S3 are {e, σ, σ , τ, στ, σ τ }.
Verify that στ = τ σ 2 and σ 2 τ = τ σ. The group table is
e σ σ2 τ στ σ 2 τ
2
e e σ σ τ στ σ 2 τ
σ σ σ2 e στ σ 2 τ τ
σ2 σ 2
e 2
σ σ τ τ στ
τ τ σ 2 τ στ e σ2 σ
2
στ στ τ σ τ σ e σ2
σ2τ 2
σ τ στ τ σ 2
σ e
4. (a) Prove that a group of order 4 have to be abelian.
soln:
Let G = {e, a, b, c} where e is the identity. The only non-trivial product where
we have to check commutativity is ab = c or permutations of the symbols a, b, c
in this equation.
(Note: If any of a, b, c is replaced by e in the above eqn. then the product has to
be commutative by the def of identity and inverse elements.)
If ba = a, or, b then either b = e or a = e which is not true.
If ba = e then ab = e which contradicts the above eqn.
Therefore the only possibility is ba = c ⇒ G is abellian.
(b) Write down all possible group table of order 4.
soln:
To make the group table we observe that if a and c pair up as inverses then b is
its own inverse.
The other possibility is that every element is its own inverse. Now based on the
discussion in part (a) and using the fact that each row and each column of a group
table contains an element exactly once, we can fill up the group table for the two
cases uniquely.
∗ e a b c ∗ e a b c
e e a b c e e a b c
a a b c e a a e c b
b b c e a b b c e a
c c e a b c c b a e
5. Let G = {(a, b)|a, b ∈ Q} and G∗ = G − {(0, 0)}.
4
(a) Let the binary operation in G∗ be defined as (a1 , b1 ) × (a2 , b2 ) = (a1 a2 , b1 b2 ). Is
G∗ along with this binary operation a group?
soln:
G∗ is not closed under the given group operation. Elements like (a, 0); a ̸= 0 and
(0, b); b ̸= 0 belongs to G∗ but (a, 0) × (0, b) = (0, 0) ̸∈ G∗ . Hence G∗ is not a
group under the given operation.
(b) Now let the binary operation be defined as (a1 , b1 )×(a2 , b2 ) = (a1 a2 +2b1 b2 , a1 b2 +
b1 a2 ). Is G∗ along with this binary operation a group?
soln:
We have to first verify whether G∗ is closed under the given operation. Let
(a1 , b1 ) and (a2 , b2 ) ∈ G∗ To ensure that (a1 , b1 ) × (a2 , b2 ) ∈ G∗ we must ensure
that (a1 , b1 ) × (a2 , b2 ) = (a1 a2 + 2b1 b2 , a1 b2 + b1 a2 ) ̸= (0, 0) .
Let if possible the product be (0, 0).
Then a1 a2 + 2b1 b2 = 0 and a1 b2 + b1 a2 = 0.
Suppose we know (a1 , b1 ). Then we can treat the above equations as a system of
two linear equations to evaluate (a2 , b2 ) as follows:
a1 2b1 a2 0
=
b1 a1 b2 0
This system will have a non-zero solution only if the determinant of the coeficient
matrix is 0, i.e , a21 − 2b21 = 0. √
This will happen only if a1 /b1 = 2 or√ a1 = b1 = 0.
G∗ doesn’t contain (0, 0) and a1 /b1 ̸= 2 since a1 , b1 ∈ Q. So the only possible
solution for (a2 , b2 ) is (0, 0). So the product of elements of G∗ is not (0, 0).
∴ G∗ is closed under the given operation.
The product is associative (check).
(1, 0) is the identity element (verify).
Let us find inverse of (a, b) ∈ G∗ . Let the inverse be (x, y).
Then ax + 2by = 1 and ay + bx = 0.
a 2b x 1
∴ =
b a y 0
a −b
The unique solution to these system of equations is x = a2 −2b 2 , y = a2 −2b2 since
2 2
a − 2b ̸= 0.
Hence G∗ is a group with this operation.
(c) Redefine G∗ so that G∗ along with the binary operation in part (a) forms a group.
soln:
If we modify the set G∗ as G∗ = {(a, b)|a, b ∈ Q∗ } then G∗ forms a group under
the multiplication as defined in part (a).
6. If G is a group of even order , prove that it has an element a ̸= e such that a2 = e.
soln:
Every element in a group has an inverse. An element and its inverse pairup to make
inverses of each other.
5
If we collect elements which are not its own inverse we get an even number of group
elements.
The identity in a group is its own inverse. So we have collected an odd number of
elements. So the number of elements left in the group which are its own inverse is odd.
a b
7. Let G be the set of all 2×2 matrices where a, b, c, d are integers modulo 2 and
c d
the determinant is non zero. Prove that G is a group of order 6 under the operation
of matrix multiplication.
soln:
The only
matrices withad − bc̸= 0 in Gare
1 0 1 1 1 0
a0 = , a1 = , a2 = ,
0 1 0 1 1 1
0 1 0 1 1 1
a3 = , a4 = , a5 = ,
1 0 1 1 1 0
a0 is the identity and matrix multiplication is associative.
We have to check for closure.
Determinant of a product of matrices is equal to the product of the determinants.
Since the determinant of any matrix in G is 1 or −1 ≡ 1(mod 2) determinant of any
of products will also be 1. So the product will be one of the above six matrices.
Now we perform the Group multiplication and verify that each element has an inverse.
a21 = a22 = a23 = a0 =1
1 0 1 0
a4 a5 = = (mod2) = a0 ,
2 1 0 1
1 2 1 0
a5 a4 = = (mod2) = a0
0 1 0 1
The Group table is
∗ a0 a1 a2 a3 a4 a5
a0 a0 a1 a2 a3 a4 a5
a1 a1 a0 a4 a5 a2 a3
a2 a2 a5 a0 a4 a3 a1
a3 a3 a4 a5 a0 a1 a2
a4 a4 a3 a1 a2 a5 a0
a5 a5 a2 a3 a1 a0 a4
