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Discrete mathematics is the study of mathematical structures that can be considered "discrete rather than "continuous" (analogously to continuous functions)
Typology: Schemes and Mind Maps
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PROGRAM TITLE: Btec-Computing
UNIT TITLE: Unit 18 - Discrete Maths
ASSIGNMENT NAME: ASSIGNMENT 4- Abstract algebra.
SUBMISSION DATE: Dec 29, 2021
TUTORIAL LECTURER: Lưu Thị Hương Giang
STUDENT NAME: Nguyễn Bích Hạnh
Summative Feedback:
Internal verification:
LO4: Explore applicable concepts within abstract algebra.
Part 1 When considering the set of all the natural numbers ( ℕ), show whether the mathematical
operations of addition, subtraction, multiplication and division are:
(a)Associative
We have ∀ a,b,c ∈ ℕwe have the operation of addition association on the set of N natural numbers as:
a + (b + c) = (a + b) +c = a + b + c
We have ∀ a,b,c ∈ ℕwe have the operation of multiplication on the set of N natural numbers as:
a ¿ (b ¿ c) = a ¿ b ¿ c = (a ¿ b) ¿ c
We have: 7- (4 - 2) = 5 ≠ (7 - 4) -2 =1 we infer that the operation of subtraction does not involve the set
of N natural numbers.
We have: 27/ (9 / 3) = 9 ≠ (27 / 9) / 3 =1 we infer that the operation of division does not involve the set of
N natural numbers.
(b) Commutative.
We have ∀ a,b ∈ ℕwe have the operation of addition association on the set of N natural numbers as:
a + b = b + a
We have ∀ a,b ∈ ℕwe have the operation of multiplication on the set of N natural numbers as:
ab = ba
We have: 8 - 4= 4 ≠ -4 = 4 - 8 we infer that the operation of subtraction does not involve the set of N
natural numbers.
We have: 9/3= 3 ≠
=3 / 9 we infer that the operation of division does not involve the set of N natural
numbers.
Part 2
the identity element in an appropriate way.
We represent group G by using *. We show that group G cannot have elements of order 3. That is, if
b
3
=e then the activity is defined wrongly. We show that the number of possible common definitions of
the operation * on group G is less than 9 *. Or for example * could be defined in the following way:
e e a b c
a a e c b
b b c a e
c c b e a
The next table is for group of order 4. We consider only first row and first column, we obtain a table for
the group of order 1 and if we remain first two rows and columns we get a table of order 2. We get a table
to collect claim 1 and assuming that we keep the original two lines and parts we get a table to collect
claim 4 (see Lagrange's conjecture). Therefor we present a table for a gathering of third request
independently:
e e a b
a a b e
b b e a
a. State the Lagrange's theorem of group theory.
What is Lagrange’s theorem?
Lagrange’s theorem was given by Joseph-Louis Lagrange. Lagrange theorem states that in group theory,
for any finite group say G, the order of subgroup H (of group G) is the divisor of the order of G i.e.,
O(G)/O(H). The order of the group represents the number of elements. In this lesson, let ú discuss the
statement and proof of the Lagrange theorem in Group theory. Lagrange theorem is one of the important
theorems of abstract algebra. We will also have a look at the three lemmas used to prove this theorem
with the solved examples. The understanding of a coset is also essential before you can fully understand
the Lagrange theorem.
g
1
g
1
h: h ∈ H} = {
g
3
h: h ∈ H} =
g
3
Or
g
1
g
3
¿ H. That is,
g
1
g
3
, and
g
1
g
2
, g
3
G are arbitrary, is transitive.
To prove the lemma, we show that A
B and B
A. As A and B are arbitrarily labeled, it suffices to
show the former.
Let a ∈ A. For A ∩ B ≠ 0, there is a c ∈ {A ∩ B}. As A is an equivalence class of and both
a and c are in A, it follows that a c. But as a c, c ∈ B and B is an equivalence
class of , it follows that a ∈ B.
With this three’s complement, we proceed to the main result.
Let H be a subgroup of order n of a finite group G of degree m. Let us consider the cost analysis of G in
relation to H.
Let us now consider that each coset of aH consists of n different elements.
Given H = {
h
1
h
2
h
n
}, then a
h
1
, a
h
2
, …, a
h
n
are n distinct elements of aH.
Suppose, a
h
i
= a
h
j
h
i
h
j
is the cancellation law of G.
Since G is a finite group, the left discrete coset will aslo be finite, say p. So, the sum of the elements of all
the cosets is np which is equal to the sum of the elements of G.
Hence, m = np → p =
m
n
This shows that n, the degree of H, is a divisor of m, the order of the finite group G. We also see that the
index p is also a divisor of the order of the group.
Therefore, it is proved, |G| = |H|.
c. Discuss whether a group H with order 6 can be a subgroup of a group with order 13 or not. Clearly
state the reasons.
Non empty subset H of a group G is a subgroup of G if and only if:
a
− 1
For example:
G = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7} / order 13
H = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7} / order 6
Property follows:
− 1
So H with order 6 is a subgroup G with order 13.
Part 3 1. Given that 𝐺 = {𝑎 ∈ ℝ| 𝑎 ≠ −1} and 𝑎 ∗ 𝑏 = 𝑎 + 𝑏 + 𝑎𝑏, show that (𝐺, ∗) is indeed a group.
We the very first thing to check is that this set G is operation ∗. We have a,b ∈ G then a ∗ b is always a
real number, but we need to see that a ∗ b ≠ −1. To see this, let us simply write down what it means for
a,b ∈ G to have a ∗ b = −1. This means
a + b + ab = − 1 ,
or equivalently
ab + a + b + 1 = 0_._
But:
ab + a + b +1 = ( a +1)( b +1) so if this is zero, then one of a +1 or b +1 is zero. This is impossible since a,b ∈
G so neither are −1. This G is really closed under ∗. Now let us check that ∗ is associative. Let a,b,c ∈
G. We have:
( a ∗ b )∗ c = ( a + b + ab )∗ c = a + b + ab + c +( a + b + ab ) c = a + b + ab + c + ac + bc + abc.
We also have:
a ∗ ( b ∗ c ) = a ∗ ( b + c + bc ) = a + b + c + bc + a ( b + c + bc ) = a + b + c + bc + ab + ac + abc.
we see that both expressions are equal. So ∗ is associative. Have 0 is an identity element for ( G, ∗). We
have
0 ∗ a = 0 + a + 0 a = a
and
a ∗ 0 = a + 0 + a 0 = a,
so 0 is an identity element.
