MTL105: Algebra - Groups

Binary Operation

Let G be a set. A binary operation on G is a function that assigns each ordered pair of elements of G an element of G.

$${\color{Red} \circ} : G \times G \rightarrow G$$ $$(a, b) \mapsto a {\color{Red} \circ} b$$

for e.g. : addition, subtraction and multiplication operation on integers.

Group

Let G be a set, i.e. $G (\neq \phi)$ together with a binary operation ${\color{Red} \circ} : G \times G \rightarrow G$.
We say that G is a group under this operation if the foll. 3 conditions are satisfied:

1. Associativity: $a {\color{Red} \circ} (b {\color{Red} \circ} c) = (a {\color{Red} \circ} b) {\color{Red} \circ} c, \ \forall a, b, c \in G$
2. Identity: $\exists e \in G \ s.t. \ a {\color{Red} \circ} e = e {\color{Red} \circ} a = a, \ \forall a \in G$
3. Inverse: $\forall a \in G, \ \exists b \in G \ s.t. \ a {\color{Red} \circ} b = b {\color{Red} \circ} a = e$

Here, b is the inverse of a and e is known as the identity element.

The notation to denote that G is a group under the operation ${\color{Red} \circ}$ is $(G, {\color{Red} \circ})$.

Order of a group

It is the number of elements in the group denoted by $\left | G \right |$. If $\left | G \right |$ is finite, we say the group is finite, otherwise the group is infinite.

Some examples:

• The set of integers under ordinary multiplication is not a group sice inverse of every element doesn’t exist.
• Group of integers modulo n:
  Let n be a +ve int, and $a \neq 0$ such that $a = nq + r, 0 \leq r < n$. Then, $a \equiv r \ (mod \ n)$. The congruent modulo n is an equivalence relation and the set of its equivalence classes is known as $\mathbb{Z}_{n}$.

$$\mathbb{Z}_{n} = \left \{ \bar{0}, \bar{1}, \bar{2}, ... , \bar{(n-1)} \right \}$$

$\mathbb{Z}_{n}$ for $n \geq 1$ is a group under addition modulo n.

But, we cannot say that $\mathbb{Z}_{n} \setminus \left \{ 0 \right \}$ is a group under multiplication modulo n, since there can be some elements without any inverse.

$\mathbb{Z}_{n} \setminus \left \{ 0 \right \}$ is denoted by $\mathbb{Z}_{n}^{\star}$.

• $\mathbf{(\mathbb{Z}_{n}^{\star}, \cdot_{m})}$ where $\cdot_{m}$ represents multiplication modulo m, is a group when n is prime. So, if it is a group, can we say that n must be prime?
• General linear group of $2 \times 2$ matrices over $\mathbb{R}$:
  $G L ( 2 , \mathbb { R } ) = \left\{ \left[ \begin{array} { l l } { a } & { b } \\ { c } & { d } \end{array} \right] | \ a , b , c , d \in \mathbb{ R } , a d - b c \neq 0 \right\}$
  This is a group under the operation matrix multiplication.
• Set of integers under subtraction is not a group since it does not follow Associativity property.
• $\mathbf{U(n)}$ is defined to be a set of all positive integers less than n and relatively prime to n. It is a group under multiplication modulo n.

$$U(n) = \left \{ \bar{a} \in {\mathbb{Z}_{n}}^{\star} \mid gcd(a, n) = 1 \right \}$$

Abelian Group

Let $(G, {\color{Red}\circ})$ be a group. It is said to be Abelian iff:
$a {\color{Red}\circ} b = b {\color{Red}\circ} a, \ \forall \ a, b \in G$.

Question: Which element must always exist in a group?
Answer: The identity element.

The following theorems lead to some elementary properties of the groups:

Theorem 1: Uniqueness of the Identity

In a group G, there is only one identity element.

Proof: Let $(G, {\color{Red}\circ})$ be the group and let there be two identity elements $e_{1}$ and $e_{2}$

Then:

1. $$a {\color{Red}\circ} e_{1} = e_{1} {\color{Red}\circ} a = a, \ \forall \ a \in G$$
2. $$a {\color{Red}\circ} e_{2} = e_{2} {\color{Red}\circ} a = a, \ \forall \ a \in G$$

Using $a = e_{2}$ in equation 1, we get $e_{2} {\color{Red}\circ} e_{1} = e_{2}$ and using $a = e_{1}$ in equation 2, we get $e_{2} {\color{Red}\circ} e_{1} = e_{1}$, which means $e_{1}$ and $e_{2}$ are both equal.

Since typing handwritten notes takes a lot of time, directly read handwritten notes for remaining part of this lecture here. (Pages 1-2)