# Subgroups

Let the set S be a group (S,*) with respect *, any non-empty subset S' of S is a **subgroup of S** if S' is itself a group with respect *.

Every subgroup (S',*) contains the same identity element (u) of group (S,*)

There are two types of subgroups

**Improper subgroups**

Subgroups (S ', *) in which the subset S' = {u} or S '= S are called improper subgroups.**Proper subgroups**

All other subgroups that are different from improper subgroups are called proper subgroups.

## Example

Let S={1,-1,i,-i} be a group (S,·) with respect to multiplication (·). The identity element of group (S,·) is u=1

A __proper subgroup__ with respect to multiplication is S'={1,-1}

$$ (S',·) \ where \ S' = \{1,-1\}$$

- a) The subset S 'is closed with respect to multiplication. For example $$ 1 \cdot (- 1) = - 1 $$ $$ (-1) \cdot (- 1) = 1 $$ $$ 1 \cdot 1 = 1 $$
- b) S' includes the identity element u = 1 of the group (S,⋅).
- c) All elements of S' have an opposite element in S' with respect to multiplication. For example $$ 1 \cdot 1 = 1 = u $$ $$ -1 \cdot (- 1) = 1 = u $$
- d) The associative property is satisfied in S' with respect to multiplication. All group properties are satisfied.

**Verify**

A __improper subgroup__ is S'={1}

$$ (S',·) \ \ \text{where} \ S' = \{ u \} = \{1\}$$

Another __improper subgroup__ is S'={1,-1,i,-i}

$$ (S',·) \ \ \text{where} \ S' = S = \{ 1,-1,i,-i \}$$

**Verify**. All group properties are satisfied in improper subgroups. The proof is trivial and omitted. For this reason improper subgroups are also called **trivial subgroups**.

These subsets S' __are not subgroups__ of (S,⋅)

The subsets {i, -i, -1}, {-1,i}, {i, -1}, {-i, -1}, {i, -i}, {i}, {-i}, {-1}, {} are not subgroups because they do not include the identity element u = 1 of the group (S,⋅).

The subset {1, i} is not subgroup because it's not closed with respect to multiplication. For example

$$ i \cdot i = i ^ 2 = -1 \notin \{1, i \} $$

The subset {1, -i} is not subgroup because it's not closed with respect to multiplication. For example

$$ -i \cdot (-i) = i ^ 2 = -1 \notin \{1, -i \} $$

The subset {1, i, -i} is not subgroup because it's not closed with respect to multiplication. For example

$$ i \cdot i = i ^ 2 = -1 \notin \{1, i, -i \} $$

The subset {1, i, -1} is not subgroup because there is no opposite element of i with respect to multiplication. For example

$$ i \cdot 1 = i \ne u \\ i \cdot -1 = -i \ne u \\ i \cdot i = i^2 = -1 \ne u $$

The subset {1, -i, -1} is not subgroup because there is no opposite element of -i with respect to multiplication. For example

$$ -i \cdot 1 = -i \ne u \\ -i \cdot -1 = i \ne u \\ -i \cdot (-i) = i^2 = -1 \ne u $$

## Theorems of subgroups

- Let a∈G be an element of group (G,*) then the set of all integral powers S={a
^{n}:n di Z} is a subgroup of (G,*) - The intersection of subgroups of G is also a subgroup of G.

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