# 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}

$$(S',·) \ where \ S' = \{1,-1\}$$

Verify
• 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.

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={an:n di Z} is a subgroup of (G,*)
• The intersection of subgroups of G is also a subgroup of G.

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Group theory

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