Subgroups

A subgroup of a group (G,*) is a group (S,*) that is contained within (G,*), closed under the same operation * of the group (G,*), $$ *:S \rightarrow S $$ and satisfies all the group properties (identity element, inverse elements, associativity).

The set S of the subgroup (S,*) is a subset of the set G:

$$ S \subseteq G $$

Thus, a subgroup (S,*) restricts the operations * allowed in the group (G,*).

Note: A group (G,*) can have one or more subgroups, or none at all. Subgroups are called improper subgroups if the subset S contains the same elements as G (S=G) or only the identity element S={u} of the group (G,*). All other subgroups are referred to as proper subgroups.

Every subgroup (S,*) satisfies the associative property and includes the same identity element (u) as the group (G,*).

$$ u \in G, S $$

Additionally, every element of the subgroup has an associated inverse element:

$$ s, s^{-1} \in S $$

Note: From a more abstract perspective, subgroups are characterized by an injective homomorphism of groups $$ F: (S,*) \rightarrow (S',*) $$ As a result, subgroups inherit the same properties as the group they belong to. For example, if (G,*) is an abelian group, then the subgroup (S,*) is also abelian.

A Practical Example

Consider the multiplicative group (Q₀,·) consisting of the set of non-zero rational numbers Q₀=Q-{0}:

$$ (Q_0,·) $$

The identity element of the group is u=1:

$$ u=1 $$

The set of positive rational numbers Q⁺ is a multiplicative subgroup of (Q₀,·):

$$ (Q^+,·) $$

This is because:

Q⁺ is a subset of Q₀: $$ Q^+ \subseteq Q_0 $$
Q⁺ includes the same identity element as (Q₀,·), which is u=1: $$ u=1 \in Q^+ $$
Q⁺ is closed under multiplication. For any two elements a and b in Q⁺, the product a·b is always in Q⁺: $$ a \cdot b \in Q^+ \ \ \ \forall \ a,b \in Q^+ $$
Q⁺ satisfies the associative property of multiplication: $$ a \cdot (b \cdot c) = (a \cdot b) \cdot c \ \ \ \forall \ a,b,c \in Q^+ $$
Every element in Q⁺ has an inverse element within Q⁺. For example, the inverse of 2 is 1/2: $$ 2 \cdot \frac{1}{2} = 1 = u $$

Therefore, the set of positive rational numbers Q⁺ is a multiplicative subgroup (Q⁺,·) of the group (Q₀,·).

Example 2

Consider a group (A,·) consisting of a finite set A={1, -1} with the operation of multiplication (·):

$$ (A,\cdot) $$

Why is (A,·) a group? It qualifies as a group because it satisfies all the group properties. The set A includes an identity element: $$ e=1 $$ Every element in A has an inverse element within A: $$ 1 \cdot 1 = e $$ $$ -1 \cdot -1 = e $$ The multiplication operation is closed within the set A and satisfies the associative property.

Once it’s established that (A,·) is indeed a group, we can check for the existence of subgroups.

A subgroup (S,·) consists of the subset S={1} with the same operation as the group (·):

$$ (S,\cdot) $$

On the other hand, the subset T={-1} is not a subgroup under multiplication (·).

Why is S={1} a subgroup, but T={-1} is not?

To explain, let’s construct the multiplication table for the subset S={1} under multiplication (·):

a·b	1
1	1

Next, we check whether the subset S={-1} meets all the conditions to be a subgroup of (A,·):

It is a subset S⊆A of the set A in the group (A,·).
It is closed under multiplication (·) because 1∈S.
It includes the identity element (1) of the group (A,·).
Every element in the subset s∈S has an inverse element s^-1∈S.

Note: The subset S has only one element {1}, and multiplying it by itself results in the group’s identity element (1): $$ 1 \cdot 1 = 1 $$ In this case, the number 1 is its own inverse, as it is the identity element.
The operation (·) applied to the subset S satisfies the associative property:
Note: According to the associative property: $$ (1·1)·1 = 1·(1·1) $$ Both sides of the equation are equal to 1. Thus, the associative condition is satisfied.

All conditions are met.

Therefore, the subset S under the multiplication operation (·) is a subgroup of (A,·).

$$ (S,·) ⊂ (A,·) $$

Note: The subgroup (S,·) inherits all the properties of the group (A,·). For example, the group (A,·) is an abelian group because it satisfies the commutative property. The subgroup (S,·) is also abelian.

Now, let’s check whether the subset T={-1} is a subgroup of (A,·).

We construct the multiplication table for the subset S={1} under multiplication (·):

a·b	-1
-1	1

Then, we verify whether the subset T={-1} meets all the conditions to be a subgroup of (A,·):

It is a subset T⊆A of the set A in the group (A,·).
It is not closed under multiplication (·) because (- 1)·(-1)=1∉T.

Since one condition is not met, we can conclude that T is not a subgroup of (A,·).

Example 3

Consider a group consisting of the finite set of integers Z₄={0,1,2,3} in modular arithmetic with the addition operation +:

$$ (Z_4, +) $$

This is a cyclic group and an abelian group.

Let’s construct the operation table for the group:

a+b	0	1	2	3
0	0	1	2	3
1	1	2	3	0
2	2	3	0	1
3	3	0	1	2

The proper subsets of Z₄ are {0}, {1}, {2}, {3}, {0,1}, {0,2}, {0,3}, {1,2}, {1,3}, {2,3}, {1,2,3}, {0,2,3}, {0,1,3}, {0,1,2}.

We check if any of these proper subsets can form subgroups of (Z₄,+) under the addition operation (+):

The subsets {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3} are not subgroups because they do not include the identity element e=0.
The subset {0,1} does not form a subgroup because it is not closed. For example, 1+1=2∉{0,1}.
The subset {0,3} does not form a subgroup because it is not closed. For example, 3+3=2∉{0,3}.
The subset {0,2,3} does not form a subgroup because it is not closed. For example, 2+3=1∉{0,2,3}.
The subset {0,1,3} does not form a subgroup because it is not closed. For example, 0+3=2∉{0,2,3}.
The subset {0,1,2} does not form a subgroup because it is not closed. For example, 2+1=3∉{0,1,2}.

The subset {0,2} is, however, a subgroup because:

{0,2} is closed under addition. For example, 0+2=2∈{0,2}, 2+2=0∈{0,2}, 0+0=0∈{0,2}.
It includes the identity element e=0.
Every element of {0,2} has an inverse. For example, 0+0=e, 2+2=e.

To confirm, let’s construct the table for the subset {0,2} under addition:

a+b	0	2
0	0	2
2	2	0

This subset is indeed a subgroup of (Z₄,+).

$$ ( \{ 0,2 \} \ , \ + ) ⊆ (Z_4, +) $$

Furthermore, it inherits the same properties as (Z₄,+).

For example, the subgroup ({0,2},+) is also a cyclic group and an abelian group.

Note: The trivial subset {0} also forms a subgroup: $$ ( \{ 0 \} \ , \ + ) ⊆ (Z_4, + ) $$ It is closed under addition since 0+0=0∈{0}. It includes the identity element e=0. Every element has an inverse. In this case, 0 is its own inverse because 0+0=e.

And so on.