subgroup (original) (raw)

In addition the notion of a subgroup of a semigroup can be defined in the following manner. Let (S,*) be a semigroup and H be a subset of S. Then H is a subgroup of S if H is a subsemigroup of S and H is a group.

Properties:

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  The set {e} whose only element is the identity is a subgroup of any group. It is called the trivial subgroup.
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  Every group is a subgroup of itself.
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  The null set {} is never a subgroup (since the definition of group states that the set must be non-empty).

There is a very useful theorem that allows proving a given subset is a subgroup.

Theorem:
If K is a nonempty subset of the group G. Then K is a subgroup of G if and only if s,t∈K implies that s⁢t-1∈K.

Proof:First we need to show if K is a subgroup of G then s⁢t-1∈K. Since s,t∈K then s⁢t-1∈K, because K is a group by itself.
Now, suppose that if for any s,t∈K⊆G we have s⁢t-1∈K. We want to show that K is a subgroup, which we will accomplish by proving it holds the group axioms.

Since t⁢t-1∈K by hypothesis, we conclude that the identity element is in K: e∈K. (Existence of identity)

Now that we know e∈K, for all t in K we have that e⁢t-1=t-1∈K so the inverses of elements in K are also in K. (Existence of inverses).

Let s,t∈K. Then we know that t-1∈K by last step. Applying hypothesis shows that

so K is closed under the operation. Q⁢E⁢D

Example:

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  Consider the group (ℤ,+). Show that(2⁢ℤ,+) is a subgroup.
  The subgroup is closed under addition since the sum of even integers is even.
  The identity 0 of ℤ is also on 2⁢ℤ since 2 divides 0. For every k∈2⁢ℤ there is an -k∈2⁢ℤ which is the inverse under addition and satisfies -k+k=0=k+(-k). Therefore (2⁢ℤ,+) is a subgroup of (ℤ,+).
  Another way to show (2⁢ℤ,+) is a subgroup is by using the proposition stated above. If s,t∈2⁢ℤ then s,t are even numbers and s-t∈2⁢ℤ since the difference of even numbers is always an even number.

See also:

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  Wikipedia, http://www.wikipedia.org/wiki/Subgroupsubgroup