PlanetMath: subgroup

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subgroup

(Definition)

Definition:
Let be a group and let be subset of . Then is a subgroup of defined under the same operation if is a group by itself (with respect to ), that is:

is closed under the operation.
There exists an identity element $e\in K$ such that for all $k \in K$ , .
Let $k \in K$ then there exists an inverse $k^{-1} \in K$ such that $k^{-1}*k = e = k*k^{-1}$ .

The subgroup is denoted likewise

. We denote

being a subgroup of

by writing $K\leq G$ .

In addition the notion of a subgroup of a semigroup can be defined in the following manner. Let be a semigroup and be a subset of . Then is a subgroup of if is a subsemigroup of and is a group.

Properties:

The set $\{e\}$ whose only element is the identity is a subgroup of any group. It is called the trivial subgroup.
Every group is a subgroup of itself.
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 is a nonempty subset of the group . Then is a subgroup of if and only if $s,t \in K$ implies that $st^{-1} \in K$ .

Proof: First we need to show if is a subgroup of then $st^{-1} \in K$ . Since $s,t \in K$ then $st^{-1} \in K$ , because is a group by itself.
Now, suppose that if for any $s,t\in K\subseteq G$ we have $st^{-1}\in K$ . We want to show that is a subgroup, which we will accomplish by proving it holds the group axioms.

Since $tt^{-1}\in K$ by hypothesis, we conclude that the identity element is in : $e\in K$ . (Existence of identity)

Now that we know $e\in K$ , for all in we have that $et^{-1}=t^{-1}\in K$ so the inverses of elements in are also in . (Existence of inverses).

Let $s,t\in K$ . Then we know that $t^{-1}\in K$ by last step. Applying hypothesis shows that

$\displaystyle s(t^{-1})^{-1} = st\in K$

is closed under the operation.

Example:

Consider the group $(\mathbb{Z}, +)$ . Show that $(2\mathbb{Z}, +)$ is a subgroup.
The subgroup is closed under addition since the sum of even integers is even.

The identity 0 of $\mathbb{Z}$ is also on $2\mathbb{Z}$ since divides 0. For every $k \in 2\mathbb{Z}$ there is an $-k \in 2\mathbb{Z}$ which is the inverse under addition and satisfies . Therefore $(2\mathbb{Z}, +)$ is a subgroup of $(\mathbb{Z}, +)$ .

Another way to show $(2\mathbb{Z}, +)$ is a subgroup is by using the proposition stated above. If $s,t \in 2\mathbb{Z}$ then are even numbers and $s-t\in 2\mathbb{Z}$ since the difference of even numbers is always an even number.