subgroup

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

• $K$ is closed under the $*$ operation.

• There exists an identity element $e\in K$ such that for all $k\in K$, $k*e=k=e*k$.

• 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 $(K,*)$. We denote $K$ being a subgroup of $G$ by writing $K\le G$.

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:

• 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 $K$ is a nonempty subset of the group $G$. Then $K$ is a subgroup of $G$ if and only if $s,t\in K$ implies that $s{t}^{-1}\in K$.

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

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

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

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

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

so $K$ is closed under the operation. $QED$

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 $2$ 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 $-k+k=0=k+(-k)$. 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 $s,t$ are even numbers and $s-t\in 2\mathbb{Z}$ since the difference of even numbers is always an even number.

See also:

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

Title	subgroup
Canonical name	Subgroup
Date of creation	2013-03-22 12:02:10
Last modified on	2013-03-22 12:02:10
Owner	Daume (40)
Last modified by	Daume (40)
Numerical id	18
Author	Daume (40)
Entry type	Definition
Classification	msc 20A05
Related topic	Group
Related topic	Ring
Related topic	FreeGroup
Related topic	Cycle2
Related topic	Subring
Related topic	GroupHomomorphism
Related topic	QuotientGroup
Related topic	ProperSubgroup
Related topic	SubmonoidSubsemigroup
Related topic	ProofThatGInGImpliesThatLangleGRangleLeG
Related topic	AbelianGroup2
Related topic	EssentialSubgroup
Related topic	PGroup4
Defines	trivial subgroup