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:

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

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

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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:
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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 nonempty).
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:

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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 $st\in 2\mathbb{Z}$ since the difference^{} of even numbers is always an even number.
See also:

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Wikipedia, http://www.wikipedia.org/wiki/Subgroupsubgroup
Title  subgroup 
Canonical name  Subgroup 
Date of creation  20130322 12:02:10 
Last modified on  20130322 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 