# 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 $K\leq G$.

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 $st^{-1}\in K$.

Proof: First we need to show if $K$ is a subgroup of $G$ then $st^{-1}\in K$. Since $s,t\in K$ then $st^{-1}\in K$, because $K$ 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 $K$ 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 $K$: $e\in K$. (Existence of identity)

Now that we know $e\in K$, for all $t$ in $K$ we have that $et^{-1}=t^{-1}\in K$ 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