# 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.