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 $K\\leq G$ .

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:

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The set $\\{e\\}$ whose only element is the identity is a subgroup of any group. It is called the trivial subgroup.
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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 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. $Q E D$

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 subgroupis 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.