is closed under the operation.
There exists an identity element such that for all , .
Let then there exists an inverse such that .
The subgroup is denoted likewise . We denote being a subgroup of by writing .
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.
If is a nonempty subset of the group . Then is a subgroup of if and only if implies that .
First we need to show if is a subgroup of then . Since then , because is a group by itself.
Now, suppose that if for any we have . We want to show that is a subgroup, which we will accomplish by proving it holds the group axioms.
Now that we know , for all in we have that so the inverses of elements in are also in . (Existence of inverses).
Let . Then we know that by last step. Applying hypothesis shows that
so is closed under the operation.
|Date of creation||2013-03-22 12:02:10|
|Last modified on||2013-03-22 12:02:10|
|Last modified by||Daume (40)|