Let .
Here is how to start off.
If then and .
Therefore there is such that .
And thus, and .
Thus, thus .
Thus is closed.
Continue with the other properties.
Show that the union of a nonempty directed family of subgroup of a group G is a subgroup of G.
Proof so far:
Let G be a group, and suppose that be a directed family of subgroups of G.
Then is the union.
Since it is nonempty, the subgroup of {e}, act as the identity of this subgroup.
Now, suppose that x and y are elements of this union, then there exist i,j in A such that and . And since this family is directed, there exist some k in A such that and . Implies that both x,y are in And since the union is consist of all subgroups, that means is a subgroup, implies that , thus proves the claim.
Is this right? Thank you.