00:02
Okay, so the question here is that you're given a group g and a family of subgroups, h .i, where i is, capital i is the indexing set.
00:12
And you have an element a of g.
00:15
And you're trying to compare these two co -sets, where you have, on the left side, the intersection of all his, and then you multiply by a.
00:26
On the right side, you have the intersection of all of cosets, hia.
00:32
So i wrote in here, want to show this, because i think you tried to type it in latex and it didn't quite work out.
00:40
Anyway, here's the proof.
00:42
The proof is basically a straightforward mutual containment.
00:44
That means that you start with the left -hand side.
00:48
You show that it's contained the right -hand side, and then you start with the right -hand side, and it shows you it's contained in the left -hand side.
00:55
And it's always going to be on an element by element basis.
00:58
So you start with an element of the left -hand side and show that it's part of the right -hand side and vice versa.
01:04
So to start, we have an element of the left -hand side.
01:08
We constructed by picking some x in the intersection of all h -i's, and then we multiply that x with a so that it's an element of the left -hand side.
01:18
And then since this x is in the intersection, this x is in each h -i.
01:25
And therefore xa is in each coset, hia...