00:01
Hello everyone.
00:02
So let g be the group and let ht, t belongs to capital t.
00:21
Be the family of normal subgroups and t here it is an index set, is an index set such that for all t belongs to t.
00:44
H -t is normal subgroup of normal subgroup of.
00:58
So, now let h is equal to t belongs to t.
01:10
H -t is equal to a belongs to g and a belongs to h -t for all, t belongs to t.
01:23
This be the intersection of this family of normal subgroups of g.
01:27
So then to prove h is a normal subgroup of g, we know that intersection of this collection of subgroups is also subgroup.
01:37
Therefore, h is a subgroup of g.
01:44
H is a subgroup of g.
01:49
Therefore, now to prove that h is a normal subgroup of g, let x be an any element of g, element of g, and h be any element of h.
02:07
Element of h of h is equal to t belongs to t h t so we have h belongs to t belongs to t ht which means h belongs to capital h t for all t belongs to t since h t is a normal subgroup of g we can say that x belongs to g and h ht so x hedge x inverse belongs to ht so from here we have x belongs to g so h belongs to t this means x x inverse belongs to ht for all t belongs to t so from here we can say that x, h, x inverse belongs to h t...