Let H1,H2,H3,... be a (possible infinite) sequence of subgroups of a group with the property that . Prove that the union of the sequence is also a subgroup.
Would I use a subgroup test on this or what? I will be very grateful for any help.
I am still lost. Here is what I have so far:
Let H= the union of subgroups H1,H2,H3,...Hn. Obviously H is a subset of the group G containing all the subgroups H1,H@,H3,... THe defining property of H is that H1 is contained in H2, H2 is contained in H3,....
I don't know how I could prove that the identity is in my group, or that for all elements a,b in H that a(b^-1) is in H.
Do I use the fact that since H1,H2,H3,... are all subgroups, and that H1 is contained in all the subgroups H2 to Hn, that the identity element of H1 is the same for all H2 to Hn? How do I show that for all a,b in H, that a(b^-1) is in H. Is it due to closure of the subgroups H1,H3,H3,....?
Thanks again for your help
would your lemma work due to the fact that since each H1, H2,... Hn is a subgroup,then a(b^-1) will be in each H1,H2,..,Hn by definition of a subgroup. Then taking the union of all these subgroups, we get all a(b^-1) that were in H1,H2,H3,...Hn in are new subgroup H?