for all so the only thing you need to prove is that if we have a bunch of normal subgroups of a group G, then we will have:
which is very easy to prove: since are normal, is a subgroup, which contains every hence it contains the union of them. but by definition
is the smallest subgroup which contains the union of hence on the other hand, clearly every is contained in since is
a subgroup, it must also contain the product of , i.e. this completes the proof.