About 1 to 2:

Assume 1 (i.e. ). Let be such that . Hence .

Notice that , so that there are such that .

Now, if , for some , hence (because is a subgroup), and , so that . We've proved that . Let's prove the reverse inclusion. If , then for some (because ), hence (because is a subgroup, again). This proves that . Finally, .

Laurent.