The Unit and Automorphism Group of Z/nZ (Pt. I)
Point of Post: In this post we find the general form of and in terms of products of cyclic groups.
There are two groups naturally associated to one which is important for ring theory and one for group theory. Namely, when dealing with as a ring the unit group plays a pivotal role, encompassing a large part of the important information about the multiplicative structure of . Similarly, when thinking of as an additive group we know that a structure paramount to understanding it is its automorphism group . Thus, it would be nice to find a nice description for each of these groups. The first step in this process is to prove that finding either or is sufficient to finding both since, as we shall see, they are isomorphic. Thus, it suffices to find a nice description of . The pivotal piece of this puzzle was recently discussed in a post on ring theory. Namely, from information from recent posts, in particular the Chinese Remainder Theorem we have that thus reducing the problem to finding for primes . This post picks up where the CRT leaves off by doing just this, finding an explicit description for .
The Unit Group of
We begin out study with a seemingly simple lemma:
Lemma: Let be an odd prime. Then, and for all .
Proof: By the Binomial Theorem we have that
What we claim now is that for each . To see this first note that we may rewrite this as
Now, if we can prove that the fraction in this last expression is an integer we’ll be done. But, since is a prime the only way that factoring out the power of could make this not an integer is if the resulting fraction has a numerator which is divisible by a higher power of than the denominator. To see why this isn’t true we first note that if for then provided . But, evidently for otherwise and thus and is never true for any prime and . Thus, we see that if we have that divides the numerator, thus if we can show that for any we’ll be done. But, this is clear since for every prime . Thus, and so
To show that we again expand by the Binomial Theorem to get
and, similar to before, we claim that . Indeed, using the same idea it suffices to show that the maximal power of dividing is greater than that of . Now, using the same idea we see that implies that assuming that . That said, if this were not true then which is impossible for and primes . Thus, by an analogous argument to before we see that we may conclude that as long as we can prove that for . That said, it’s trivial to check that for all and odd primes (this is where we need , because this is false otherwise since ). Thus, we have that and so
Corollary: For odd primes and one has that has order in
Proof: We merely note that by the previous lemma we have that but that (since otherwise would be modulo ) and so evidently .
We are now able to complete the first ‘half’ of identifying
Theorem: Let be an odd prime, then .
Proof: The result has already been proven for and so it suffices to prove this for . To see this note that there is a natural morphism which is just the reduction map. This is an epimorphism since for each (where obviously the two ‘s are ‘different’). We thus have by the first isomorphism theorem that and since and we must have that . Now, choose such that has order in (we can do this since is cyclic of order and is surjective). We then have from elementary group theory that . So, note that since we have that for some . Now, since we have that is invertible in , so let and let . Note then that
since . So, but since we have by the same argument as before that and so . Note then that since and and commute we have, from basic group theory, that . And, since (where is Euler’s totient function). The conclusion follows.
1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.
2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.