## The Unit and Automorphism Group of Z/nZ (pt. II)

**Point of Post: **This is a continuation of this post.

We now have to deal with the case separately.

**Theorem: ***Let be arbitrary. Then, , and for one has that *

**Proof: **The cases for are trivial by just counting the order of the respective groups (since there is only one group of order and likewise for ). So, let , we claim that . Indeed, it’s trivially true for and if we assume it’s true for then we see that

where we used the easily verifiable fact that implies that for primes (just apply the Binomial Theorem). Thus, the induction is complete. But, this clearly implies that and and so (using similar logic to the case for odd primes) we have that . So, now consider the reduction map . Note that since this is obviously surjective we have that and since we have that . Now, note that are a subgroup of which are disjoint from (they are not in ) and so the product is a subgroup of of order and so equal to . It follows then from basic group theory that as desired.

With this and the Chinese Remainder Theorem we may thus finally conclude that

**Theorem: ***Let be a prime number and . Then,*

and if has prime decomposition then

*The Automorphism Group of *

So, how does this help us figure out what is? Well, the following theorem gives us a hint:

**Theorem: ***Let . Then, .*

**Proof: **Define by for . We begin by showing that this is a well-defined mapping (in the sense that for each . To see that is an automorphism we first note that so that is a homomorphism, it’s an injection since implies and since is invertible this implies , and it’s surjective since for any we have that (note, we already knew it was surjective since is an injection from a finite set to itself, from where the rest follows from basic set theory).

Now, to show that itself is an isomorphism. To see that is a homomorphism we just compute that and since was arbitrary we thus have that and so a homomorphism. Now, injectivity of is trivial since implies . To prove surjectivity let we claim that . Indeed, (where we’ve used to distinguish the -fold sum of an element) from where the conclusion follows.

*Remark: *The more sophisticated, shorter proof of the above (which we spared for those who don’t know as much ring theory) is that, by definition, for an abelian group , where is the endomorphism ring but it is easy to show that if is cyclic.

**References:**

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.

No comments yet.

## Leave a Reply