# Math Help - Automorphism groups of cyclic groups

1. ## Automorphism groups of cyclic groups

I was just playing with the automorphism groups of cyclic groups, and I'm finding that perhaps they're not as straightforward as I thought they might be. Does anyone know if there's a formula to determine the automorphism group of $\mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z}$ in general?

For the prime case, how is this 'conjecture' of mine?
$\mbox{Aut}(\mathbb{Z}_p) \cong \mathbb{Z}_{p-1}$, $p$ prime

Also...I'd like to put a little plug in for my thread on automorphisms of subfields, as I'd really like an answer and it seems to have fallen out of interest...

2. ## Re: Automorphism groups of cyclic groups

Originally Posted by AlexP
I was just playing with the automorphism groups of cyclic groups, and I'm finding that perhaps they're not as straightforward as I thought they might be. Does anyone know if there's a formula to determine the automorphism group of $\mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z}$ in general?

For the prime case, how is this 'conjecture' of mine?
$\mbox{Aut}(\mathbb{Z}_p) \cong \mathbb{Z}_{p-1}$, $p$ prime

Also...I'd like to put a little plug in for my thread on automorphisms of subfields, as I'd really like an answer and it seems to have fallen out of interest...
$Aut(Z_n) \cong (\mathbb{Z}/n \mathbb{Z})^{\times}$

3. ## Re: Automorphism groups of cyclic groups

That's what I thought...and it's in the book now that I think about it...so I must have screwed up when I did my calculations. Oh well. Thanks.

4. ## Re: Automorphism groups of cyclic groups

Originally Posted by AlexP
That's what I thought...and it's in the book now that I think about it...so I must have screwed up when I did my calculations. Oh well. Thanks.
Well, $\left(\mathbb{Z}/p\mathbb{Z}\right)^\times\cong\mathbb{Z}_{p-1}$.

5. ## Re: Automorphism groups of cyclic groups

Yes, I know I was right in that case. I worked it all out for $\mathbb{Z}_{20}$ and it did not turn out correctly.

6. ## Re: Automorphism groups of cyclic groups

Originally Posted by AlexP
Yes, I know I was right in that case. I worked it all out for $\mathbb{Z}_{20}$ and it did not turn out correctly.
Well, from the CRT you know that, as rings, $\mathbb{Z}/n\mathbb{Z}\cong\left(\mathbb{Z}/p_1^{a_1}\mathbb{Z}\right)\times\cdots\times\left( \mathbb{Z}/p_m^{a_m}\right)$ (where $n=p_1^{a_1}\cdots p_m^{a_m}$ is the prime factorization) which then induces a group isomorphism $\left(\mathbb{Z}/n\mathbb{Z}\right)^\times\cong\left(\mathbb{Z}/p_1^{a_1}\mathbb{Z}\right)^\times\cdots\times\left (\mathbb{Z}/\mathbb{Z}p_m}^{a_m\right)$. Thus, the problem of finding $\text{Aut}\left(\mathbb{Z}/n\mathbb{Z}\right)^\times$ reduces to finding $\left(\mathbb{Z}/p^a\mathbb{Z}\right)^\times$ for primes $p$. That said, from the above result you have the (interesting) result that if $n$ is square free (i.e. that $n=p_1\cdots p_m$ where $p_k\ne p_j$) then $\text{Aut}\left(\mathbb{Z}_n\right)\cong\left( \mathbb{Z}/n\mathbb{Z}\right)^\times\cong\mathbb{Z}_{p_1-1}\times\cdots\times\mathbb{Z}_{p_m-1}$. For more information on $\left(\mathbb{Z}/n\mathbb{Z}\right)^\times$ you can see the entire chapter devoted to it in this book.