# group of automorphisms

• May 27th 2009, 12:56 PM
georgel
group of automorphisms
Let $\displaystyle p$ be odd prime number such that $\displaystyle 2p+1$ is prime as well. Let$\displaystyle n=4p+2.$ Identify $\displaystyle Aut(\mathbb{Z}_n).$

Here's what I did.

$\displaystyle Aut(\mathbb{Z}_n)$ is isomorphic to the group $\displaystyle U_{\phi(n)}=($numbers relatively prime to $\displaystyle n, ._n)$, where $\displaystyle \phi(n)$ denotes the number of these numbers relatively prime to $\displaystyle n$.
But what is this group..?

Also, $\displaystyle \mathbb{Z}_n=\mathbb{Z}_{2(2p+1)}$ which is isomophic to $\displaystyle \mathbb{Z}_2 \oplus \mathbb{Z}_{2p+1}$, but that doesn't get me anywhere either.

• May 27th 2009, 01:20 PM
ThePerfectHacker
Quote:

Originally Posted by georgel
Let $\displaystyle p$ be odd prime number such that $\displaystyle 2p+1$ is prime as well. Let$\displaystyle n=4p+2.$ Identify $\displaystyle Aut(\mathbb{Z}_n).$

Here's what I did.

$\displaystyle Aut(\mathbb{Z}_n)$ is isomorphic to the group $\displaystyle U_{\phi(n)}=($numbers relatively prime to $\displaystyle n, ._n)$, where $\displaystyle \phi(n)$ denotes the number of these numbers relatively prime to $\displaystyle n$.
But what is this group..?

Also, $\displaystyle \mathbb{Z}_n=\mathbb{Z}_{2(2p+1)}$ which is isomophic to $\displaystyle \mathbb{Z}_2 \oplus \mathbb{Z}_{2p+1}$, but that doesn't get me anywhere either.

Notice that $\displaystyle \phi(n) = \phi(4p+2) = \phi(2)\phi(2p+1) = \phi(2p+1) = 2p$.
• May 31st 2009, 07:58 AM
georgel
Quote:

Originally Posted by ThePerfectHacker
Notice that $\displaystyle \phi(n) = \phi(4p+2) = \phi(2)\phi(2p+1) = \phi(2p+1) = 2p$.

Thank you!

So now the group is isomorphic to either $\displaystyle \mathbb{Z}_2\otimes\mathbb{Z}_p$ or $\displaystyle \mathbb{Z}_{2p}$. I check the order of the elements to determine which one.
Is this okay?
• May 31st 2009, 09:57 AM
NonCommAlg
Quote:

Originally Posted by georgel
Thank you!

So now the group is isomorphic to either $\displaystyle \mathbb{Z}_2\otimes\mathbb{Z}_p$ or $\displaystyle \mathbb{Z}_{2p}$. I check the order of the elements to determine which one.
Is this okay?

note that for any prime $\displaystyle q$ and $\displaystyle n \geq 1: \ Aut(\mathbb{Z}/q^n\mathbb{Z}) \cong \mathbb{Z}/(q^n-q^{n-1})\mathbb{Z}.$ in particular $\displaystyle Aut(\mathbb{Z}/q\mathbb{Z}) \cong \mathbb{Z}/(q-1)\mathbb{Z}.$ also, if $\displaystyle \gcd(a,b)=1,$ then: $\displaystyle (\mathbb{Z}/ab\mathbb{Z})^{\times} \cong (\mathbb{Z}/a\mathbb{Z})^{\times} \times (\mathbb{Z}/b\mathbb{Z})^{\times}.$ thus:

$\displaystyle Aut(\mathbb{Z}/n\mathbb{Z}) \cong (\mathbb{Z}/n\mathbb{Z})^{\times} \cong (\mathbb{Z}/2\mathbb{Z})^{\times} \times (\mathbb{Z}/(2p+1)\mathbb{Z})^{\times} \cong \mathbb{Z}/2p \mathbb{Z}.$