# Thread: group of automorphisms

1. ## group of automorphisms

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

Here's what I did.

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

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

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

Here's what I did.

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

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

Notice that $\phi(n) = \phi(4p+2) = \phi(2)\phi(2p+1) = \phi(2p+1) = 2p$.

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

Thank you!

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

4. Originally Posted by georgel
Thank you!

So now the group is isomorphic to either $\mathbb{Z}_2\otimes\mathbb{Z}_p$ or $\mathbb{Z}_{2p}$. I check the order of the elements to determine which one.
Is this okay?
note that for any prime $q$ and $n \geq 1: \ Aut(\mathbb{Z}/q^n\mathbb{Z}) \cong \mathbb{Z}/(q^n-q^{n-1})\mathbb{Z}.$ in particular $Aut(\mathbb{Z}/q\mathbb{Z}) \cong \mathbb{Z}/(q-1)\mathbb{Z}.$ also, if $\gcd(a,b)=1,$ then: $(\mathbb{Z}/ab\mathbb{Z})^{\times} \cong (\mathbb{Z}/a\mathbb{Z})^{\times} \times (\mathbb{Z}/b\mathbb{Z})^{\times}.$ thus:

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