# Automorphism

• Dec 11th 2009, 02:21 PM
mms
Automorphism
let p be a prime. Show that $
Aut\left( {\mathbb{Z}_p } \right) \simeq \mathbb{Z}_{p - 1}

$

and show that $
Aut\left( {\mathbb{Z}_8 } \right) \simeq \mathbb{Z}_2

$
• Dec 11th 2009, 02:58 PM
Bruno J.
In general, $\mbox{Aut } \mathbb{Z}_n \cong \mathbb{Z}_n^\times$.

It's not trivial at all that $\mathbb{Z}_p^\times$ is cyclic. It's not very difficult to show once you know it, but you are probably not expected to come up with the proof. Instead you are probably expected to use this as a fact.

The second one is false - you have $\mbox{Aut } \mathbb{Z}_8 \cong \mathbb{Z}_8^\times \cong \mathbb{Z}_2 \times \mathbb{Z}_2$.
• Dec 11th 2009, 03:04 PM
mms
thanks for taking the time for explaining...
but when you write the x on top of Zn you mean its the multiplicative group?

also, how could you prove that aut(Z8) is isomorphic to Z2 X Z2?
• Dec 11th 2009, 04:18 PM
Bruno J.
Yes that is what I mean! Sorry for not being more clear.

To see that $
\mbox{Aut } \mathbb{Z}_8 \cong \mathbb{Z}_8^\times \cong \mathbb{Z}_2 \times \mathbb{Z}_2
$
, just notice that every one of the four elements of $\mathbb{Z}_8^\times$ has order 2. There is only one group of order 4 in which every element has order 2; it is the Klein 4-group.
• Dec 11th 2009, 04:56 PM
mms
i see.. thank you

i can't prove the first problem though :/
• Dec 11th 2009, 05:55 PM
Jose27
Notice that an homomorphism $f:G \rightarrow H$ when $G$ is cyclic is completely determined once you know the image of a generator. Knowing this $g: \mathbb{Z}_p \rightarrow \mathbb{Z}_p$ is determined by $g(1)$, but if $g$ is to be an automorphism then $g(1)=a$ must be a generator ie. a nonzero element, but we have $p-1$ such elements. Now take $g,h : \mathbb{Z}_p \rightarrow \mathbb{Z}_p$ automorphisms then if $a=h(1), b=g(1)$ then $gh(1)=ba$ so we can identify an automorphism $g$ with it's value at $1$ and this defines an isomorphism between $Aut( \mathbb{Z}_p ) \cong \mathbb{Z}_p^{\times }$
• Dec 11th 2009, 06:14 PM
Bruno J.
Yes, that is how one goes about showing that $
\mbox{Aut } \mathbb{Z}_p \cong \mathbb{Z}_p^\times
$
. However, it is showing that $\mathbb{Z}_p^\times \cong \mathbb{Z}_{p-1}$ which is difficult (and which I doubt has to be proven by mms - it is probably to be assumed).
• Dec 11th 2009, 06:26 PM
mms
For every $a \in \mathbb{Z}_p^\times$, let $\sigma_a : \mathbb{Z}_p \rightarrow \mathbb{Z}_p$ be the automorphism $x \mapsto ax$ (prove that it is an automorphism). Then the map $\pi : \mathbb{Z}_p^\times \rightarrow \mbox{Aut }\mathbb{Z}_p$ defined by $a \mapsto \sigma_a$ is the isomorphism you are looking for. (Prove it!)