Automorphism

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December 11th 2009, 01:21 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 $
December 11th 2009, 01:58 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 $\mbox{Aut } \mathbb{Z}_8 \cong \mathbb{Z}_8^\times \cong \mathbb{Z}_2 \times \mathbb{Z}_2$ .
December 11th 2009, 02:04 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?
December 11th 2009, 03:18 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.
December 11th 2009, 03:56 PM
mms

i see.. thank you

i can't prove the first problem though :/
December 11th 2009, 04:55 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 $Aut( \mathbb{Z}_p ) \cong \mathbb{Z}_p^{\times }$
December 11th 2009, 05:14 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).
December 11th 2009, 05:26 PM
mms

thanks for your time...

i understand how you show that you have p-1 automorphisms, but i dont get hoy you define the isomorphism -.- plz help this dumb man (Itwasntme)
December 11th 2009, 05:31 PM
Bruno J.

Jose explained it, but I can explain some more.

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!)
December 11th 2009, 06:16 PM
mms

i think i understand now...thank you :D