let p be a prime. Show that $\displaystyle

Aut\left( {\mathbb{Z}_p } \right) \simeq \mathbb{Z}_{p - 1}

$

and show that $\displaystyle

Aut\left( {\mathbb{Z}_8 } \right) \simeq \mathbb{Z}_2

$

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- Dec 11th 2009, 01:21 PMmmsAutomorphism
let p be a prime. Show that $\displaystyle

Aut\left( {\mathbb{Z}_p } \right) \simeq \mathbb{Z}_{p - 1}

$

and show that $\displaystyle

Aut\left( {\mathbb{Z}_8 } \right) \simeq \mathbb{Z}_2

$ - Dec 11th 2009, 01:58 PMBruno J.
In general, $\displaystyle \mbox{Aut } \mathbb{Z}_n \cong \mathbb{Z}_n^\times$.

It's not trivial at all that $\displaystyle \mathbb{Z}_p^\times$ 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 $\displaystyle \mbox{Aut } \mathbb{Z}_8 \cong \mathbb{Z}_8^\times \cong \mathbb{Z}_2 \times \mathbb{Z}_2 $. - Dec 11th 2009, 02:04 PMmms
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, 03:18 PMBruno J.
Yes that is what I mean! Sorry for not being more clear.

To see that $\displaystyle

\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 $\displaystyle \mathbb{Z}_8^\times$ 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, 03:56 PMmms
i see.. thank you

i can't prove the first problem though :/ - Dec 11th 2009, 04:55 PMJose27
Notice that an homomorphism $\displaystyle f:G \rightarrow H$ when $\displaystyle G$ is cyclic is completely determined once you know the image of a generator. Knowing this $\displaystyle g: \mathbb{Z}_p \rightarrow \mathbb{Z}_p$ is determined by $\displaystyle g(1)$, but if $\displaystyle g$ is to be an automorphism then $\displaystyle g(1)=a$ must be a generator ie. a nonzero element, but we have $\displaystyle p-1$ such elements. Now take $\displaystyle g,h : \mathbb{Z}_p \rightarrow \mathbb{Z}_p$ automorphisms then if $\displaystyle a=h(1), b=g(1)$ then $\displaystyle gh(1)=ba$ so we can identify an automorphism $\displaystyle g$ with it's value at $\displaystyle 1$ and this defines an isomorphism between $\displaystyle Aut( \mathbb{Z}_p ) \cong \mathbb{Z}_p^{\times }$

- Dec 11th 2009, 05:14 PMBruno J.
Yes, that is how one goes about showing that $\displaystyle

\mbox{Aut } \mathbb{Z}_p \cong \mathbb{Z}_p^\times

$. However, it is showing that $\displaystyle \mathbb{Z}_p^\times \cong \mathbb{Z}_{p-1}$ which is difficult (and which I doubt has to be proven by mms - it is probably to be assumed). - Dec 11th 2009, 05:26 PMmms
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) - Dec 11th 2009, 05:31 PMBruno J.
Jose explained it, but I can explain some more.

For every $\displaystyle a \in \mathbb{Z}_p^\times$, let $\displaystyle \sigma_a : \mathbb{Z}_p \rightarrow \mathbb{Z}_p$ be the automorphism $\displaystyle x \mapsto ax$ (prove that it is an automorphism). Then the map $\displaystyle \pi : \mathbb{Z}_p^\times \rightarrow \mbox{Aut }\mathbb{Z}_p$ defined by $\displaystyle a \mapsto \sigma_a$ is the isomorphism you are looking for. (Prove it!) - Dec 11th 2009, 06:16 PMmms
i think i understand now...thank you :D