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I'm studying field theory on my own, and I am stuck in the definition of an automorphism and automorphism group. Could you give me examples, like the $Aut(\mathbb Z_7)$? How they are computed?

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Are you familiar with isomorphisms? – Alex Becker Sep 11 '12 at 20:31
Where are you stuck in the definition? – Qiaochu Yuan Sep 11 '12 at 20:35
Since an automorphism must take $1$ to itself, there are no nontrivial automorphisms of ${\mathbb Z}_p$ for any prime $p$. – Robert Israel Sep 11 '12 at 20:35
On the other hand, the field ${\mathbb F}_4$ (consisting of $0, 1, a, 1+a$ where $a^2 = 1+a$) has an automorphism that interchanges $a$ and $1+a$. – Robert Israel Sep 11 '12 at 20:39
up vote 2 down vote accepted

An automorphism of a group $G$ is a group isomorphism from $G$ onto $G$.

The set of automorphisms on a group forms a group itself, where the product is composition of homomorphisms.

Now everywhere that I boldfaced "group", you can replace it with "ring" or "module" or "field" or "field extension". They will all produce automorphism groups.

In the case of Galois theory, you are usually in the last case. You have an extension of fields $F\subseteq E$, and a field extension automorphism (which is, by definition, an isomorphism from $E$ to $E$ that leaves $F$ fixed.)

Computing them depends heavily on the context. In the case of $\mathrm{Aut}\mathbb{Z}_7$, you know that an isomorphism would be completely determined by where $1$ goes, and you know that the image of 1 would to have the same order as 1. If you check, all the nonzero elements have the same order as 1. So, you have 6 different places to send 1, corresponding to 6 different automorphisms.

In the case of a field extension, another interesting thing happens: if $F$ is a field and $E$ is a field formed by adjoining roots of a polynomial over $F$, then any automorphism of the extension $F\subseteq E$ has to permute the roots of the polynomial. So, the game then is to figure out which roots go to which roots, and in the process you come up with all possible automorphisms. That sounds a bit advanced for you right now, though, so I'll wait for you to reach it in your reading.

Let's do $\mathbb{Z}_6$'s automorphisms, so that I can say we did something new. Again, a group homomorphism will be completely determined by where 1 goes, and 1 has order 6, here. Now 2 and 4 have order 3, and 3 has order 2, but 1 and 5 have order 6. That means there are only two isomorphisms, the one where you send 1 to 1, and the one where you send 1 to 5. That makes the automorphism group of $\mathbb{Z}_6$ the group with two elements. (There is only one!)

Sorry, from reading other answers I could see you might be asking about $\mathbb{Z}_7$ as a ring too. My stuff above considers it only as an abelian group. You will find good explanations explaining its group of ring isomorphisms above, I'm sure.

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I will first give a very general definition. Let $X$ be an object in a category. Then the automorphism group of $X$ (noted $\mathrm{Aut}(X)$)is the set of invertible morphisms from $X$ to itself. Since we have a composition law, we see that this is indeed a group.

In concrete settings, this has the following interpretations:

  • If $X$ is a group, then $\mathrm{Aut}(X)$ is the group of all group isomorphisms from $X$ to itself.
  • If $X$ is a topological space, then $\mathrm{Aut}(X)$ is the group of all homeomorphisms from $X$ to itself.
  • If $X$ is a vector space, then $\mathrm{Aut}(X)$ is the group of all linear isomorphisms from $X$ to itself.
  • If $X$ is a topological group, then $\mathrm{Aut}(X)$ is the group of all continuous group isomorphisms from $X$ to itself.

In some sense, the automorphism group is measuring how much the labelling of the elements in your group (say) matters.

Now, going back to your particular example $\mathrm{Aut}(\mathbb Z_7)$, if you are looking at all field isomorphisms, then as Robert Israel noted, this group is trivial (i.e. $\mathrm{Aut}(\mathbb Z_7)\cong 1$), since $1$ has to go to $1$, and this determines the isomorphism. On the other hand, if you are looking at all group isomorphisms, then remember that $\mathbb Z_7$ is generated (as a group) by $1$, but also by all nonzero residue class. Since you can define a group homomorphism by specifying where a generator goes, you have an isomorphism for each nonzero residue class (namely $1\mapsto p$, where $p\in\{1,\ldots,7\}$). Hence, we have $\mathrm{Aut}(\mathbb Z_7)\cong\mathbb Z_7^\times$.

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