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Can anyone please tell me what is the "order" of an automorphism of a group? I skim through some books but can't find the clear definition for this notion.

Thanks so much and I really appreciate if you give me some material about automorphism...

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    @BenMillwood Thats a perfectly fine preference, but I do not believe your view reflects community norms on the proper usage of commons. With almost 20,000 rep, I *am* a fairly frequent user, so I think I have a fairly good idea of how the site operates.2012-09-02

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If $G$ is a group and $g \in G$, then the order of $g$, denoted $|g|$, is the least nonnegative $n$ such that $g^n = e$, where $e$ is the identity.

The collection of automorphism of a group $G$, denoted $\text{Aut}(G)$, is a group under function composition. The order $\sigma \in \text{Aut}(G)$ is $|\sigma|$, i.e. order of $\sigma$ as an element of the group $\text{Aut}(G)$.

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    Oh thanks so much, it's just simple if we consider an automorphism as an element of Aut(G).Thanks :-D2012-09-01
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The order of a group is the cardinality of its underlying set. In the case of an automorphism group, it is the cardinality of the set of all automorphisms. I.E. (finitely many automorphisms) the number of isomorphisms from a particular group to its self. There is a special kind of automorphism (self isomorphism) called an inner automorphism. An automorphism is an inner automorphism iff it's conjugation by a group element. As it turns out, the set of inner automorphisms is a normal subgroup of the set of automorphisms, and, as a group, is isomorphic to $\frac{G}{Z}$, where Z is the center of the group. The center of a group is the group of elements that commute with everything.

Exercises:

  1. Prove that Inn(G) (group of inner automorphisms) is normal in Aut(G). (just grind it out)

  2. Prove that the center of a group is a (normal) subgroup of G

  3. Prove that Inn(G) is isomorphic to G mod the center by considering a the natural map: $g\rightarrow \chi_g$, where $\chi_g$ is conjugation by g. (prove it's a homomorphism, prove that it's onto (trivial), find its kernel, use the 1st isomorphism theorem.

Bonus: (If you don't know the 1st isomorphism theorem) Let G, H be groups and $\phi: G\rightarrow H$ be a surjective homomorphism. Prove that H is isomorphic to G mod the kernel.

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    Okay, sure, but discussing one meaning o$f$ the word order when the questioner asks for a different one runs the risk of misleading them. You should at least answer the question /as well/ as giving your extended discussion.2012-09-02
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There is an amazing beautiful Theorem of Horosevskii: Let $\sigma \in Aut(G)$, where $G$ is a non-trivial finite group. Then the order $|\sigma|$ of $\sigma$, is smaller than $|G|$.
The proof is far from trivial and can be found in Marty Isaacs' book Finite Group Theory.

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    Sir, $\sigma_1 , \sigma_2 \in Aut(G) $, can $\sigma_1 , \sigma_2$ have same order ? in genral, can we divide/classify set $Aut(G)$ in to sets of same order ($G$ is not a cyclic group)?2015-08-07