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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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The smallest $n\geq 1$ such that $f^n$ is the identity function, where $f^n$ denotes $f$ composed with itself $n$ times. –  Alex Becker Sep 1 '12 at 5:05
    
@AlexBecker: it's usually considered bad practice to post answers as comments. –  Ben Millwood Sep 1 '12 at 9:36
    
@BenMillwood Only when there are no answers, so that the question will not be automatically bumped by the community user. I see no point in bumping the question by adding another answer when there were already two perfectly good ones. I just wanted to give a slightly different explanation in case it helped things click more for the OP. Different perspectives are good IMHO. –  Alex Becker Sep 1 '12 at 18:25
    
@AlexBecker: if you think you have something to add, imo it's worth an answer. Comments are meant to be for asking for clarification, etc. - I feel like I should be able to read a question and answer ignoring the comments and not lose any mathematical content. –  Ben Millwood Sep 2 '12 at 13:29
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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. –  Alex Becker Sep 2 '12 at 17:49

3 Answers 3

up vote 1 down vote accepted

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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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 :-D –  le duc quang Sep 1 '12 at 6:46

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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I thought this stuff would help you out with understanding automorphisms. –  Chris Dugale Sep 1 '12 at 5:16
    
Yes, it do help me so much. Really thanks for this:-D –  le duc quang Sep 1 '12 at 6:47
    
@leducquang: If the answer was helpful, then you could vote it up (click on the up-arrows). Also don't forget to accept an answer if one of them really helped you out. You can upvote as much as you like, but accept only one answer per question. –  Marc van Leeuwen Sep 1 '12 at 7:56
    
I didn't downvote, but I can see why you might in that this doesn't actually answer the question - it discusses the order of a group, rather than the order of elements. –  Ben Millwood Sep 1 '12 at 9:38
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I wrote this answer because he is obviously having a hard time with the automorphism concept. So, I thought that I would give him some information on it, and some exercises to help reinforce the knowledge. I would much rather have someone do that for me, then to simply give a definition that had already been given by a previous person. –  Chris Dugale Sep 1 '12 at 18:03

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