What are the automorphism groups of S3?

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The discussion focuses on determining the automorphism group of S3, the symmetric group of three elements. It is established that Aut(S3) equals Inn(S3), with Inn(S3) being a subgroup containing six automorphisms. The challenge lies in proving that all automorphisms in Aut(S3) are indeed inner, which involves showing that there are no more than six automorphisms. The conversation suggests using the properties of conjugacy classes and the structure of S3 to demonstrate this relationship. Ultimately, the conclusion is that Inn(S3) encompasses all automorphisms of S3, confirming that Aut(S3) is complete.
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The question is to determine the group of automorphisms of S3 (the symmetric group of 3! elements).

I know Aut(S3)=Inn(S3) where Inn(S3) is the inner group of the automorphism group.

For a group G, Inn(G) is a conjugation group (I don't fully understand the definition from class and the book doesn't give one).

I also know that Inn(S3) is a subgroup of Aut(S3), so Inn(S3) is contained in
Aut(S3).

I'm not sure how to show the other way thought, that everything in Aut(S3) is in Inn(S3).

I talked to my professor, and he gave me the hint that Inn(S3) has only 6 automorphisms, so I should show there are no more than 6 automorphisms in
Aut(G). The problem is I'm not sure how to do this.
 
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A basic result on the subject, which I am sure would be found in any book or article pertaining to such, is that for n not equal to 6, Inn(Sn) exhausts the Aut(Sn) i.e. they are the same. I think its called complete (one of the many uses of the word). In other words for n not equal to 6, Out(Sn) is trivial. I recall briefly the proof. The idea is to show that all automorphisms that preserve the conjugacy class of transpositions is inner. Then you show for that for all n not 6, there does not exsist a conjugacy class of order 2 with the same order as the class of tranpositions. After this the proof is trivial since Inn(S3) is comprised of the mappings which conjugate all elements of S3 by a fixed element in S3. There are only 3!=6 such elements for which to fix, and therefore the order of Inn(S3)=6.
 
S_3 is the permutation group on 3 elements, it has order 3!.consider where an automorphism sends the 2-cycles (there are 3; what groups of order 3! do you know?)

there is no need to invoke the aut/inn general S_n argument for this case, it can be done by inspection.

oh, and inner automorphism of G is a map f_x from G to G f_x(g)=xgx^{-1}

this is an automorphism, and f_x=f_y means that x=y, thus there are distinct automorphisms for each element of G
 
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Hints:

Find a set of two generators for S3. Actually, find all pairs of generators for the group. What do all these pairs have in common? You should find two types of pairs. Specifically, you will find 9 pairs of generators, 3 will be of one type, 6 will be of another. Pick one type.

Inn(S3) = Aut(S3).

Determine the relationship between whether or not two permutations have the same cycle structure and whether or not they are conjugate.

Having picked your type in the first hint, if {x,y} is a pair of elements, and {x',y'} is any pair of the same type, is x -> x', y -> y' an automorphism?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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