Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the number of automorphisms (including identity) for permutation group $S_3$ on 3 letters?

I believe the answer for this is 6. As we can write the group elements as below

  1. (a)(b)(C)
  2. (ab)(c)
  3. (ac)(b)
  4. (bc)(a)
  5. (abc)
  6. (bac)

Can we generalize that for any $S_n$ onto $n$ the number of automorphisms will be $n!$ Also I cannot find this anywhere in my text, can we have a permutation $S_n$ onto $m$ where $n\neq m$? (for eg: is it possible to have a permutation group say $S_5$ on say 2 elements?)

share|cite|improve this question
Are you asking about the number of elements, or the number of automorphisms? An automorphism of $S_3$ is a group homomorphism $f\colon S_3\to S_3$ that is one-to-one and onto. (It just so happens that there is a bijection between this group and $S_3$, but that is non-trivial, certainly not a matter of just listing elements). I don't understand your last question. – Arturo Magidin Mar 25 '12 at 4:41
By definition, $S_n$ means the group of all permutations of $\{1,2,\ldots,n\}$. "A group $S_5$ on 2 elements" is unclear at best. You are still not making much sense, and you are still not clarifying whether you are trying to count elements or automorphisms. This suggests that you are very confused, regardless of anything else. – Arturo Magidin Mar 25 '12 at 4:48
I updated the question. Yes I am talking of automorphisms. So this number is the number of bijections we can have on $S_n$ ? – codejammer Mar 25 '12 at 4:49
If you are talking about "automorphisms", then what you wrote is useless. Your list is giving the elements of $S_3$; how are you relating an element to an automorphism, and how you are you showing that what you wrote are all the automorphisms? It is a theorem of Otto Holder that $\mathrm{Aut}(S_n)\cong S_n$ for all $n\neq 2,6$. – Arturo Magidin Mar 25 '12 at 4:51
"All possible mappings of $S_n$ to $n$ elements" is nonsense, and it does not help you in finding the automorphism. An automorphism is a group homomorphism from $S_n$ to itself. Not a "map from $S_n$ to $n$ elements" (which would mean a set theoretic function $S_n\to \{1,2,\ldots,n\}$). Nothing you wrote in the last comment makes any sense in the context of this question. – Arturo Magidin Mar 25 '12 at 5:15
up vote 5 down vote accepted

What you did doesn't really work: you are listing elements of $S_3$, instead of automorphisms of $S_3$. While there is a natural isomorphism between the two, I suspect that if you are confusing elements and automorphisms, you are not expected to know this yet.

Try this:

  1. If $f\colon S_3\to S_3$ is a group homomorphism, and you know what $f(12)$ and $f(123)$ are, then you know $f(\sigma)$ for all $\sigma\in S_3$.

  2. If $f,g\colon S_3\to S_3$ are group homomorphisms and have the same values at $(12)$ and at $(123)$, then $f=g$.

  3. $f(12)$ must be an element of order $2$. So there are, at most, three possibilities.

  4. $f(123)$ must be an element of order $3$, so there are at most two possibilities.

  5. Conclude that there are at most six automorphisms of $S_3$.

  6. Exhibit six different automorphisms of $S_3$ (Hint: Conjugation...)

share|cite|improve this answer
@codejammer: Please unaccept the answer. From your comments, it is clear you do not understand what is going on. Accepting the answer when you are clearly still so confused seems... dishonest. – Arturo Magidin Mar 25 '12 at 5:16
Plus one for this comment. – Rudy the Reindeer Mar 25 '12 at 9:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.