> I know that > G/Z(G) = Inn(G), Out(G) = Aut(G)/Inn(G).
> But I can't figure out why Aut(S_6) = S_6 \semidirect C_2 > Aut(S_n) = S_n, for n>7.
Should be n>6.
This is nontrivial. That Aut(S_n)=S_n for n>7 follows by looking at the conjugacy classes: any automorphism must send conjugacy classes to conjugacy classes. A simple count shows that any automorphism of S_n with n>6 must fix the conjugacy class of the transpositions, and then you can leverage that to a proof. It will also show that there is a possible non-inner automorphism for S_6. Constructing it is not obvious, however. -- Arturo Magidin
> > I know that > > G/Z(G) = Inn(G), Out(G) = Aut(G)/Inn(G).
> > But I can't figure out why Aut(S_6) = S_6 \semidirect C_2 > > Aut(S_n) = S_n, for n>7.
> Should be n>6.
> This is nontrivial. That Aut(S_n)=S_n for n>7 follows by looking at > the conjugacy classes: any automorphism must send conjugacy classes to > conjugacy classes. A simple count shows that any automorphism of S_n > with n>6 must fix the conjugacy class of the transpositions, and then > you can leverage that to a proof. It will also show that there is a > possible non-inner automorphism for S_6. Constructing it is not > obvious, however. > -- > Arturo Magidin
I have wondered about this inactively since grad school. Do you have a reference for this result? I also noticed that the OP left out S_3, S_4, and S_5. I suspect they are not difficult cases as it is quite easy to get one's hands on all the elements, but for completeness, it would be nice to know.
> > > I know that > > > G/Z(G) = Inn(G), Out(G) = Aut(G)/Inn(G).
> > > But I can't figure out why Aut(S_6) = S_6 \semidirect C_2 > > > Aut(S_n) = S_n, for n>7.
> > Should be n>6.
> > This is nontrivial. That Aut(S_n)=S_n for n>7 follows by looking at > > the conjugacy classes: any automorphism must send conjugacy classes to > > conjugacy classes. A simple count shows that any automorphism of S_n > > with n>6 must fix the conjugacy class of the transpositions, and then > > you can leverage that to a proof. It will also show that there is a > > possible non-inner automorphism for S_6. Constructing it is not > > obvious, however.
> I have wondered about this inactively since grad school. Do you have > a reference for this result? I also noticed that the OP left out S_3, > S_4, and S_5. I suspect they are not difficult cases as it is quite > easy to get one's hands on all the elements, but for completeness, it > would be nice to know.
Aut(S_n) = S_n for n=/=6, and Aut(S_6) is a (non-direct) semidirect product of S_6 by a cyclic group of order 2.
Rotman (Introduction to the Theory of groups, 4th Edition), Lemma 7.4 shows that an automorphism of S_n is inner if and only if it preserves transpositions. From there, you can just count how many elements of order 2 there are in each conjugacy class in S_n to get that in all cases except for n=6 and n=2, there is no conjugacy class of elements of order 2 with the same number of elements as the class of transpositions, which yields the result. He then explicitly constructs an outer automorphism for S_6, but for that I like "Combinatorial structure on the automorphism group of S_6", by T.Y. Lam and David Leep, Expositiones Mathematicae 11 (1993), no. 4, pp. 289-308.
A sketch of the proof of the lemma: let phi be an automorphism that preserves transpositions. It maps (1,2) to some (i,j); let g_2 be conjugation by (1,i)(2,j); then g_2^{-1}phi fixes (1,2). Proceed by induction to assume you have g_r^{-1}...g_2^{-1}phi fixing (1,2),..., (1,r). This map still preserves transpositions, and must sned (1,r+1) to some (t,v). But (1,2) cannot be disjoint from (t,v), because then (1,2)(1,r+1) would map to (1,2)(t,v) of order 2, instead of something of order 3. So (1,r+1) maps to either (1,k) or to (2,k) for some k. Then k>r; now let g_{r+1} be conjugation by (k,r+1), so g_{r+1]^ {-1}...g_2^{-1}phi fixes (1,2),...,(1,r+1). Inductively, you can compose phi with enough conjugations so that you get a map that fixes every transposition, and hence is the identity. Thus, phi is a conjugation.
> Aut(S_n) = S_n for n=/=6, and Aut(S_6) is a (non-direct) semidirect > product of S_6 by a cyclic group of order 2.
Note that S_2 has two elements, hence Aut(S_2) = {1}, so there is another exception missing there. The OP misreported what Wikipedia says: it says Aut(S_2) is trivial, but that S_2 is isomorphic to C_2.
> > > > I know that > > > > G/Z(G) = Inn(G), Out(G) = Aut(G)/Inn(G).
> > > > But I can't figure out why Aut(S_6) = S_6 \semidirect C_2 > > > > Aut(S_n) = S_n, for n>7.
> > > Should be n>6.
> > > This is nontrivial. That Aut(S_n)=S_n for n>7 follows by looking at > > > the conjugacy classes: any automorphism must send conjugacy classes to > > > conjugacy classes. A simple count shows that any automorphism of S_n > > > with n>6 must fix the conjugacy class of the transpositions, and then > > > you can leverage that to a proof. It will also show that there is a > > > possible non-inner automorphism for S_6. Constructing it is not > > > obvious, however.
> > I have wondered about this inactively since grad school. Do you have > > a reference for this result? I also noticed that the OP left out S_3, > > S_4, and S_5. I suspect they are not difficult cases as it is quite > > easy to get one's hands on all the elements, but for completeness, it > > would be nice to know.
> Aut(S_n) = S_n for n=/=6, and Aut(S_6) is a (non-direct) semidirect > product of S_6 by a cyclic group of order 2.
> Rotman (Introduction to the Theory of groups, 4th Edition), Lemma 7.4 > shows that an automorphism of S_n is inner if and only if it preserves > transpositions. From there, you can just count how many elements of > order 2 there are in each conjugacy class in S_n to get that in all > cases except for n=6 and n=2, there is no conjugacy class of elements > of order 2 with the same number of elements as the class of > transpositions, which yields the result. He then explicitly > constructs an outer automorphism for S_6, but for that I like > "Combinatorial structure on the automorphism group of S_6", by T.Y. > Lam and David Leep, Expositiones Mathematicae 11 (1993), no. 4, pp. > 289-308.
> A sketch of the proof of the lemma: let phi be an automorphism that > preserves transpositions. It maps (1,2) to some (i,j); let g_2 be > conjugation by (1,i)(2,j); then g_2^{-1}phi fixes (1,2). Proceed by > induction to assume you have g_r^{-1}...g_2^{-1}phi fixing (1,2),..., > (1,r). This map still preserves transpositions, and must sned (1,r+1) > to some (t,v). But (1,2) cannot be disjoint from (t,v), because then > (1,2)(1,r+1) would map to (1,2)(t,v) of order 2, instead of something > of order 3. So (1,r+1) maps to either (1,k) or to (2,k) for some k. > Then k>r; now let g_{r+1} be conjugation by (k,r+1), so g_{r+1]^ > {-1}...g_2^{-1}phi fixes (1,2),...,(1,r+1). Inductively, you can > compose phi with enough conjugations so that you get a map that fixes > every transposition, and hence is the identity. Thus, phi is a > conjugation.
> -- > Arturo Magidin-
Thanks for the info. I will study it when I get a chance, hopefully soon. I was always a big fan of Professor Lam, but I am surprised that wrote an article on group theory. Of course I was long gone from Berkeley by 1993.
A first step to understanding is being careful; the first line is incorrect, and does not appear in Wikipedia: Aut(S_2)={1}, but S_2 is isomorphic to C_2. Either you misread, or you misunderstood what you read. You have shown a tendency to carelessness in several threads: that is something you need to guard against if you want to understand and succeed.
I borrowed the book Rotman's group theory 3rd edition from my school library.
I am currently reading the lemma you mentioned.
"... there is no conjugacy class of elements of order 2 with the same number of elements as the class of transpositions, which yields the result"
I am having hard time understanding the above.
What I understand is,
1. S_n can be generated by transpostions. 2. Inner automorphisms preserve cycle types of an element in S_n. 3. S_n can be partitioned to conjugacy classes. Therefor a conjugacy class can be represented by its cycle type. 3. There is a conjugacy class of S_n, which consist of disjoint transpositions only.
In Rotman's book, "Let C_1 be the conjugacy class of S_n consisting of all transpostions,,,; every element of C_1 has order 2" ........ "Let C_k be the conjugacy class consisting of all products of k disjoint transpostions",
So S_n for large n, C_2 = {(i_1 i_2)(i_3 i_4), (i_5 i_6)(i_7 i_8),,,,,} C_3 = {(j_1 j_2)(j_3 j_4)(j_5 j_6), (j_7 j_8)(j_9 j_10)(j_11 j_12) ,,,
Intuitively, for some k, C_k = C_1, but that is not the case for n=6.
Any simple counter example showing that why C_k \neq C_1 ?
> A first step to understanding is being careful; the > first line is > incorrect, and does not appear in Wikipedia: > Aut(S_2)={1}, but S_2 is > isomorphic to C_2. Either you misread, or you > misunderstood what you > read. You have shown a tendency to carelessness in > several threads: > that is something you need to guard against if you > want to understand > and succeed.
> -- > Arturo Magidin
Thanks for your advice :) I checked wiki again, and you are correct. Hope I dont make this kind of mistakes again.
> I borrowed the book Rotman's group theory 3rd edition from my school library.
> I am currently reading the lemma you mentioned.
> "... there is no conjugacy class of elements of order 2 with the same number of elements as the class of transpositions, which yields the result"
> I am having hard time understanding the above.
> What I understand is,
> 1. S_n can be generated by transpostions. > 2. Inner automorphisms preserve cycle types of an element in S_n. > 3. S_n can be partitioned to conjugacy classes. Therefor a conjugacy class can be represented by its cycle type. > 3. There is a conjugacy class of S_n, which consist of disjoint transpositions only.
Wby do you have "disjoint" here? It is false. The conjugacy class of S_n that contains (1,2) will also contain (1,3).
> In Rotman's book, > "Let C_1 be the conjugacy class of S_n consisting of all transpostions,,,; every element of C_1 has order 2" > ........ > "Let C_k be the conjugacy class consisting of all products of k disjoint transpostions",
> So S_n for large n, > C_2 = {(i_1 i_2)(i_3 i_4), (i_5 i_6)(i_7 i_8),,,,,}
No; C_2 would contain (1,2)(3,4), and (1,3)(2,4), and (1,4)(2,3), and lots of other things. You notation here is bad because it is likely to confuse you.
If you understand what you are writing, then you realize how utterly silly what you have just written is. C_k consists of all elements of C_k that are the product of EXACTLY k disjoint transpositions. C_k = C_r if and only if k=r.
The point Rotman makes is that for n>2, the SIZE of C_1 is different from the size of every *other* C_k, except in the case of n=6 (when C_1 and C_3 happen to have the same size).
> Any simple counter example showing that why C_k \neq C_1 ?
You are being careless again. No wonder you are having trouble understanding! THEY ARE ALL DIFFERENT! C_i contains *exactly* the elements that can be written as a product of i disjoint transpositions. They are ALL DIFFERENT from elements that are the product of k disjoint transpositions for k=/=i. I mean: if sigma is a product of k disjoint transpositions, then how many elements does it "move"?
I also find your lack of initiative here troubling. Why did you not try some small values of n? They would have quickly revealed to you your carelessness and confusion.
In case someone is interested, here is some data of conjugacy classes of S_6,
In S_6,
* the class of (a,b) has 15 elements, * the class of (a,b)(c,d) has 45 elements, * the class of (a,b)(c,d)(e,f) has 15 elements, * the class of (a,b,c) has 40 elements, * the class of (a,b,c)(d,e,f) has 40 elements * the class of (a,b,c,d) has 90 elements * the class of (a,b,c,d)(e,f) has 90 elements * the class of (a,b,c,d,e) has 144 elements * the class of (a,b,c,d,e,f) has 120 elements
Notice that the class of (a,b) and the class of (a,b)(c,d)(e,f) have same 15 elements, and an automorphism takes a transposition to a product of three disjoint transpostions (order two element--->order two element).
> In case someone is interested, here is some data of conjugacy classes of S_6,
> In S_6,
> * the class of (a,b) has 15 elements, > * the class of (a,b)(c,d) has 45 elements, > * the class of (a,b)(c,d)(e,f) has 15 elements, > * the class of (a,b,c) has 40 elements, > * the class of (a,b,c)(d,e,f) has 40 elements > * the class of (a,b,c,d) has 90 elements > * the class of (a,b,c,d)(e,f) has 90 elements > * the class of (a,b,c,d,e) has 144 elements > * the class of (a,b,c,d,e,f) has 120 elements
> Notice that the class of (a,b) and the class of (a,b)(c,d)(e,f) have same 15 elements, and an automorphism takes a transposition to a product of three disjoint transpostions (order two element--->order two element).
Notice that you are (yet again!) being horribly careless.
No, it's not true that "an automorphism takes a tranposition to a product of three disjoint transpositions". For example, *none* of the inner automorphisms do that.
What is true is that an automorphism *must* send the class of a transposition to some conjugacy class that has (i) the same number of elements; and (ii) the orders of the elements of that conjugacy class is equal to the orders of the elements of the original.
Now, in the case of S_6, you have three conjugacy classes of elements of order 2: the transpositions, the products of two disjoint transpositions, and the products of three disjoint transpositions. So an automorphism of S_6 must permute these three conjugacy classes.
Since the class of products of two disjoint transpositions has a different number of elements from the other two, that class must be fixed by any automorphism (as a set). So an automorphism of S_6 *either* sends transpositions to transpositions, or else it sends each transposition to a product of three disjoint tranpositions (and any product of three disjoint transpositions to a transposition). The latter would necessarily be a non-inner automorphism.
This observation, by itself, does *not* establish that any automorphism that sends tranpositions to transpositions must be inner (that's what the Lemma from Rotman's book does); *nor* does it establish that there *is* a non-inner automorphism (or how many there are). It only leaves the possiblity open. This possibility does not exist in any other S_n because there is no conjugacy class of elements of order 2 that has the same number of elements as the conjugacy class of tranpositions.
So at this point, you have not established *anything*, let alone the *false* statement that "an automorphism takes a transposition to a product of three disjoint transpostions."
On Nov 3, 3:05 pm, Al2009 <algebra_whate...@yahoo.ca> wrote:
> I am trying to understand some automorphism groups of symmetric groups.
Oh, what the hell; I need to kill 10 minutes before lunch....
First: let G be a group; two elements x and y are "conjugate" if and only if there exists g in G such that gxg^{-1} = y. It is easy to check that being conjugate is an equivalence relation on G, so that it induces a partition of G into equivalence classes, called "conjugacy classes."
If G is a group, and x is conjugate to y, then the order of x is the same as the order of y: for (gxg^{-1})^n = gx^ng^{-1}.
If G is a group, and f:G-->G is an automorphism, then x is conjugate to y if and only if f(x) is conjugate to f(y): for if gxg^{-1}=y, then f(g)f(x)f(g)^{-1}=f(y), so f(x) is conjugate to f(y). Applying the argument to f(x) and f(y) using f^{-1}:G-->G gives the converse.
Thus, an automorphism of a group must permute the conjugacy classes of G (as sets).
Second: The conjugacy classes of S_n are determined by their disjoint cycle structure. This follows because if sigma=(a_1,...,a_r) is a cycle, and tau is a permutation, then
tau*sigma*tau^{-1} = (tau(a_1),...,tau(a_r))
(we compose permutations right to left, so tau*sigma means "do sigma first, then do tau").
If n=1, then S_1 is trivial, so Aut(S_1) is trivial.
If n= 2, then S_2 = C_2, so Aut(S_2) is trivial.
Assume n>2.
We know that S_n is generated by the transpositions (permutations of the form (i,j), with i=/=j). Thus, an automorphism of S_n is completely determined by what it does to the transpositions.
So, let n>2, let f be an automorphism of S_n, and let us try to figure out what kind of permutation f(i,j) can be.
Since f maps conjugacy classes to conjugacy classes, the image of (i,j) will have the same disjoint cycle structure as the image of (r,s), for any i=/=j and any r=/=s. Moreover, the image of (i,j) must be of order 2, so f(i,j) will necessarily be an element of order 2; the only elements of order 2 in S_n are the products of disjoint transpositions. So there exists a k (that depends only on f) such that f(i,j) is a product of k disjoint transpositions.
How many elements in S_n are the product of k disjoint tranpositions? We must select 2 elements out of n for the first transposition; then 2 elements out of the remaining n-2; then 2 out of the remaining n-4;... and finally 2 out of the remaining (n-2k+2). But the order in which we select the pairs is irrelevant, so there are k! ways in which we can pick the pairs. Thus, the total number of elements of S_n that are the product of k disjoint transpositions is
Now, if k>=4, then (2k-2)! > 2^{k-1}k!: this holds for k=2; and if it holds for k, then (2(k+1)-2)! = (2k)! = 2k(2k-1)(2k-2)!>2k(2k-1)2^{k-1} k! = 2^k(2k-1)k! > 2^k(k+1)(k!) = 2^k(k+1)!. Thus, for k>=4, we cannot have 2^{k-1}*k! = (n-2)(n-3)(n-4)...(n-2k+2)(n-2k+1).
So any solutions with k>1 will necessarily be k=2 or k=3. k=2 gives
4 = (n-2)(n-3),
or n^2 - 5n+6 = 4, or n^2-5n+2 = 0. This has no integer solutions, so there is no solution with k=2.
If k=3, we get 24 = (n-2)(n-3)(n-4)(n-5). If n>6, then the right hand side is greater than 4! = 24. So the only possible solution is n=6, k=3.
Thus: if n>2 and n=/=6, then the only conjugacy class of elements of order 2 of S_n that has the same number of elements as the class of the transpositions is the class of the transpositions itself. Therefore, if n>2 and n=/=6, then any automorphism must send transpositions to transpositions.
If n=6, then there are two conjugacy classes of elements of order 2 that have the same number of elements: the class of the transpositions, and the class of the products of three disjoint transpositions. Thus, an automorphism of S_6 will *either* send transpositions to transpositions, or will send transpositions to products of three disjoint transpositions, we don't know which.
We have not established that there is an automorphism of S_6 that does the latter; only that there *could* be an automorphism of S_6 that does not send transpositions to transpositions. This is the *only* value of n for which this *could* happen.
Arturo Magidin wrote: > On Nov 4, 10:23 am, Al2009 <algebra_whate...@yahoo.ca> wrote: >> In case someone is interested, here is some data of conjugacy classes of S_6,
>> In S_6,
>> * the class of (a,b) has 15 elements, >> * the class of (a,b)(c,d) has 45 elements, >> * the class of (a,b)(c,d)(e,f) has 15 elements, >> * the class of (a,b,c) has 40 elements, >> * the class of (a,b,c)(d,e,f) has 40 elements >> * the class of (a,b,c,d) has 90 elements >> * the class of (a,b,c,d)(e,f) has 90 elements >> * the class of (a,b,c,d,e) has 144 elements >> * the class of (a,b,c,d,e,f) has 120 elements
>> Notice that the class of (a,b) and the class of (a,b)(c,d)(e,f) have same 15 elements, and an automorphism takes a transposition to a product of three disjoint transpostions (order two element--->order two element).
> Notice that you are (yet again!) being horribly careless.
> No, it's not true that "an automorphism takes a tranposition to a > product of three disjoint transpositions". For example, *none* of the > inner automorphisms do that.
> What is true is that an automorphism *must* send the class of a > transposition to some conjugacy class that has (i) the same number of > elements; and (ii) the orders of the elements of that conjugacy class > is equal to the orders of the elements of the original.
> Now, in the case of S_6, you have three conjugacy classes of elements > of order 2: the transpositions, the products of two disjoint > transpositions, and the products of three disjoint transpositions. So > an automorphism of S_6 must permute these three conjugacy classes.
> Since the class of products of two disjoint transpositions has a > different number of elements from the other two, that class must be > fixed by any automorphism (as a set). So an automorphism of S_6 > *either* sends transpositions to transpositions, or else it sends each > transposition to a product of three disjoint tranpositions (and any > product of three disjoint transpositions to a transposition). The > latter would necessarily be a non-inner automorphism.
> This observation, by itself, does *not* establish that any > automorphism that sends tranpositions to transpositions must be inner > (that's what the Lemma from Rotman's book does); *nor* does it > establish that there *is* a non-inner automorphism (or how many there > are). It only leaves the possiblity open. This possibility does not > exist in any other S_n because there is no conjugacy class of elements > of order 2 that has the same number of elements as the conjugacy class > of tranpositions.
> So at this point, you have not established *anything*, let alone the > *false* statement that "an automorphism takes a transposition to a > product of three disjoint transpostions."
What one -can- immediately conclude from this is that (if Aut(S_6) is different from Int(S_6), then) the index of Int(S_6) in Aut(S_6) is 2. Because any non-inner automorphism must also map the (12)(34)(56) conjugacy class to that of (12). So the composition of the automorphism with itself maps transpositions to transpositions.
(BTW, Rotman's lemma is rather trivial, not? By composition with inner automorphisms, we can easily see, first that, without loss of generalization, we can assume that f((12)) = (12), then in a similar way that f((23)) = (23), and so on, up to f((56)) = (56), and we know that S_6 is generated by these elements.)
It's also not soooo terribly difficult to come up with a non-inner automorphism. If we define: f((12)) = (12)(34)(56) f((23)) = (23)(45)(61) f((34)) = (13)(24)(56) f((45)) = (16)(25)(34) f((56)) = (14)(23)(56) Then we can easily check that (to put it a bit loosely) all elements that should commute do commute, and relations of the form "order(ab) = 3" hold where they should hold. Now note that S_6 can be characterized as a group generated by five elements of order two plus these relations, and the rest is easy.
But this is not why i post. Here comes my question.
It is claimed not only that Aut(S_6) / Int(S_6) = C_2, but moreover that Aut(S_6) = Int(S_6) x C_2. That implies (correct me if i'm wrong) that there exists one very special non-inner automorphism, namely one (and no more than one) that commutes with all inner automorphisms.
And that 'smells' like there is a very natural construction of this single, very special non-inner automorphism. Perhaps something involving cube, tetraeder, or some other regular polyeder, or so?
Anyone got any ideas on that?
(Of course, it can also be that Aut(S_6) is not this direct product at all, and that the OP (or his source) is simply wrong about this.)
> Arturo Magidin wrote: > > On Nov 4, 10:23 am, Al2009 <algebra_whate...@yahoo.ca> wrote: > >> In case someone is interested, here is some data of conjugacy classes of S_6,
> >> In S_6,
> >> * the class of (a,b) has 15 elements, > >> * the class of (a,b)(c,d) has 45 elements, > >> * the class of (a,b)(c,d)(e,f) has 15 elements, > >> * the class of (a,b,c) has 40 elements, > >> * the class of (a,b,c)(d,e,f) has 40 elements > >> * the class of (a,b,c,d) has 90 elements > >> * the class of (a,b,c,d)(e,f) has 90 elements > >> * the class of (a,b,c,d,e) has 144 elements > >> * the class of (a,b,c,d,e,f) has 120 elements
> >> Notice that the class of (a,b) and the class of (a,b)(c,d)(e,f) have same 15 elements, and an automorphism takes a transposition to a product of three disjoint transpostions (order two element--->order two element).
> > Notice that you are (yet again!) being horribly careless.
> > No, it's not true that "an automorphism takes a tranposition to a > > product of three disjoint transpositions". For example, *none* of the > > inner automorphisms do that.
> > What is true is that an automorphism *must* send the class of a > > transposition to some conjugacy class that has (i) the same number of > > elements; and (ii) the orders of the elements of that conjugacy class > > is equal to the orders of the elements of the original.
> > Now, in the case of S_6, you have three conjugacy classes of elements > > of order 2: the transpositions, the products of two disjoint > > transpositions, and the products of three disjoint transpositions. So > > an automorphism of S_6 must permute these three conjugacy classes.
> > Since the class of products of two disjoint transpositions has a > > different number of elements from the other two, that class must be > > fixed by any automorphism (as a set). So an automorphism of S_6 > > *either* sends transpositions to transpositions, or else it sends each > > transposition to a product of three disjoint tranpositions (and any > > product of three disjoint transpositions to a transposition). The > > latter would necessarily be a non-inner automorphism.
> > This observation, by itself, does *not* establish that any > > automorphism that sends tranpositions to transpositions must be inner > > (that's what the Lemma from Rotman's book does); *nor* does it > > establish that there *is* a non-inner automorphism (or how many there > > are). It only leaves the possiblity open. This possibility does not > > exist in any other S_n because there is no conjugacy class of elements > > of order 2 that has the same number of elements as the conjugacy class > > of tranpositions.
> > So at this point, you have not established *anything*, let alone the > > *false* statement that "an automorphism takes a transposition to a > > product of three disjoint transpostions."
> What one -can- immediately conclude from this is that (if Aut(S_6) is > different from Int(S_6), then) the index of Int(S_6) in Aut(S_6) is 2. > Because any non-inner automorphism must also map the (12)(34)(56) > conjugacy class to that of (12). So the composition of the automorphism > with itself maps transpositions to transpositions.
> (BTW, Rotman's lemma is rather trivial, not? > By composition with inner automorphisms, we can easily see, first that, > without loss of generalization, we can assume that f((12)) = (12), then > in a similar way that f((23)) = (23), and so on, up to f((56)) = (56), > and we know that S_6 is generated by these elements.)
> It's also not soooo terribly difficult to come up with a non-inner > automorphism. If we define: > f((12)) = (12)(34)(56) > f((23)) = (23)(45)(61) > f((34)) = (13)(24)(56) > f((45)) = (16)(25)(34) > f((56)) = (14)(23)(56) > Then we can easily check that (to put it a bit loosely) all elements > that should commute do commute, and relations of the form "order(ab) = > 3" hold where they should hold. > Now note that S_6 can be characterized as a group generated by five > elements of order two plus these relations, and the rest is easy.
> But this is not why i post. > Here comes my question.
> It is claimed not only that Aut(S_6) / Int(S_6) = C_2, but moreover that > Aut(S_6) = Int(S_6) x C_2.
No: it's not a direct product, but a semidirect product; the action of C_2 on Inn(S_6) is not trivial.
> That implies (correct me if i'm wrong) that there exists one very > special non-inner automorphism, namely one (and no more than one) that > commutes with all inner automorphisms.
Rather, it means that there is a non-inner automorphism that is of order 2, so that the projection Aut(S_6) --> Aut(S_6)/Inn(S_6) splits.
> And that 'smells' like there is a very natural construction of this > single, very special non-inner automorphism. Perhaps something involving > cube, tetraeder, or some other regular polyeder, or so? > Anyone got any ideas on that?
> (Of course, it can also be that Aut(S_6) is not this direct product at > all, and that the OP (or his source) is simply wrong about this.)
The OP correctly quoted his source (for a change!) saying that Aut (S_6) = S_6\semidirect C_2.
> On Nov 4, 1:14 pm, Herman Jurjus <hjm...@hetnet.nl> wrote:
> > Arturo Magidin wrote: > > > On Nov 4, 10:23 am, Al2009 <algebra_whate...@yahoo.ca> wrote: > > >> In case someone is interested, here is some data of conjugacy classes of S_6,
> > >> In S_6,
> > >> * the class of (a,b) has 15 elements, > > >> * the class of (a,b)(c,d) has 45 elements, > > >> * the class of (a,b)(c,d)(e,f) has 15 elements, > > >> * the class of (a,b,c) has 40 elements, > > >> * the class of (a,b,c)(d,e,f) has 40 elements > > >> * the class of (a,b,c,d) has 90 elements > > >> * the class of (a,b,c,d)(e,f) has 90 elements > > >> * the class of (a,b,c,d,e) has 144 elements > > >> * the class of (a,b,c,d,e,f) has 120 elements
> > >> Notice that the class of (a,b) and the class of (a,b)(c,d)(e,f) have same 15 elements, and an automorphism takes a transposition to a product of three disjoint transpostions (order two element--->order two element).
> > > Notice that you are (yet again!) being horribly careless.
> > > No, it's not true that "an automorphism takes a tranposition to a > > > product of three disjoint transpositions". For example, *none* of the > > > inner automorphisms do that.
> > > What is true is that an automorphism *must* send the class of a > > > transposition to some conjugacy class that has (i) the same number of > > > elements; and (ii) the orders of the elements of that conjugacy class > > > is equal to the orders of the elements of the original.
> > > Now, in the case of S_6, you have three conjugacy classes of elements > > > of order 2: the transpositions, the products of two disjoint > > > transpositions, and the products of three disjoint transpositions. So > > > an automorphism of S_6 must permute these three conjugacy classes.
> > > Since the class of products of two disjoint transpositions has a > > > different number of elements from the other two, that class must be > > > fixed by any automorphism (as a set). So an automorphism of S_6 > > > *either* sends transpositions to transpositions, or else it sends each > > > transposition to a product of three disjoint tranpositions (and any > > > product of three disjoint transpositions to a transposition). The > > > latter would necessarily be a non-inner automorphism.
> > > This observation, by itself, does *not* establish that any > > > automorphism that sends tranpositions to transpositions must be inner > > > (that's what the Lemma from Rotman's book does); *nor* does it > > > establish that there *is* a non-inner automorphism (or how many there > > > are). It only leaves the possiblity open. This possibility does not > > > exist in any other S_n because there is no conjugacy class of elements > > > of order 2 that has the same number of elements as the conjugacy class > > > of tranpositions.
> > > So at this point, you have not established *anything*, let alone the > > > *false* statement that "an automorphism takes a transposition to a > > > product of three disjoint transpostions."
> > What one -can- immediately conclude from this is that (if Aut(S_6) is > > different from Int(S_6), then) the index of Int(S_6) in Aut(S_6) is 2. > > Because any non-inner automorphism must also map the (12)(34)(56) > > conjugacy class to that of (12). So the composition of the automorphism > > with itself maps transpositions to transpositions.
> > (BTW, Rotman's lemma is rather trivial, not? > > By composition with inner automorphisms, we can easily see, first that, > > without loss of generalization, we can assume that f((12)) = (12), then > > in a similar way that f((23)) = (23), and so on, up to f((56)) = (56), > > and we know that S_6 is generated by these elements.)
> And that would be the proof that Rotman gives...
> > It's also not soooo terribly difficult to come up with a non-inner > > automorphism. If we define: > > f((12)) = (12)(34)(56) > > f((23)) = (23)(45)(61) > > f((34)) = (13)(24)(56) > > f((45)) = (16)(25)(34) > > f((56)) = (14)(23)(56) > > Then we can easily check that (to put it a bit loosely) all elements > > that should commute do commute, and relations of the form "order(ab) = > > 3" hold where they should hold. > > Now note that S_6 can be characterized as a group generated by five > > elements of order two plus these relations, and the rest is easy.
> > But this is not why i post. > > Here comes my question.
> > It is claimed not only that Aut(S_6) / Int(S_6) = C_2, but moreover that > > Aut(S_6) = Int(S_6) x C_2.
> No: it's not a direct product, but a semidirect product; the action of > C_2 on Inn(S_6) is not trivial.
In fact, this is trivial: remember that if x is an element of G, and varphi_x is conjugation by x, and f is an automorphism, then
f*varphi_x*f^{-1} (g) = f(xf^{-1}(g)x^{-1}) = f(x) g f(x)^{-1} = varphi_{f(x)}(g).
So if f is in the centralizer of Inn(G) in Aut(G), then for all g in G you must have x congruent to f(x) modulo Z(G). That is, f must induce the identity map on G/Z(G).
In the case of S_6, since Z(S_6) = {1}, it would follow that x=f(x) for all x, hence f is the identity. So the only automorphism that commutes with every inner automorphism of S_6 is the identity. The same holds for any centerless group.
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