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From: Al2009 on 4 Nov 2009 01:23
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

I got this data from
http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=3017.0001.0001.0001

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).
From: Arturo Magidin on 4 Nov 2009 12:30
On Nov 4, 10:23 am, Al2009 <algebra_whate...(a)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
>
> I got this data fromhttp://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg...
>
> 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."

--
Arturo Magidin
From: Arturo Magidin on 4 Nov 2009 13:15
On Nov 3, 3:05 pm, Al2009 <algebra_whate...(a)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

(n choose 2)*(n-2 choose 2)* .... * (n-2k+2 choose 2)/k!

(if 2k>n, then one of the choice coefficients will be equal to 0, so
you get 0 for the product, which is correct).

That means that if f maps a transposition to a product of k disjoint
transpositions, then we must have that

(n choose 2) = (n choose 2)*(n-2 choose 2)* .... * (n-2k+2 choose 2)/
k!

If k=1, you have a solution. So assume k>1. Then the equation is
equivalent to

1 = (n-2 choose 2)* .... * (n-2k+2 choose 2)/k!,

or that k! = (n-2 choose 2)* .... * (n-2k+2 choose 2).

which gives 2^{k-1}*k! = (n-2)(n-3)(n-4)...(n-2k+2)(n-2k+1).

Since 2^{k-1}k! is positive, we must have 2k<=n. So

(n-2)(n-3)...(n-2k+1) >= (2k-2)(2k-3)...(2k-2k+1) = (2k-2)!.

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
From: Herman Jurjus on 4 Nov 2009 14:14
Arturo Magidin wrote:
> On Nov 4, 10:23 am, Al2009 <algebra_whate...(a)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
>>
>> I got this data fromhttp://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg...
>>
>> 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.)

--
Cheers,
Herman Jurjus

From: Arturo Magidin on 4 Nov 2009 14:29
On Nov 4, 1:14 pm, Herman Jurjus <hjm...(a)hetnet.nl> wrote:
> Arturo Magidin wrote:
> > On Nov 4, 10:23 am, Al2009 <algebra_whate...(a)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
>
> >> I got this data fromhttp://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg...
>
> >> 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.

> 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.

--
Arturo Magidin
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