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From: Arturo Magidin on 4 Nov 2009 14:39
On Nov 4, 1:29 pm, Arturo Magidin <magi...(a)member.ams.org> wrote:
> 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.

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.

--
Arturo Magidin
From: Herman Jurjus on 4 Nov 2009 14:48
Arturo Magidin wrote:
>
> The OP correctly quoted his source (for a change!) saying that Aut
> (S_6) = S_6\semidirect C_2.

Ah; that makes more sense.
Excuse the interruption - please proceed.

--
Cheers,
Herman Jurjus
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