From: Dave Rusin
Subject: Re: Q:group extensions
Date: Fri, 7 May 1999 03:54:20 -0500 (CDT)
Newsgroups: [missing]
To: jheckman@my-dejanews.com
Keywords: semidirect products, automorphisms
>> and H has a (normal? I forget) subgroup isomorphic to a subgroup of
>> out(G) (the outer automorphism group of G).
>
>No. Since G is normal in the extension, H, like *any* subgroup of the
>extension, has an 'action' on G, i.e., a homomorphism into aut(G) - not
>necessarily a subgroup isomorphic to one in aut(G). (The smallest
>counter-example I could find is a group of order 48, which is a split
>extension of C2xC2 [of order 4] by the [unique] non-abelian split extension
>of C3 by C4 [of order 12]. This latter group has a homomorphism into
>aut(C2xC2) [=L2(2)=S3=D3, of order 6], but does not contain it as a subgroup.
>The order-48 group in question can perhaps be most easily constructed as the
>[unique] non-trivial split extension [i.e., not a direct product] by C4 of
>A4, the alternating group of degree 4 [of order 12, which is itself the
>unique non-abelian split extension of C2xC2 by C3]. Whew!) Note that a direct
>product is the trivial case of a semi-direct product in which the
>homomorphism of H into aut(G) is onto the identity group.
>
>Also, FWIW, although it seems clear to me that the homomorphism of H into
>aut(G) must in fact be into out(G) (since the intersection of G and H is
>trivial), I haven't yet been able to actually prove it (not that I've tried
>that hard); I'm sure it's somewhere in one of my dusty books on group theory,
>which I've really got to crack open again one of these days...
Careful: H itself is not uniquely defined; there can be several complements
to G in the semidirect product. From G being normal we have maps to
Aut(G) in every case, and thus maps to Out(G) too. While all the complements
have the same image in Out(G), they need not have the same image in Aut(G).
Trivial example: Let G have a noncentral involution t and form the
direct product G x H where H = has order 2. Then h' = (t,h) is
an element of order 2 in GxH not lying in G, and so generates a complement
H' to G in GxH. But the images of H and H' in Aut(G) are different
(even though their images in Out(G) are the same, namely trivial).
In short: don't focus on the image in Aut(G) unless you really have to.
(You _do_ have to, when constructing the group, but the complement soon
fades to be merely one of many after the group is built.)
>Another
>question that all this brings up in my mind is whether aut(G) is necessarily
>a split extension of inn(G) by out(G) - it seems likely, and I can't think of
>any obvious counter-examples, but again I haven't been able to actually prove
>it, and again I'm sure it's somewhere in one of my books...
Oh, I don't think so. Probably there's an example among the groups of order
2^n for say n=5 or 6. What about that funny autopmorphism of Sym(6)?
How abou the sporadic groups; In know Out(G) is solvable then, but does
it split off? I don't see why it should.
[deletia --djr]
==============================================================================
From: Dave Rusin
Subject: Re: group theory question
Date: Mon, 21 Jun 1999 16:44:39 -0500 (CDT)
Newsgroups: [missing]
To: jheckman@my-deja.com
>In article <7jremt$1mv$1@holly.csv.warwick.ac.uk>,
> mareg@lily.csv.warwick.ac.uk (Dr D F Holt) wrote:
>> I think that a [finite] group G is the normal closure of a single element if
>> and only if G/[G,G] is cyclic.
>
>Derek (or someone else), could I ask you to step through the following
>a little more slowly? I'm having trouble keeping up with parts of it. :(
In general I think requests of this type are better directed via email
to the original poster, rather than to the group. Let me try to comment
on a few of your points to see if this helps you.
>> Let R(G) be the intersection of all maximal normal subgroups of a finite
>> group G. Then G/R(G) is a direct product of simple groups - to see
>> this, note that any minimal normal subgroup of G/R(G) has trivial
>> intersection with some maximal normal subgroup,
>
>Understood, though it's not trivial to prove for someone at my (admittedly
>not advanced) level.
If K is a minimal normal subgroup, then for any normal subgroup
K\intersect N must either be N itself (if K is in N) or {1}.
So the only way K could escape having "trivial intersection with some
maximal normal subgroup" would be if it were contained in every maximal
normal subgroup of G/R(G). But the max-normals of G/R(G) are naturally
in 1-1 correspondence with those of G, whose intersection is R(G),
so the max-normals of G/R(G) meet in {1}. Therefore, assuming K > {1}
we would have a contradiction.
>> and so must be a direct factor of G/[G,G].
>
>Whoa! *What* must be a direct factor of G/G'?
He goofed; your example (A_5 x C_4) shows this.
But it's actually quite trivial that G/G' must be cyclic, since if G is
generated by the set { x^g ; g in G } for some single x in G, then
any homomorphic image phi(G) will be generated by the set
{ phi(x) ^ phi(g) : g in G }. Taking in particular the natural map
phi : G -> G/G' we see G/G' must be generated by the coset of x,
i.e., cyclic.
>> Now, if G/[G,G] is cyclic, then G/R(G) is a direct product of zero or
>> more nonabelian simple groups and a cyclic group.
>
>You've lost me again.
Let's think generally about what G/R(G) looks like in any group G.
It's simply the largest homomorphic image of G in which the set of
maximal normal subgroups meets in {1}. I claim that this is the same as
saying it's the largest homomorphic image which is the direct product of
simple groups, say G/S(G) for short. On the one hand, if G/S(G) is
the product of some simple groups S_i then N_i = prod(S_j, j <> i) is
a maximal normal subgroup of G/S(G) and the set of all these N_i
meets in {1}; in particular, the set of _all_ maxmal normals meets in {1},
so G/R(G) is at least as large as this, that is, R(G) \subseteq S(G).
On the other hand, if N is any minimal normal subgroup of G/R(G),
then as above N meets some maximal subgroup M of G/R(G) trivially,
giving a decomposition G/R(G) = N x M. Now, the maximal normals of
N x M project to M to give either M itself or a maximal normal of M,
and conversely if M' is a maximal normal of M then N x M is a
maxnormal of G/R(G); this correspondence lets you prove that M also
has the property that the set of _its_ maxnormals has trivial intersection.
So, by induction on |M| if you like, you may conclude that M is a
product of simple groups, as is N likewise, so that G/R(G) is a product
of simple groups. Thus G/S(G) is at least as large as this, that is,
S(G) \subseteq R(G).
Comparing the two paragraphs shows R(G) = S(G), that is, R(G) is simply
the smallest normal subgroup of G giving a quotient G/R(G) which is
a direct product of simple groups.
Now, if G/G' is cyclic, then G has NO noncyclic abelian factor groups.
So a decomposition G/R(G) = C_r x C_r x etc. is impossible as this would
make C_r x C_r be a noncyclic abelian factor group of G. So this
decomposition of G/R(G) into a product of simples has no repeats among
the cyclic factors (necessarily of prime order) and so all the cyclic factors
multiply to a single cyclic (usu. non-simple) one.
>> Hence G/R(G) is the normal closure of a single element gR(G) - simply
>> take an element with nontrivial component in each direct factor, and
>> where the component in the cyclic factor generates that factore.
>> Then the normal closure of g in G must be all of G, because otherwise
>> it would be contained in some maximal normal subgroup that would
>> both comtain R(G) and map onto G/R(G), which is impossible.
>
>*Sigh*. I'm having trouble making headway with all of this, too -- but
>maybe I'll grok it once I: (1) play around with the cosets of R(G); and (2)
>understand better the things you lost me on earlier...
Suggestion: forget cosets and focus on homomorphic images. They're the
same up to natural isomorphism, but it's a lot easier to me thinking about
a group of, um, elements than a group of cosets. He's just observing that
if Gbar ( = G/R(G) ) is a group which happens to have the form
Gbar = x S_1 x S_2 x ... with each S_i simple, then the element
g = (a, x1, x2, ...) in this direct product has the property that you want
(it and its conjugates generate all of Gbar) as long as each x_i isn't
the identity element. Indeed, since S_i has trivial center, there's a
y_i which doesn't commute with x_i, so let h_i = (1, 1, ..., y_i, 1, ...,1)
and observe that with just g and g^h_i we can generate the element
(1, 1, ..., [x_i, y_i], 1, ..., 1), that is, a non-trivial element of S_i.
But as you know the set generated by _all_ cojugates of g will be
normal in Gbar, so its intersection with S_i is normal in S_i, and
now not just the trivial subgroup of S_i, so it contains all of S_i.
Since this is true for every i, it contains all of Gbar.
>> > Also, what happens if I ask whether a group is generated by all the
>> > automorphic images of a single element, i.e. whether there exists
>> > g in G such that G = < {f(g) | f in Aut(G)} >. Are there any groups
>> > for which this is not the case?
>>
>> Yes there are groups for which this is not the case - for example
>> S_5 x S_6.
>>
>> Derek Holt
>
>OK, let me go out on a limb again. I took Cris's question to be equivalent
>to: Is there any finite group G, *all* of whose elements are contained in
>*proper*, *characteristic* (i.e., Aut(G)-invariant) subgroups? If this is
>what you thought he meant, too, I'd appreciate it if you (or someone)
>could fill in more detail on how S_5 x S_6 satisfies this criterion, since
>I'm having trouble seeing it for myself. (Actually, I'm having trouble
>believing the answer to his question isn't "no".)
Write this "S_5 x S_6" as G = G1 x G2 where G1 = S_5, G2 = S_6.
If phi is an automorphism of G, write it as phi=(phi1,phi2) with
phi_i = homomorphism from G to G_i. (That is, for every g in G,
phi(g) = (g1,g2) for some g_i, but you easily check that the mapping
g |-> g1 is a homomorphism, as is g |-> g2.) Observe that G1 and G2
commute with each other, so their images under any homomorphism also commute.
In particular, phi1(G1) and phi1(G2) have to be two commuting subgroups
of G1 = S_5. Now, phi1(G_i) is also a homomorphic image of G_i, and
so in this case must be either {1}, Z/2Z, or G_i itself. Clearly
phi1(G_2)=G_2 is impossible as |G2| > |G1|. If phi1(G_1)=G_1 then
phi1(G_2)=Z/2Z is also impossible because there's no element of order 2
in the center of G1. Likewise there is no subgroup of G2 which is
isomorphic to G1 and which commutes with an element of order 2 in G2.
So only these combinations remain for homormorphisms G -> G:
phi1(G1) phi1(G2)
1 1
1 Z/2Z
Z/2Z 1
Z/2Z Z/2Z
G1 1
phi2(G1) phi2(G2)
1 1
1 Z/2Z
1 G2
Z/2Z 1
Z/2Z Z/2Z
G1 1
Since the kernels of phi1 and phi2 must intersect trivially if phi is
to be an automorphism, we see from the second columns that phi2 must be
an injection on G2 and thus (reading the rest of that row) that phi2 must
send G1 to {1} and thus (reading the 1st column) phi1 is an injection on G1
and thus (reading the rest of that row) phi1 sends G2 to {1}.
In short: Aut(S_5 x S_6) = Aut(S_5) x Aut(S_6).
In particular, all automorphisms of S_5 x S_6 act trivially on its
abelianization = C_2 x C_2, which is not generated by any single element.
So no element of S_5 x S_6, together with all images under all automorphisms,
can generate any more than a subgroup of index 2 in S_5 x S_6.
>FWIW, someone else in this thread suggested C_4 x C_2, which clearly
>does *not* satisfy the criterion, since it's generated by the 4 automorphic
>images of any generator of its two subgroups isomorphic to C_4 (which
>subgroups, though normal, are mapped into each other by Out(G)).
Yes, you're right.
dave