Date: Mon, 15 Apr 1996 18:13:26 +0100
From: neumann@vax.ox.ac.uk
Subject: RE: 120 squared
Dear Roger,
I'm too ignorant of {5,3,3} to be able to answer your question.
I suppose I could look it up in my copy of Coxeter's book on
Regular Polytopes, but I don't have time just now. So can you
tell me:
(1) is its symmetry group a subgroup of O(4) ?
(2) and does the symmetry group act transitively on the set of 120
dodecahedral faces?
If the answer to my first query is yes, then I think the answer
to your first question is no. For, the group G of symmetries of
the regular dodecahedron is isomorphic to Alt(5) \times Z_2
(where the Z_2 is generated by the central reflection).
Therefore G \times G is isomorphic to Alt(5)^2 \times (Z_2)^2.
But this group has no faithful linear representation of degree 4
(over any field).
If the answer to my second query is yes, then I think the answer
to your first question is no. For, I rather think that Alt(5)^2
\times (Z_2)^2 has no faithful transitive permutation
representation of degree 120. Oh no---that is nonsense. It does
have such a representation. Namely, on the cosets of Alt(4) \times
D_10. Even so, that does not look plausible as a permutation
representation of interest in geometry.
You also ask what non-abelian groups of order 120^2 there are.
An answer, but perhaps not a very helpful one, is lots and lots
and lots. There must be something like ten or a dozen with
composition factors Alt(5), Alt(5), Z_2, Z_2. I've no time to
work those out just now. And since there are 14 possibilities
for Sylow 2-subgroups there'll be lots of soluble groups.
There'll also be lots with just one non-abelian composition
factor (which could be Alt(5) or Alt(6)).
All best wishes, $\Pi$eter
Queen's: 15.iv.96
---
Date: Tue, 16 Apr 1996 08:38:53 +1200
From: Marston Conder
[a message containing a minor error, now removed to avoid
confusion; see further on for Marston's correction.- ed.]
----
From: "Ken W. Smith"
Subject: Re: 120 squared
Date: Mon, 15 Apr 1996 21:29:54 +0000
Hi Roger,
I see that Marston Conder answered your first question. As for your
second question: I doubt if the answer is tractable. There are MANY groups
of order 120 and so VERY MANY nonabelian groups of order 120^2. There must
be thousands, probably tens of thousands.
There are 257 groups of order 64 and so there are 257 groups of the
form G x H where H is cyclic of order 225 and G has order 64. These would
presumably be very rare among the groups of order 120^2.
I'd be interest in just knowing how many groups of order 120 there are!
(BTW, thanks all for your answers to my earlier posts on nonsolvable
groups of order 120!)
Yours,
Ken
---
Date: Tue, 16 Apr 96 13:06 MET DST
Subject: Re: 120 squared
Cc: obrien@math.rwth-aachen.de
[.. in response to Ken Smith (ed.) ]
According to two independent sources, the number of groups
of order 120 is 47. Of these, three are insoluble and
three are abelian.
The first source is
J.K. Senior and A.C. Lunn (1934), Determination of groups
of order 101-161. Amer. J. Math. 56, 328--338.
They determine the number and provide sufficient
information to allow an interested reader to
write down presentations.
The second source is a teacher-trainee student,
Volker Eggers, of Joachim Neubueser. Eggers independently
determined presentations for these groups in the early
1960s in Kiel and also found 47 such groups.
By the way, there are 267 groups of order 64. Also, since
the number of groups of order 120 is 47, a lower bound
on the number of groups of order 120^2 is 1128.
There is a second paper by Senior and Lunn from 1935
in Volume 35 of the same journal which continues
to determine the number of groups for orders up to 215,
with some exceptions. Some time ago, I prepared
a bibliography on sources for group determinations
which I'm happy to supply on request.
Eamonn O'Brien
obrien@math.rwth-aachen.de
--
Date: Wed, 24 Apr 96 17:56:53 0100
From: "Robert A. Wilson"
[pointed out there was a glitch in the article excised from above - ed.]
--
Date: Fri, 26 Apr 1996 06:53:49 +1200
From: Marston Conder
To: group-pub-forum@maths.bath.ac.uk
Subject: Re: 120 squared
It's a fair cop! As Peter Neumann and Rob Wilson have both pointed out,
the group of symmetries of the regular dodecahedron is A_5 x C_2, not S_5.
Thus my email answer to the "120 squared" question should have been this:
------------------------------------------------------------------------------
Dear Roger
>Are the following two groups of order 120^2 distinct? The group of
>symmetries of the regular 4-polytope {5,3,3} which has 120 dodecahedral
>faces and GXG where G is the group of symmetries of the regular
>dodecahedron? What non abelian groups of that order are there?
No, the group of symmetries of the regular 4-polytope {5,3,3} is the
Coxeter group [3,3,5], which is an extension by C_2 of a central product of
two copies of SL(2,5), where the C_2 is generated by an involution which
interchanges the two factors.
In particular, the centre of this group has order 2, while the centre of
the direct product (A_5 x C_2) x (A_5 x C_2) has order 4.
More easily, the Coxeter group [3,3,5] has presentation
,
from which it is easy to see that its abelianisation is C_2, while clearly
the abelianisation of (A_5 x C_2) x (A_5 x C_2) is C_2 x C_2.
Best wishes
Marston Conder