Resent-Date:  Sat, 15 Jul 95 19:19:28 BST

From: Gotz Pfeiffer 

Larry Grove wrote

            Here are two groups:
    G = ,
    H = .
            Both have order 605, they are not isomorphic.  Can anyone find
    any "group-theoretical" (whatever that means) difference between the
    two? 

The structure of both groups clearly is 11^2 : 5, where the 5 acts as
the matrix (ie. element of GL(2, 11))

       [4  0]                               [4  0]
  C =  [0  3]  (mod 11)   in G   and    Z = [0  5]  (mod 11)   in H.

Note that det C = 4 * 3 = 12 = 1 (mod 11) and det Z = 4 * 5 = 20 = 9 (mod
11).  Hence C lies in SL(2, 11) while Z doesn't.  In other words:
G is a subgroup of 11^2 : SL(2, 11), and H is but a subgroup of 
11^2 : GL(2, 11).

Two more remarks on G and H: as can be readily checked in GAP, the
character tables of these groups are different.  Their tables of
marks, however, coincide.

Best wishes,
Goetz.

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Resent-Date:  Sun, 16 Jul 95 15:14:41 BST

From: Werner Nickel 

As G"otz has pointed out, the action of the generators c and z can be
represented by matrices C and Z in GL(2,11), respectively. 

The centralizer of C and the centralizer of Z in GL(2,11) is the
subgroup of GL(2,11) consisting of the diagonal matrices.  This
is also the normalizer of  in GL(2,11).  The matrix 

        [ 0  1 ]
        [ 1  0 ]

normalizes the subgroup generated by C.  This matrix together with the
centralizer of C generates the normalizer of .

A consequence of this is that the automorphism group of G is twice as
large as the automorphism group of H.  A straightforward computation
with GAP shows that  Aut(G)  has order 24200 while  Aut(H) has order
12100. 

Cheers, Werner.

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Date: Mon, 17 Jul 1995 14:32:08 +1000
Geoff Smith 

Larry Grove asked for "group-theoretical" (whatever that means) 
differences between 
G = ,
and
H = .
-- and both  G\"otz Pfieffer and Werner Nickel provided sensible
answers.

Of course there are many possible answers --  including the fatuous:
For any group X consider the subgroup of X generated by all 
subgroups isomorphic to G etc.

However, I suspect what Larry wants is a "natural" invariant"
which discriminates between G and H and in that sense G\"otz's
and Werner's postings are less silly.

If you want to see some *really* similar non-isomorphic groups
may I commend Eamonn O'Brien's list of 2-groups of order
dividing 256. A few years ago Huseyin Aydin tried to
discriminate (using natural invariants) between members of families of 
these groups which Eamonn O'Brien suggested might be very similar
(see reference).

There are some pairs of groups which are remarkably similar --
not only are the automorphism groups similar -- but
even the power maps (x -> x^2) have the same shape. 

Geoff Smith

Reference:
H. AYDIN ``On Similar Distinct Finite 2-Groups'', 
Bath Mathematics and Computer Science Technical Report 90-42 (1990) 
 
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Resent-Date:  Mon, 17 Jul 95 8:19:40 BST

From: MANN@vms.huji.ac.il

I want to draw the attention of this list to the paper `A Complete Set of
Invariants for Finite Groups and Other Results', by M.Roitman, Adv. Math. 41
(1981), 301-311. Let n(X) be the number of subgroups of X. Then one of the
results of this paper reads:
Let G and H be finite groups such that n(L >< H) = n(L >< G), for each group
L that is a subgroup of a quotient group of G. Then G and H are isomorphic.
                                                          Avinoam Mann

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Resent-Date:  Mon, 17 Jul 95 12:05:07 BST

From: "A.Camina" 

> Dear Pub-gang,
>       Here are two groups:
> G = ,
> H = .
>       Both have order 605, they are not isomorphic.  Can anyone find
> any "group-theoretical" (whatever that means) difference between the
> two? 
>                       Cheers,
>                       Larry Grove
> 

Doesn't this just say that the matrices (4,0;0,3) and (4,0;0,5) are not 
conjugate in GL(2,11)?
Alan




+++++++++++++++++++++++++++++++

Subject: not iso redux

From: Larry Grove 
Resent-Date:  Wed, 19 Jul 95 22:16:01 BST

        Many thanks for all the replies, both privately and via the
pub-forum. 
        As many noted the groups were cooked up via nonconjugate
elements of order 5 in GL(2,11).  I especially liked Werner's
transition from there to the automorphism groups via normalizers. 
        I'll append some observations made by Laci Kovacs (seemingly
within minutes of the posting), as they differ somewhat from those
already seen in replies to the forum.

> The two minimal normal subgroups of one group are characteristic,
>but those of the other are not. The automorphism group of one acts
>nontrivially on the factor group of order 5, but that of other does
>not. The multiplicator of one group is trivial, but that of the other 
>is not. Greetings, Laci.

                All the best to all,
                   Larry G