| Given a finite group N and a prime p, what is a reasonable way to
enumerate all the groups G such that N is a normal subgroup of G,
and [G:N] = p.
Let Aut(N) be the group of automorphisms of N, let Inn(N) be the
group of inner automorphisms of N (ie those which are induced by
conjugation by some element of N). Let Z(N) be the centre of N. Then
let's remember that:
Inn(N) is normal in Aut(N)
Inn(N) is IM to N/Z(N)
Pick � in Aut(N), � in N, such that:
(1) �^p is the inner AM induced by conjugation by �
(2) �(�) = �
Then let P = {0,1,...p-1} have addition defined upon it, and let
'.' denote the group operation already defined on N, then we'll define an
operation * on the set N x P, which we'll call G, such that;
(a, i)*(b, j) = (a.�(b), i+j) (if i+j < p)
= (a.�(b).� ,i+j-p) (if i+j >= p)
By consideration of the group axioms and the criterion for normality,
conditions (1) & (2) are necessary and sufficient for N to be IM to a
normal subgroup of G, evidently with index p.
The question then is, how is it possible to simplify the search for
� & p, to avoid blind alleys and repetition. Some immediate observations
use the orders of N, Aut(N) & Z(N) the centre of N. Label them n,s & z
respectively. z|n & n|zs, since Z(N) =< N & N/Z(N) == Inn(N) =< Aut(N).
Let '$' denote 'does not divide'.
(i) Wlog, � can be chosen to have order a power of p, by consideration
of the Sylow-p-subgroup of G. So this gives us another constraint:
(3) � has order a power of p.
(ii) So by the above, if p$n, then �=1. (1) & (2) reduce to two
possibilities in this case
EITHER: �=I, & G is the direct product N X Cp.
OR: � has order p, and p|s.
(iii) If � is a strictly outer AM, then since �^p is inner, have pn|zs.
(iv) If � is an inner AM, afforded by an element a of N, say, then �^p
is induced by exactly the elements in the coset Za^p, which are fixed
elementwise upon conjugation by a.
Other constraint come from considering symmetries: after investigating
�, don't try any other AM which is in the same automorphism class (under
Aut(Aut(N)). Don't consider any AM which is a power of �.
Where trying to determine all groups of a given order, different
candidate N may appear. This also offers scope for (in)efficiency.
Want also to pull out of the above the conclusion that (unless
we're dealing with N x Cp):
p | ns
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Example: up to IM how many groups of what order have a normal subgroup
of order 8 and index p?
The groups N up to IM of order 8 are C2^3, C2C4 C8, D4, Q.
Inn(N) respectively are I, I, I, C2^2, C2^2
Aut(N) respectively are GL3(2) (order 168), D4, C2^2, D4, S4
From the above, unless p|ns, G is a direct product N x Cp.
Other possibilities before considering p=2 are:
N == Q, p = 3
N == C2^3, p = 3
N == C2^3, p = 7
Note, for each of these � = 1, since p$8, so p is the *order* of �.
N == Q, p = 3 � = (i j k)(-i -j -k) is a suitable representative of
the AM class of Aut(Q) in the standard notation. So there's a nonabelian
group of order 24, it's not S4, since S4 does not have 6 elements with the
same square.
N == C2^3, p = 3 � = ( 0 1 0 ) is a suitable representative of the AM
( 0 0 1 )
( 1 0 0 )
class of Aut(Q) in the standard notation. So, there's another nonabelian
group of order 24, it's not S4, since S4 doesn't have 7 involutions. Nor is it
the previous group, since in each group only 8 elements have order coprime to
3, and these two subgroups are not IM.
N = C2^3, p = 3 � = ( 0 1 0 ) is a suitable representative of the AM
( 0 0 1 )
( 1 1 0 )
class of Aut(Q) in the standard notation. So, there's a nonabelian group of
order 56.
I'll leave the p = 2 case, which I haven't worked out, if someone
wants to play around with the theory above. I seem to remember that
there are 14 groups up to IM of order 16, if you want a check.
Regards,
Andrew.
|
| Ah! I see the confusion. My fault, for not expressing myself
in .0 unambiguously.
What I was looking for was this:
(A) Given group N, for what p is there more than one G such that
G/N = Cp? [There will always be at least one such G, for any p.]
What you interpreted was:
(B) Given group N, for what p is there a G such that G/N = Cp, and
N is not the only normal subgroup of order n of the group G?
So if we are talking about (B), you are right. A normal
Sylow-q-subgroup is always unique in a group. In A4, for instance, the
Sylow-2-subgroup is { I, (12)(34), (13)(24), (14)(23) }.
To return to the original question, it crossed my mind that p =< n,
since the degree of any element of Aut(N) will be a product of orbit
lengths of N under Aut(N), and any such orbit must have length < n.
Regards,
Andrew.
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