The Essential Rank of the Alternating Group

The Essential Rank of the Alternating Group
Antoine Nectoux
University of Auckland
August 2013
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Antoine Nectoux
The Essential Rank of the Alternating Group
A denition
Let p be a prime number, G be a nite group and P be a
p -subgroup of G .
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Antoine Nectoux
The Essential Rank of the Alternating Group
A denition
Let p be a prime number, G be a nite group and P be a
p -subgroup of G .
Denition
A conjugation family for a Sylow p -subgroup P is a set C of
subgroups of P such that whenever A and B are subgroups of P
and g ∈ G is such that Ag = B , there exist Q1 , Q2 , ..., Qn ∈ C and
gi ∈ NG (Qi ) and c ∈ CG (A), such that g = cg1 g2 ...gn , A 6 Q1
and Acg1 g2 ...g 6 Qi +1 for all i = 1, ..., n − 1.
i
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Antoine Nectoux
The Essential Rank of the Alternating Group
A denition
Let p be a prime number, G be a nite group and P be a
p -subgroup of G .
Denition
A conjugation family for a Sylow p -subgroup P is a set C of
subgroups of P such that whenever A and B are subgroups of P
and g ∈ G is such that Ag = B , there exist Q1 , Q2 , ..., Qn ∈ C and
gi ∈ NG (Qi ) and c ∈ CG (A), such that g = cg1 g2 ...gn , A 6 Q1
and Acg1 g2 ...g 6 Qi +1 for all i = 1, ..., n − 1.
i
We can think of it as a "chain of conjugation".
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Antoine Nectoux
The Essential Rank of the Alternating Group
A denition
Let p be a prime number, G be a nite group and P be a
p -subgroup of G .
Denition
A conjugation family for a Sylow p -subgroup P is a set C of
subgroups of P such that whenever A and B are subgroups of P
and g ∈ G is such that Ag = B , there exist Q1 , Q2 , ..., Qn ∈ C and
gi ∈ NG (Qi ) and c ∈ CG (A), such that g = cg1 g2 ...gn , A 6 Q1
and Acg1 g2 ...g 6 Qi +1 for all i = 1, ..., n − 1.
i
We can think of it as a "chain of conjugation".
Alperin, 1967.
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Antoine Nectoux
The Essential Rank of the Alternating Group
Alperin-Goldschmidt fusion theorem
Conjugation families exist! We need a few more denitions:
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Antoine Nectoux
The Essential Rank of the Alternating Group
Alperin-Goldschmidt fusion theorem
Conjugation families exist! We need a few more denitions:
Denition
A proper subgroup U of G is a strongly p-embedded subgroup
(s.p.e. subgroup) if p | |U | and p - |U ∩ U g | for all g ∈ G \ U .
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Antoine Nectoux
The Essential Rank of the Alternating Group
Alperin-Goldschmidt fusion theorem
Conjugation families exist! We need a few more denitions:
Denition
A proper subgroup U of G is a strongly p-embedded subgroup
(s.p.e. subgroup) if p | |U | and p - |U ∩ U g | for all g ∈ G \ U .
Denition
A a subgroup Q 6 P is essential (in G ) if and only if Z (Q ) is a
Sylow p -subgroup of CG (Q ) and NG (Q )/QCG (Q ) has a s.p.e.
subgroup.
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Antoine Nectoux
The Essential Rank of the Alternating Group
Alperin-Goldschmidt fusion theorem
Conjugation families exist! We need a few more denitions:
Denition
A proper subgroup U of G is a strongly p-embedded subgroup
(s.p.e. subgroup) if p | |U | and p - |U ∩ U g | for all g ∈ G \ U .
Denition
A a subgroup Q 6 P is essential (in G ) if and only if Z (Q ) is a
Sylow p -subgroup of CG (Q ) and NG (Q )/QCG (Q ) has a s.p.e.
subgroup.
Remark
The set of essential subgroups of G is closed under conjugacy by
elements of G .
Antoine Nectoux
The Essential Rank of the Alternating Group
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The essential rank
Theorem (Alperin - Goldschmidt, 1970)
Let C be a set of subgroup of P. Then C is a conjugation family for
P if and only if P ∈ C and C intersects non-trivially with every
conjugacy class of essential subgroups.
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Antoine Nectoux
The Essential Rank of the Alternating Group
The essential rank
Theorem (Alperin - Goldschmidt, 1970)
Let C be a set of subgroup of P. Then C is a conjugation family for
P if and only if P ∈ C and C intersects non-trivially with every
conjugacy class of essential subgroups.
It is then natural to look for the conjugation family with minimal
cardinality.
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Antoine Nectoux
The Essential Rank of the Alternating Group
The essential rank
Theorem (Alperin - Goldschmidt, 1970)
Let C be a set of subgroup of P. Then C is a conjugation family for
P if and only if P ∈ C and C intersects non-trivially with every
conjugacy class of essential subgroups.
It is then natural to look for the conjugation family with minimal
cardinality.
Denition
We call the number of conjugacy classes of essential subgroups the
essential rank rke (G ) of G .
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Antoine Nectoux
The Essential Rank of the Alternating Group
Results
The essential rank for the following groups has been determined in
the case p > 3 (J. An and H. Dietrich, 2012 ):
GL(d , q) , GU(d , q) , Sp(2d , q) , O0(2d + 1, q) , O±(2d , q),
sporadic simple groups,
and the symmetric group S (n) even for p = 2.
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Antoine Nectoux
The Essential Rank of the Alternating Group
Results
The essential rank for the following groups has been determined in
the case p > 3 (J. An and H. Dietrich, 2012 ):
GL(d , q) , GU(d , q) , Sp(2d , q) , O0(2d + 1, q) , O±(2d , q),
sporadic simple groups,
and the symmetric group S (n) even for p = 2.
My goal: Find the essential rank of the Frobenius category of the
alternating group A(n) by extending the results on S (n).
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
Let p be an odd prime number, n > 3. Then
Sylp (S (n)) = Sylp (A(n)), and S (n) and A(n) have the same
p -subgroups. So it makes sense to try to extend the results on S (n)
to A(n).
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
Theorem
If p is odd, then a subgroup R 6 S (n) is essential in S (n) if and
only if it is essential in A(n).
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
Theorem
If p is odd, then a subgroup R 6 S (n) is essential in S (n) if and
only if it is essential in A(n).
We have to prove two things:
• Z (R ) ∈
Sylp (CS (n)(R )) ⇐⇒ Z (R ) ∈ Sylp (CA(n)(R )).
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
Theorem
If p is odd, then a subgroup R 6 S (n) is essential in S (n) if and
only if it is essential in A(n).
We have to prove two things:
• Z (R ) ∈
Sylp (CS (n)(R )) ⇐⇒ Z (R ) ∈ Sylp (CA(n)(R )).
• NS (n) (R )/RCS (n) (R )
has a s.p.e. subgroup
⇐⇒ NA(n) (R )/RCA(n) (R ) has a s.p.e. subgroup.
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
The rst point is relatively easy.
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
The rst point is relatively easy.
The second point requires the use of the following theorem and
lemma:
Theorem (Aschbacher, 1993)
Let q be a prime number.
A nite group G has a strongly q-embedded subgroup if and only if
its commuting graph Λq (G ) is disconnected.
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
Lemma
Let q be a prime number.
If N is a normal subgroup of a nite group G such that q - |G /N |,
then Λq (G ) = Λq (N ).
We apply this lemma to G = NS (n) (R )/CS (n) (R )R and
N = NA(n) (R )/CA(n) (R )R and we deduce that
NA(n) (R )/CA(n) (R )R is isomorphic to a subgroup of
NS (n) (R )/CS (n) (R )R of index 1 or 2. In any case N C G and
p - |G /N |.
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
To nd the essential rank of A(n) we still have to nd the number
of conjugacy classes of essential subgroups.
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
To nd the essential rank of A(n) we still have to nd the number
of conjugacy classes of essential subgroups.
Since this result is known for S (n), which has the same essential
subgroups as A(n), we just have to nd
|NS (n) (Q ) : NA(n) (Q )|
for all essential subgroup Q 6 A(n).
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
Denition
We say that a subgroup H 6 G is radical in G if H = Op (NG (H )).
We write Rp (G ) the set of all radical p -subgroups of G .
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
Denition
We say that a subgroup H 6 G is radical in G if H = Op (NG (H )).
We write Rp (G ) the set of all radical p -subgroups of G .
Remark
Every essential subgroup is radical.
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
Denition
We say that a subgroup H 6 G is radical in G if H = Op (NG (H )).
We write Rp (G ) the set of all radical p -subgroups of G .
Remark
Every essential subgroup is radical.
Theorem
If p is odd then
Rp (A(n)) = Rp (S (n)).
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
Theorem (Alperin - Fong, 1989)
If a subgroup R 6 S (n) is radical,then there exist decompositions:
{1, ..., n} = V0 ∪ V1 ∪ ... ∪ Vs
and
R = R0 × R1 × ... × Rs
where R0 is the trivial subgroup of S (V0 ) and for i > 1, Ri is a
basic subgroup of S (Vi ), i.e. for each i > 1 there exist a sequence
of positive integers c i = (ci ,1 , ci ,2 , ..., ci ,t ) such that
Ri = Ac = Ac ,1 o Ac ,2 o ... o Ac , , where Am is the elementary
abelian group of order p m acting regularly on its underlying set, and
|Vi | = p c ,1 +c ,2 +...+c , for all i > 1.
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
Theorem
If p is odd and R is a subgroup of S (n) and A(n) having a
decomposition R = R0 × R1 × ... × Rs where R0 is a trivial group
and Ri = Ai1 o Ai2 o ... o Ai and ik = 1 or 2 for all k = 1, ..., ti and
all i = 1, ..., s, then
t
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|NS (n) (R ) : NA(n) (R )| = 2.
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
Theorem
If p is odd and R is a subgroup of S (n) and A(n) having a
decomposition R = R0 × R1 × ... × Rs where R0 is a trivial group
and Ri = Ai1 o Ai2 o ... o Ai and ik = 1 or 2 for all k = 1, ..., ti and
all i = 1, ..., s, then
t
i
|NS (n) (R ) : NA(n) (R )| = 2.
In particular, if R is an essential subgroup of S (n) then
|NS (n) (R ) : NA(n) (R )| = 2.
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
Theorem
If p is odd and R is a subgroup of S (n) and A(n) having a
decomposition R = R0 × R1 × ... × Rs where R0 is a trivial group
and Ri = Ai1 o Ai2 o ... o Ai and ik = 1 or 2 for all k = 1, ..., ti and
all i = 1, ..., s, then
t
i
|NS (n) (R ) : NA(n) (R )| = 2.
In particular, if R is an essential subgroup of S (n) then
|NS (n) (R ) : NA(n) (R )| = 2.
We can then deduce that the conjugacy classes of essential
subgroups have the same cardinality under conjugacy by S (n) and
A( n ) .
Antoine Nectoux
The Essential Rank of the Alternating Group
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For an odd prime number p
Conclusion:
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Antoine Nectoux
The Essential Rank of the Alternating Group
For an odd prime number p
Conclusion:
Theorem
If p is odd and P is a Sylow p-subgroup of A(n), then
rke (A(n)) = rke (S (n))
= l (l + 1)/2 + H (m) + Y (m) − 1.
where m is the number of points moved by P, and m =
l
P
mi p i is
i =0
the p-adic expansion of m and H (m) = |{1 ∈ {1, ..., l } | mi = 0}|
and Y (m) = −1 if l > 2, ml −1 = 0 and ml = 1, and Y (m) = 0
otherwise.
Antoine Nectoux
The Essential Rank of the Alternating Group
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If p = 2
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Antoine Nectoux
The Essential Rank of the Alternating Group
If p = 2
Then A(n) and S (n) have dierent Sylow 2-subgroups.
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Antoine Nectoux
The Essential Rank of the Alternating Group
If p = 2
Then A(n) and S (n) have dierent Sylow 2-subgroups.
However, we can simplify the problem :
Lemma
• A(2n) and A(2n + 1) have the same essential 2-subgroups
(similarly for S (2n) and S (2n + 1)).
• An essential 2-subgroup of S (2n) has no xed-point.
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Antoine Nectoux
The Essential Rank of the Alternating Group
If p = 2
Moreover, we can obtain the radical 2-subgroups of A(n) from the
following lemma
Lemma (An - Huang, 2013)
Let H be a normal subgroup of G such that |G : H |p = p,
Q ∈ Rp (G ) and R ∈ Rp (H ).
(a) There exists Q 0 ∈ Rp (G ) such that |Q 0 : R | = 1 or p,
R = Q 0 ∩ H and NG (Q 0 ) = NG (R ). In particular, if Q 0 6= R
then Q 0 = Op (NG (R )) and hence it is uniquely determined.
(b) The group R 0 = Q ∩ H is radical in H, NG (Q ) 6 NG (R 0 ) and
Q ∈ (R )p (NG (R 0 )). If NG (R 0 ) = NH (R 0 ) then R 0 = Q. If
NG (R 0 ) > NH (R 0 ) and Op (NG (R 0 )) 6= R 0 then
Q = Op (NG (R 0 )). If NG (R 0 ) > NH (R 0 ) and Op (NG (R 0 )) = R 0
then Q = R 0 or Q = hR 0 , y i for some y ∈ NG (R 0 ) \ NH (R 0 ).
Antoine Nectoux
The Essential Rank of the Alternating Group
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If p = 2
Hence for any essential 2-subgroup R of A(n) there exists a radical
2-subgroup Q 6 S (n) such that R = Q ∩ A(n). We also know that
Q has a decomposition Q = Q1 × ... × Qs where each Qi is a
wreath product of elementary abelian subgroups.
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Antoine Nectoux
The Essential Rank of the Alternating Group
If p = 2
Let Q = Q1 × ... × Qs be a radical 2-subgroup of S (2n) having no
xed-point.
Lemma
• CS (2n) (R ) = CS (2n) (Q ) if and only if Q is not of the form
•
•
•
•
A1 × A1 × Q + or A1 o A1 × Q + where Q + only contains even
permutations.
CS (2n) (R ) = Z (R ) if and only if Qi 6= A1 , for all i = 1, ..., s.
NS (2n) (R ) = NS (2n) (Q ) if and only if Q is not of the form
A1 o A1 × Q+ where Q+ contains only even permutations.
If Q 6 A(2n) then NS (2n) (R ) = NA(2n) (R ) if and only if R has
no direct factor A2 .
NA(4k ) (A2 k )/A2 k has no s.p.e. subgroup.
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Antoine Nectoux
The Essential Rank of the Alternating Group
If p = 2
Conclusion :
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Antoine Nectoux
The Essential Rank of the Alternating Group
If p = 2
Conclusion :
Theorem
If 2n > 8 then
• if Q is an essential 2-subgroup of S (2n) then Q ∩ A(2n) is an
essential 2-subgroup of A(2n).
• if R is an essential 2-subgroup of A(2n) then there exists an
essential 2-subgroup Q of S (2n) such that R = Q ∩ A(2n).
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Antoine Nectoux
The Essential Rank of the Alternating Group
If p = 2
Conclusion :
Theorem
If 2n > 8 then
• if Q is an essential 2-subgroup of S (2n) then Q ∩ A(2n) is an
essential 2-subgroup of A(2n).
• if R is an essential 2-subgroup of A(2n) then there exists an
essential 2-subgroup Q of S (2n) such that R = Q ∩ A(2n).
Moreover
• if 2n 6= 2i or then rke (A(2n)) = rke (S (2n)).
• if 2n = 2i for some i > 3 then rke (A(2n)) = rke (S (2n)) + 1.
• rke (A(2)) = 0, rke (A(4)) = 0, rke (A(8)) = 3.
Antoine Nectoux
The Essential Rank of the Alternating Group
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Thanks !
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Antoine Nectoux
The Essential Rank of the Alternating Group

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