The Essential Rank of the Alternating Group Antoine Nectoux University of Auckland August 2013 artlogo 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 . artlogo 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 artlogo 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". artlogo 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. artlogo Antoine Nectoux The Essential Rank of the Alternating Group Alperin-Goldschmidt fusion theorem Conjugation families exist! We need a few more denitions: artlogo 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 . artlogo 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. artlogo 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 artlogo 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. artlogo 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. artlogo 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 . artlogo 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. artlogo 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). artlogo 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). artlogo 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). artlogo 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 )). artlogo 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. artlogo Antoine Nectoux The Essential Rank of the Alternating Group For an odd prime number p The rst point is relatively easy. artlogo 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. artlogo 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 |. artlogo 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. artlogo 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). artlogo 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 . artlogo 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. artlogo 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)). artlogo 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. i i i i i i i t i t i i artlogo 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. artlogo 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. artlogo 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 artlogo For an odd prime number p Conclusion: artlogo 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 artlogo If p = 2 artlogo Antoine Nectoux The Essential Rank of the Alternating Group If p = 2 Then A(n) and S (n) have dierent Sylow 2-subgroups. artlogo 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. artlogo 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 artlogo 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. artlogo 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. artlogo Antoine Nectoux The Essential Rank of the Alternating Group If p = 2 Conclusion : artlogo 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). artlogo 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 artlogo Thanks ! artlogo Antoine Nectoux The Essential Rank of the Alternating Group

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