The kernels of the linear maps of finite Abelian group algebras
aa r X i v : . [ m a t h . R A ] J a n The kernels of the linear maps of finite Abeliangroup algebras
Dan Yan ∗ MOE-LCSM,School of Mathematics and Statistics,Hunan Normal University, Changsha 410081, China
E-mail: [email protected]
Abstract
In our paper, we give a necessary and sufficient conditions for the kernelsof the linear maps of finite Abelian group algebras to be Mathieu-Zhaospaces of K [ G ] if G is a finite Abelian group and K is a split field for G . Hence we classify all Mathieu-Zhao spaces of the finite Abelian groupalgebras if K is a split field for G . Keywords.
Mathieu-Zhao spaces, Regular representations, Finite Abelian GroupAlgebras
MSC(2010).
Throughout this paper, we will write K for a field without specific note and K [ G ]for the group algebra of G over K . V G is the K -subspace of the group algebra K [ G ]consisting of all the elements of K [ G ] whose coefficient of the identity element 1 G of G is equal to zero. L is a linear map from K [ G ] to K and L | H means restricting L on H , where H is a subgroup of G . H is a p ′ -subgroup of G if p ∤ | H | .The Mathieu-Zhao space was introduced by Wenhua Zhao in [6], which is anatural generalization of ideals. The notion was named after A. van den Essen’ssuggestion and Mathieu conjecture. In [1], J. J. Duistermaat and W. van derKallen proved the Mathieu conjecture for the case of tori, which can be re-statedas follows. ∗ The author is supported by the NSF of China (Grant No. 11871241; 11601146), the ChinaScholarship Council and the Construct Program of the Key Discipline in Hunan Province. heorem 1.1. Let z = ( z , z , . . . , z m ) be m commutative free variables and V the subspace of the Laurent polynomial algebra C [ z − , z ] consisting of the Laurentpolynomials with no constant term. Then V is a Mathieu-Zhao space of C [ z − , z ] . Let G be the free Abelian group Z m ( m ≥ C [ z − , z ] can be identified with the group algebra C [ G ]. Under this iden-tification, the subspace of V in the theorem is V G . In [7], Wenhua Zhao and R.Willems proved that V G is a Mathieu-Zhao space of K [ G ] if G is a finite group andchar K = 0 or char K = p > | G | . For finite Abelian group, they proved that if K contains a primitive d -th root of unity, then V G is a Mathieu-Zhao space of K [ G ]if and only if char K = p > d , where | G | = p a d , p ∤ d . In [8], Wenhua Zhao andthe author give a sufficient and necessary condition for V G to be a Mathieu-Zhaospace of K [ G ] if G is a finite group and K is a split field for G . Hence it’s naturalto ask the following question. Problem 1.2.
Let G be a finite group with | G | = n and L = ( L , L , . . . , L r ) , L i is a linear map from K [ G ] to K such that L i ( g j ) = l i,j for all ≤ i ≤ r , ≤ j ≤ n . Suppose that L , L , . . . , L r are linearly independent over K . Thenunder what conditions on L and K , Ker L forms a Mathieu-Zhao space of thegroup algebra K [ G ] ? It’s easy to see that if r ≥ n , then Ker L = 0. If r ≤ n −
1, then dim K Ker L = n − r and every codimension r subspace of K [ G ] is Ker L for some linear map L .Hence Ker L are all the codimension r subspaces of K [ G ].In our paper, we first prove some properties of Ker L in section 2. Then weclassify all Mathieu-Zhao spaces of K [ G ] if G is a finite Abelian group and K isa split field for G in section 3. Thus, we solve Problem 1.2 if G is a finite Abeliangroup. L Proposition 2.1.
Let K , G , L be as in Problem 1.2 and g is the identity G of G . Then we have the following statements: (1) If all the l i,j are equal for all ≤ i ≤ r , ≤ j ≤ n , then Ker L is an idealof K [ G ] . (2) If Ker L is a Mathieu-Zhao space of K [ G ] , then there exists i ∈ { , , . . . , r } such that l i , = 0 .Proof. (1) Let l := l i,j for all 1 ≤ i ≤ r , 1 ≤ j ≤ n . Then Ker L = { α = P nj =1 c j g j ∈ K [ G ] | l · P nj =1 c j = 0 } . Since l = 0, we have that Ker L = { α = P nj =1 c j g j ∈ K [ G ] | P nj =1 c j = 0 } . It’s easy to check that Ker L is an ideal of K [ G ]. 22) If l , = · · · = l r, = 0, then 1 G ∈ Ker L . If Ker L is a Mathieu-Zhao spaceof K [ G ], then Ker L = K [ G ]. That is, L = 0, which is a contradiction. Then theconclusion follows. Remark 2.2.
We can see from Proposition 2.1 that we can assume l i , = 0 forsome i ∈ { , , . . . , r } in the following arguments. If r = 1 and l , = l , = · · · = l ,n = 0, l , = 0, then Ker L = V G , which is discussed in [7] and [8]. Proposition 2.3.
Let R be any commutative ring and G any group. Suppose that L = ( L , L , . . . , L r ) is a linear map from R [ G ] to R . If Ker L is a Mathieu-Zhaospace of R [ G ] , then Ker( L | H ) is a Mathieu-Zhao space of R [ H ] , where H is anysubgroup of G .Proof. Assume otherwise. Then there exist u, v , v ∈ R [ H ] such that u m ∈ Ker( L | H ) for all m ≥
1, but v u m v / ∈ Ker( L | H ) for infinitely many m ≥ R [ H ] ⊆ R [ G ], we have u, v , v ∈ R [ G ] and u m ∈ Ker L for all m ≥ v u m v / ∈ Ker L for infinitely many m ≥
1. Otherwise, v u m v ∈ Ker L ∩ R [ H ] = Ker( L | H ), which is a contradiction. Hence Ker L is not a Mathieu-Zhaospace of R [ G ], which is a contradiction. Then the conclusion follows. Corollary 2.4.
Let L , G be as in Problem 1.2 and K be a field of characteristic p , H a normal subgroup of G . If H is a p ′ -subgroup and Ker L is a Mathieu-Zhaospace of K [ G ] , then Ker( L | G/H ) is a Mathieu-Zhao space of K [ G/H ] .Proof. Let ϕ be the natural surjective homomorphism from K [ G ] to K [ G/H ]and E H = | H | P | H | j =1 h j . Then (1 − E H ) K [ G ] = Ker ϕ and E H K [ G ] ∼ = K [ G/H ].Thus, we have K [ G ] ∼ = (1 − E H ) K [ G ] ⊕ K [ G/H ]. Therefore, K [ G/H ] can be seenas a subalgebra of K [ G ]. It follows from the arguments of Proposition 2.3 thatKer( L | G/H ) is a Mathieu-Zhao space of K [ G/H ]. Lemma 2.5.
Let L and G be as in Problem 1.2. Then Ker L = { β ∈ K [ G ] | Tr βα i = 0 f or all ≤ i ≤ r } , where α i = P nj =1 l i,j g − j for ≤ i ≤ r .Proof. Let β = P nj =1 c j g j . Then L i ( β ) = P nj =1 c j l i,j = Tr βα i for all 1 ≤ i ≤ r .Hence the conclusion follows. Theorem 2.6.
Let L and G be as in Problem 1.2 and K a field of characteristiczero or a field of characteristic p and p ∤ | G | . If K is a split field for G , then Ker L ∼ = { ( A , . . . , A s ) ∈ A | s X j =1 n j Tr( C i,j A j ) = 0 f or all ≤ i ≤ r } , where A = M n ( K ) × · · · × M n s ( K ) is the product of matrices and C i,j = ρ j ( α i ) ∈ M n j ( K ) for ≤ j ≤ s and for all ≤ i ≤ r , where α i be as in Lemma 2.5 for ≤ i ≤ r , ρ j is an irreducible representation of G and n j = ρ j (1) for ≤ j ≤ s and s is the number of distinct (up to isomorphism) irreducible representationsof G . roof. Since char K = 0 or char K = p and p ∤ | G | , we have that K [ G ] is semi-simple. Since K is a split field for G , we have that K [ G ] ∼ = M n ( K ) × M n ( K ) × · · · × M n s ( K ) , where M n j ( K ) is the ring of n j × n j matrices over K for 1 ≤ j ≤ s . Let ˜ ρ be theregular representation of K [ G ]. Then Tr( β ) = 0 if and only if Tr( ˜ ρ ( β )) = 0 forall β ∈ K [ G ]. Let ρ = ( ρ , ρ , . . . , ρ s ). Then ρ is a ring isomorphism from K [ G ]to A . Let β be any element in K [ G ]. Then ρ ( α i β ) = ( ρ ( α i β ) , ρ ( α i β ) , . . . , ρ s ( α i β )) = ( ρ ( α i ) ρ ( β ) , . . . , ρ s ( α i ) ρ s ( β )) . Suppose that ρ ( α i ) = ( ρ ( α i ) , . . . , ρ s ( α i )) = C i, · · · C i, · · · · · · C i,s ∈ A and ρ ( β ) = ( ρ ( β ) , . . . , ρ s ( β )) = A · · · A · · · · · · A s ∈ A for all 1 ≤ i ≤ r . Then we have that ρ ( α i β ) = C i, A · · · C i, A · · · · · · C i,s A s ∈ A. Thus, we have the following commutative diagram: K [ G ] ∼ = −−−→ M n ( K ) × M n ( K ) × · · · × M n s ( K ) ˜ ρ ( α i β ) y φ ( ρ ( α i β )) y K [ G ] ∼ = −−−→ M n ( K ) 0 · · · M n ( K ) · · · · · · M n s ( K ) , where φ is the natural isomorphism between the two algebras. Thus, we havethat Tr( ˜ ρ ( α i β )) = 0 if and only if Tr( φ ( ρ ( α i β ))) = 0. Since Tr( φ ( ρ ( α i β ))) =4 Tr( C i, A ) + n Tr( C i, A ) + · · · + n s Tr( C i,s A s ), we have that Tr( α i β ) = 0 ifand only if n Tr( C i, A )+ n Tr( C i, A )+ · · · + n s Tr( C i,s A s ) = 0 for all 1 ≤ i ≤ r .Thus, we have that Ker L ∼ = V , where V = { ( A , A , . . . , A s ) ∈ A | s X j =1 n j Tr C i,j A j = 0 for all 1 ≤ i ≤ r } . Corollary 2.7.
Let L and G be as in Problem 1.2 and K a field of characteristiczero or a field of characteristic p and p ∤ | G | . If K is a split field for G and r = 1 ,then Ker L is a Mathieu-Zhao space of K [ G ] if and only if n λ d + n λ d + · · · + n t λ t d t = 0 for all non-zero vectors ˜ d = ( d , . . . , d t ) , d j ∈ { , , . . . , n j } for ≤ j ≤ t , where n j λ j = Tr ρ j ( α ) and α be as in Lemma 2.5, ρ j is anirreducible representation of G for ≤ j ≤ s and s is the number of distinct (upto isomorphism) irreducible representations of G and t ∈ { , , . . . , s } .Proof. It follows from Theorem 2.6 that Ker L ∼ = V , where V = { ( A , . . . , A s ) ∈ A | P sj =1 n j Tr( C ,j A j ) = 0 } and C ,j = ρ j ( α ) ∈ M n j ( K ) for 1 ≤ j ≤ s . Let ρ beas in Theorem 2.6. Since α = 0 and ρ is an isomorphism, we have that ρ ( α ) = 0.We can assume that C , , . . . , C ,t are not equal to zero and C ,t +1 = · · · = C ,s = 0for some t ∈ { , , . . . , s } by reordering the ρ j for 1 ≤ j ≤ s . It follows fromTheorem 5.8.1 in [2] or Theorem 4.4 in [4] that V is a Mathieu-Zhao space of A if and only if C ,j = λ j I n j and n λ d + · · · + n t λ t d t = 0 for all nonzero vectors˜ d = ( d , . . . , d t ) and d j ∈ { , , . . . , n j } for 1 ≤ j ≤ t . Then the conclusionfollows. Corollary 2.8.
Let L and G be as in Problem 1.2 and K a field of characteristiczero or a field of characteristic p and p ∤ | G | . If K is a split field for G and r = 1 ,then the following two statements are equivalent: (1) Ker L is a Mathieu-Zhao space of K [ G ] . (2) There exists µ , . . . , µ t ∈ K such that L = µ χ + µ χ + · · · + µ t χ t and µ d + · · · + µ t d t = 0 for all nonzero vectors ˜ d = ( d , d , . . . , d t ) , d j ∈ { , , . . . , n j } for ≤ j ≤ t , where χ , χ , . . . , χ s are the non-isomorphic irreducible charactersof G and µ j = n − n j λ j , n j = χ j (1) , n j λ j = Tr ρ j ( α ) , α be as in Lemma 2.5and ρ j is an irreducible representation of G with character χ j for ≤ j ≤ s , s isthe number of distinct (up to isomorphism) irreducible representations of G and t ∈ { , , . . . , s } .Proof. (1) ⇒ (2) Since L ( β ) = Tr( α β ) for any β ∈ K [ G ], where α be as inLemma 2.5, we have that n Tr( α β ) = Tr ˜ ρ ( α β ) = Tr φ ( ρ ( α β ))5y following the arguments of Theorem 2.6, where ˜ ρ be as in Theorem 2.6. SinceKer L is a Mathieu-Zhao space of K [ G ], it follows from Corollary 2.7 that C ,j = λ j I n j for λ j ∈ K and for all 1 ≤ j ≤ s . We can assume that λ · · · λ t = 0 and λ t +1 = · · · = λ s = 0 for some t ∈ { , , . . . , s } by reordering χ , χ , . . . , χ s .Thus, it follows from Lemma 2.5 that L ( β ) = Tr( α β ) = n − ( n λ Tr A + n λ Tr A + · · · + n t λ t Tr A t ) . Since Tr A j = χ j ( β ) for all 1 ≤ j ≤ s , we have that L = n − ( n λ χ + n λ χ + · · · + n t λ t χ t ) . It follows from Corollary 2.7 that n λ d + · · · + n t λ t d t = 0 for all nonzero vectors˜ d = ( d , d , . . . , d t ), d j ∈ { , , . . . , n j } for 1 ≤ j ≤ t . Let µ j = n − n j λ j for all1 ≤ j ≤ s . Then the conclusion follows.(2) ⇒ (1) Since Ker L = { β ∈ K [ G ] | L ( β ) = 0 } = { β ∈ K [ G ] | µ χ ( β ) + · · · + µ t χ t ( β ) = 0 } and there exists A j ∈ M n j ( K ) such that Tr A j = χ j ( β ) for all1 ≤ j ≤ t , we have thatKer L = { ( A , . . . , A t ) ∈ M n ( K ) × · · · × M n t ( K ) | t X j =1 µ j Tr A j = 0 } . Then the conclusion follows from Theorem 5.8.1 in [2] or Theorem 4.4 in [4].
Remark 2.9.
To prove that if n λ d + n λ d + · · · + n t λ t d t = 0 for all nonzerovectors ˜ d = ( d , d , . . . , d t ) and d j ∈ { , , . . . , n j } for 1 ≤ j ≤ t , then Ker L is aMathieu-Zhao space of K [ G ] for r = 1, we don’t need the condition that K is asplit field for G in Corollary 2.7 by following the arguments Theorem 5.8.1 in [2]because an idempotent matrix can be conjugated to a diagonal matrix with only0 and 1 on the diagonal over division rings.If L = µ j χ j for some j ∈ { , , . . . , t } , µ j ∈ K ∗ , then it follows from thearguments of Corollary 2.8 that the condition n λ d + n λ d + · · · + n t λ t d t = 0in Theorem 2.6 is equivalent to n j d j = 0 for all 1 ≤ d j ≤ n j , which is clearlytrue if char K = 0. If char K = p , then the condition is equivalent to p > n j .To see this, we can assume that p | n j d j for some d j ∈ { , , . . . , n j } , then p | n j or p | d j , which contradicts with p > n j . Thus, if p > n j , then n j d j = 0 mod p for all1 ≤ d j ≤ n j . Conversely, suppose that p ≤ n j . Then let d j = p ∈ { , , . . . , n j } ,we have that n j p = 0 mod p , which is a contradiction. Hence if n j d j = 0 mod p forall 1 ≤ d j ≤ n j , then p > n j . Therefore, the conclusion is the same as Theorem5.1 in [5] in this situation. 6 Mathieu-Zhao spaces of finite Abelian groupalgebras
Proposition 3.1.
Let B = K × · · · × K be a K -algebra and V = { ( a , a , . . . , a n ) ∈ B | n X j =1 γ i,j a j = 0 f or all ≤ i ≤ r } , where γ i,j ∈ K for all ≤ i ≤ r , ≤ j ≤ n . If at least one of γ i,j is nonzero forall ≤ i ≤ r , ≤ j ≤ n , then V is a Mathieu-Zhao space of B if and only if γ i, d + γ i, d + · · · + γ i,t i d t i = 0 for some i ∈ { , , . . . , r } for all nonzero vectors ˜ d = ( d , d , . . . , d t i ) and d j i ∈ { , } for ≤ j i ≤ t i , t i ∈ { , , . . . , n } and at leastone of t i is nonzero for ≤ i ≤ r .Proof. We can assume that γ i,j = 0 for all 1 ≤ j ≤ t for some i ∈ { , , . . . , r } and γ i,j = 0 for all 1 ≤ i ≤ r , t + 1 ≤ j ≤ n by reordering γ i,j for 1 ≤ i ≤ r ,1 ≤ j ≤ n . Then we have t columns z }| { × · · · × × K · · · × K ⊆ V and 0 × · · · × K × × · · · × n − t columns z }| { × · · · × * V, where t = max { t , t , . . . , t r } .“ ⇒ ” Suppose that γ i, d + γ i, d + · · · + γ i,t i d t i = 0 for some nonzerovector ˜ d = ( d , d , . . . , d t i ), d j i = 0 or 1 for 1 ≤ j i ≤ t i for all 1 ≤ i ≤ r , then e = ( d , . . . , d t , , . . . ,
0) is an idempotent of V . Since V is a Mathieu-Zhao spaceof B , we have that Be = Kd ×· · ·× Kd t × ×· · ·× ⊆ V , which is a contradiction.Then the conclusion follows.“ ⇐ ” Let I = t columns z }| { × · · · × × K × · · · × K . Then I is an ideal of B . We claimthat V /I contains no nonzero idempotent. Suppose that e is a nonzero idempotentof V /I . Then we have e = ( e , e , . . . , e t ), where e j = 0 or 1 for 1 ≤ j ≤ t . Let˜ d = ( d , . . . , d t ) = e = (0 , . . . , γ i, d + γ i, d + · · · + γ i,t i d t i = 0 for all1 ≤ i ≤ r , which is a contradiction. It follows from Theorem 4.2 in [5] that V /I is a Mathieu-Zhao space of
B/I . Then it follows from Proposition 2.7 in [5] that V is a Mathieu-Zhao space of B . Remark 3.2.
In Proposition 3.1, if γ i,j = 0 for all 1 ≤ i ≤ r , 1 ≤ j ≤ n , then V = B . Clearly, V is a Mathieu-Zhao space of B . Corollary 3.3.
Let L and G be as in Problem 1.2 and K a field of characteristiczero or a field of characteristic p and p ∤ | G | , If K is a split field for G and G is Abelian, then Ker L is a Mathieu-Zhao space of K [ G ] if and only if γ i, d +7 i, d + · · · + γ i,t i d t i = 0 for some i ∈ { , , . . . , r } for all nonzero vectors ˜ d =( d , d , . . . , d t i ) and d j i ∈ { , } for ≤ j i ≤ t i , t i ∈ { , , . . . , n } and at least oneof t i is nonzero for ≤ i ≤ r , where γ i,j = ρ j ( α i ) for all ≤ i ≤ r , ≤ j ≤ n and ρ j is an irreducible representation of G for ≤ j ≤ n and α i be as in Lemma2.5 for ≤ i ≤ r .Proof. Since G is Abelian, we have that all the irreducible representations of G are degree one. It follows from Theorem 2.6 that Ker L ∼ = { ( a , a , . . . , a n ) ∈ A | P nj =1 γ i,j a j = 0 f or all ≤ i ≤ r } , where A is n times product of K , γ i,j = ρ j ( α i ) = Tr ρ j ( α i ) ∈ K for all 1 ≤ i ≤ r , 1 ≤ j ≤ n . Since L = 0, we have that atleast one of γ i,j is nonzero for 1 ≤ i ≤ r , 1 ≤ j ≤ n . Then the conclusion followsfrom Proposition 3.1. Lemma 3.4.
Let R be an integral domain of characteristic p and G a finiteAbelian group with | G | = p a d , p ∤ d . Then every idempotent of R [ G ] is alsoan idempotent of R [ ˜ G ] , where G = H × ˜ G and | H | = p a . In particular, theidempotents of R [ G ] are the same as the idempotents of R [ ˜ G ] .Proof. Since G is a finite Abelian group, we have that G = H × ˜ G and | ˜ G | = d .Let e be an idempotent of R [ G ]. Then e can be written as e = X h ∈ H α h h with α h ∈ R [ ˜ G ] for each h ∈ H . Since | H | = p a , we have h q m = 1 for any m ≥ h ∈ H , where q = p a . Thus, we have e = e q m = X h ∈ H α q m h ∈ R [ ˜ G ] . Then the conclusion follows.
Theorem 3.5.
Let L and G be as in Problem 1.2 and K a field of characteristic p . If K is a split field for G and G is Abelian with | G | = p a d , p ∤ d , then thefollowing statements are equivalent: (1) Ker L is a Mathieu-Zhao space of K [ G ] . (2) γ i, d + γ i, d + · · · + γ i,t i d t i = 0 for some i ∈ { , , . . . , r } for all nonzerovectors ˜ d = ( d , d , . . . , d t i ) and d j i ∈ { , } for ≤ j i ≤ t i , t i ∈ { , , . . . , d } and at least one of t i is nonzero for ≤ i ≤ r , where γ i,j = ρ j ( α i ) = Tr ρ j ( α i ) for ≤ i ≤ r , ≤ j ≤ d and ρ j is an irreducible representation of G/H for ≤ j ≤ d , H is a Sylow p -subgroup of G and α i be as in Lemma 2.5 by replacing G with G/H for ≤ i ≤ r ; l i, , l i, , . . . , l i,n satisfy the following equations: (3.1) χ j (˜ g − ) l i, + χ j (˜ g − ) l i,p a +1 + · · · + χ j (˜ g − d ) l i, ( d − p a +1 = 0 χ j (˜ g − ) l i, + χ j (˜ g − ) l i,p a +2 + · · · + χ j (˜ g − d ) l i, ( d − p a +2 = 0 ... χ j (˜ g − ) l i,p a + χ j (˜ g − ) l i, p a + · · · + χ j (˜ g − d ) l i,dp a = 08 or all ≤ i ≤ r , t + 1 ≤ j ≤ d , where χ j is the irreducible character according to ρ j for t + 1 ≤ j ≤ d and G = ∪ dk =1 ˜ g k H with ˜ g = 1 G/H and H = { h , h , . . . , h p a } with h = 1 H , L i ( h k ) = l i,k and L i (˜ g k h q ) = l i, ( k − p a + q for all ≤ i ≤ r , ≤ k ≤ d , ≤ q ≤ p a and t = max { t , t , . . . , t r } .Proof. Since G is Abelian, we have that G = H × ˜ G , where ˜ G ∼ = G/H and | ˜ G | = d .Note that γ i,j = Tr ρ j ( α i ) = d X k =1 Tr ρ j (˜ g − k ) l i, ( k − p a +1 = d X k =1 χ j (˜ g − k ) l i, ( k − p a +1 for all 1 ≤ i ≤ r , 1 ≤ j ≤ d . Let e j = d − P dk =1 χ j (˜ g − k )˜ g k for 1 ≤ j ≤ d . Then itfollows from Theorem 2.12 in [3] that e , e , . . . , e d are the primitive orthogonalidempotents of K [ ˜ G ]. Without loss of generality, we can assume that γ i,j = 0 forall 1 ≤ i ≤ r , t + 1 ≤ j ≤ d and γ i,j = 0 for all 1 ≤ j ≤ t for some i ∈ { , , . . . , r } by reordering ρ j ( α i ) for all 1 ≤ i ≤ r , 1 ≤ j ≤ d .(1) ⇒ (2) It’s easy to see that if γ i,j = 0 for all 1 ≤ i ≤ r , t + 1 ≤ j ≤ d ,then e t +1 , . . . , e d belong to Ker( L | ˜ G ) ⊆ Ker L . Thus, the ideal I generated by e t +1 , . . . , e d belongs to Ker L . Since ˜ G is Abelian, it’s easy to check that e j ˜ g k = χ j (˜ g k ) e j for all 1 ≤ j, k ≤ d . Hence we have e j ˜ g k ∈ Ker L for all t + 1 ≤ j ≤ d ,1 ≤ k ≤ d . Note that e j h q ∈ Ker L for all t + 1 ≤ j ≤ d , 1 ≤ q ≤ p a . Then wehave equations (3.1) for all 1 ≤ i ≤ r , t + 1 ≤ j ≤ d . It follows from Proposition2.3 that Ker( L | ˜ G ) is a Mathieu-Zhao space of K [ ˜ G ]. That is, Ker( L | G/H ) is aMathieu-Zhao space of K [ G/H ]. Since p ∤ | G/H | , the conclusion follows fromCorollary 3.3.(2) ⇒ (1) If γ i,j = 0 for all 1 ≤ i ≤ r , t + 1 ≤ j ≤ d , then e t +1 , . . . , e d ∈ Ker( L | ˜ G ) ⊆ Ker L . It’s easy to check that e j ˜ g k = χ j (˜ g k ) e j and e j ˜ g k h q = χ j (˜ g k ) e j h q for all t + 1 ≤ j ≤ d , 1 ≤ k ≤ d , 1 ≤ q ≤ p a . Therefore, we have I ⊆ Ker L ,where I is an ideal generated by e t +1 , . . . , e d . Since e , . . . , e d are the primitiveorthogonal idempotents of K [ ˜ G ] and there are 2 d idempotents in K [ ˜ G ], we havethat any idempotent of K [ ˜ G ] is a sum of some of the e j for 1 ≤ j ≤ d . Notethat the condition that γ i, d + γ i, d + · · · + γ i,t i d t i = 0 for some i ∈ { , , . . . , r } for all nonzero vectors ˜ d = ( d , d , . . . , d t i ) and d j i ∈ { , } is equivalent to thatany sum of some of the e j is not in Ker( L | ˜ G ) except zero for all 1 ≤ j ≤ t .Hence any sum of some of the e j is not in Ker( L | ˜ G ) for all 1 ≤ j ≤ d if thesum contains e j for some j ∈ { , , . . . , t } . Thus, any sum of some of the e j isnot in Ker L for all 1 ≤ j ≤ d if the sum contains e j for some j ∈ { , , . . . , t } .Otherwise, the sum of e j belong to Ker L ∩ K [ ˜ G ] = Ker( L | ˜ G ) for 1 ≤ j ≤ d , whichis a contradiction. It follows from Lemma 3.4 that K [ G ] and K [ ˜ G ] have the sameidempotents. Hence Ker L/I has no nonzero idempotent. It follows from Theorem4.2 in [5] that Ker
L/I is a Mathieu-Zhao space of K [ G ] /I . Hence it follows fromProposition 2.7 in [5] that Ker L is a Mathieu-Zhao space of K [ G ].9 emark 3.6. If G is cyclic in Theorem 3.5, then all the primitive orthogonalidempotents of K [ G ] are e j = d − (1 + ( ξ d − ) j − ˜ g + · · · + ξ j − ˜ g d − ) for 1 ≤ j ≤ d ,where ξ is a d -th root of unity and ˜ G is generated by ˜ g , where ˜ G be as in Theorem3.5. Acknowledgement : The author is very grateful to professor Wenhua Zhaofor some useful suggestions. She is also grateful to the Department of Mathematicsof Illinois State University, where this paper was partially finished, for hospitalityduring her stay as a visiting scholar.
References [1] J. J. Duistermaat, W. van der Kallen,
Constant terms in powers of a Laurentpolynomial , Indag. Math. (N.S.) 9(2) (1998) 221-231.[2] A. van den Essen, S. Kuroda, A. J. Crachiola,
Polynomial Automorphismsand the Jacobian Conjecure: New Results from the Beginning of the 21stcentury , To appear.[3] I.M. Isaacs,
Characters Theory of Finite Groups , Dover, New York, 1994.[4] A. Konijnenberg,
Mathieu subspaces of finite products of matrix rings , mas-ter’s thesis, Radboud University Nijmegen, The Netherlands, 2012.[5] Wenhua Zhao,
Mathieu subspaces of associative algebras , J. Algebra350(2)(2012) 245-272.[6] Wenhua Zhao,
Generalizations of the image conjecture and Mathieu conjec-ture , J. Pure Appl. Algebra 214(7)(2010) 1200-1216.[7] Wenhua Zhao, Roel Willems,
Analogue of the Duistermaat-Van Der Kallentheorem for group algebras , Cent. Eur. J. Math. 10(3)(2012) 974-986.[8] Wenhua Zhao, Dan Yan,