A bound for the orders of centralizers of irreducible subgroups of algebraic groups

Let 𝐺 be a connected semisimple algebraic group of rank 𝑟 over an algebraically closed field 𝐾, and let 𝐻 be a 𝐺-irreducible subgroup. Then there is a constant c ≤ 197 such that

(1) Let G = SO n ⁢ ( K ) with char ⁡ ( K ) ≠ 2 , and let v 1 , … , v n be an orthonormal basis of the underlying orthogonal space. Then 𝐺 has an elementary abelian subgroup H ≅ 2 n - 1 consisting of elements that send each v i ↦ ± v i . It is easy to see that 𝐻 is 𝐺-irreducible, and | C G ⁢ ( H ) | = | H | = 2 n - 1 . When 𝑛 is odd, this is equal to 4 r , where 𝑟 is the rank of 𝐺.

(2) Let G = Sp 2 ⁢ r ⁢ ( K ) with char ⁡ ( K ) ≠ 2 . Then 𝐺 has an irreducible subgroup H = Sp 2 ⁢ ( K ) r , and | C G ⁢ ( H ) | = | Z ⁢ ( H ) | = 2 r .

(3) Let 𝑠 be a prime and H = s 1 + 2 ⁢ a an extraspecial group. There is an irreducible embedding of 𝐻 in SL n ⁢ ( K ) , where n = s a and char ⁡ ( K ) ≠ s . Hence we have H ¯ = s 2 ⁢ a < G = PGL n ⁢ ( K ) and | C G ⁢ ( H ¯ ) | = | H ¯ | = s 2 ⁢ a = n 2 . Our proof of Theorem 1 shows that n 2 is actually the correct bound for 𝐺 simple of type A n - 1 (see Lemma 2.3).

(4) Here are some examples for 𝐺 simple of exceptional type (see for example [2, Theorem 3]):

So, for example, there is an irreducible subgroup H = ( 2 5 ) l < G = E 8 l such that | C G ⁢ ( H ) | = 2 15 ⁢ l = c rk ⁡ ( G ) , where c = 2 15 / 8 .

Suppose the conclusion of Theorem 1 holds in the case where 𝐺 is simple of adjoint type. Then the conclusion holds in general.

Proof

Let G = G 1 ⁢ ⋯ ⁢ G k , a commuting product of simple algebraic groups G i . Let G ¯ = G / Z ⁢ ( G ) = G ¯ 1 × ⋯ × G ¯ k , the direct product of adjoint groups G ¯ i , and let π : G ↦ G ¯ be the natural map. Let H < G be 𝐺-irreducible and H ¯ = π ⁢ ( H ) . Then 𝜋 maps C G ⁢ ( H ) ↦ C G ¯ ⁢ ( H ¯ ) , so | C G ⁢ ( H ) | ≤ | C G ¯ ⁢ ( H ¯ ) | ⁢ | Z ⁢ ( G ) | . Moreover, we have C G ¯ ⁢ ( H ¯ ) = ∏ C G ¯ i ⁢ ( H ¯ i ) , where H ¯ i is the projection of H ¯ in G ¯ i . By hypothesis, | C G ¯ i ⁢ ( H ¯ i ) | < c r i , where r i = rank ⁡ ( G i ) , and so

and the lemma follows. ∎

The group 𝐹 consists of semisimple elements.

Proof

Suppose false, and let f ∈ F be an element with non-identity unipotent part 𝑢. Then H ≤ C G ⁢ ( u ) , which is contained in a parabolic subgroup, contradicting the irreducibility of 𝐻. ∎

Suppose G = PGL n ⁢ ( K ) . Then | C G ⁢ ( H ) | ≤ n 2 .

Proof

Let G ^ = SL n ⁢ ( K ) = SL ⁢ ( V ) , and let H ^ , F ^ and C ^ be the preimages in G ^ of 𝐻, 𝐹 and C G ⁢ ( F ) , respectively. Also let Z = Z ⁢ ( G ^ ) . Note that C G ^ ⁢ ( H ^ ) = Z since H ^ acts irreducibly on 𝑉.

Observe first that [ H ^ , F ^ , F ^ ] = [ F ^ , H ^ , F ^ ] = 1 , and therefore [ F ^ , F ^ , H ^ ] = 1 . Hence F ^ ′ ≤ C G ^ ⁢ ( H ^ ) = Z , and so 𝐹 is abelian. Let 1 ≠ f ∈ F , and let f ^ be a preimage of 𝑓. Then C G ^ ⁢ ( f ^ ) = ∏ GL m i ⁢ ( K ) ∩ G ^ , where ∑ m i = n . As C G ⁢ ( f ) is irreducible, its preimage in G ^ must permute the factors transitively, and it follows that C G ( f ) = ( GL m ( K ) r . r ∩ G ^ ) / Z for some 𝑟 dividing 𝑛, and 𝑓 has order dividing 𝑟. Hence 𝐹 is abelian of exponent dividing 𝑛.

For c ∈ C ^ , there is a map χ c ∈ Hom ⁡ ( F ^ , Z ) given by χ c ⁢ ( f ) = [ c , f ] for all f ∈ F ^ . The map π : C ^ ↦ Hom ⁡ ( F ^ , Z ) sending c ↦ χ c is a homomorphism.

Assume now that F ^ is abelian. For χ ∈ Hom ⁡ ( F ^ , Z ) , define

Then V ↓ F ^ = ⨁ 1 t V χ i , where V χ i ≠ 0 for all 𝑖. For c ∈ C ^ , we have V χ i ⁢ c = V χ i ⁢ χ c , and as C ^ is irreducible, this action of C ^ permutes the set { V χ i : 1 ≤ i ≤ t } transitively. Replacing each f ∈ F ^ by a scalar multiple, we may take χ 1 to be trivial (i.e. χ 1 ⁢ ( f ) = 1 for all f ∈ F ^ ). It follows that | F | is at most the order of the transitive group π ⁢ ( C ^ ) . Being transitive and abelian, this group has order 𝑡, and hence | F | ≤ t ≤ n , giving the conclusion in this case.

Now assume that F ^ is non-abelian. Let V ↓ F ^ = ⨁ 1 t W i , where W i are the homogeneous components. As above, { W 1 , … , W t } is permuted transitively by π ⁢ ( C ^ ) . The action of F ^ on the homogeneous component W 1 has order at most ( dim ⁡ W 1 ) 2 ⁢ | Z | by [3, Theorem 2.31], and hence

This completes the proof. ∎

Suppose G = PGSp n ⁢ ( K ) or PGO n ⁢ ( K ) . Then | C G ⁢ ( H ) | ≤ 4 2 ⁢ r , where r = rank ⁡ ( G ) .

Proof

Let G ^ = Sp n ⁢ ( K ) or SO n ⁢ ( K ) , Z = Z ⁢ ( G ^ ) , and V = K n . Let F ^ be the preimage in G ^ of F = C G ⁢ ( H ) .

If 𝐹 contains an element 𝑓 of odd prime order, then C G ⁢ ( f ) is connected and has a central torus, hence cannot be irreducible. It follows that 𝐹 is a 2-group.

Next we show that F ^ has exponent dividing 4. Suppose then that F ^ contains an element 𝑓 of order 8, with image f ¯ ∈ F . Let ω ∈ K be a primitive 8-th root of unity, and let E ω j be the ω j -eigenspace of 𝑓 on 𝑉 for 0 ≤ j ≤ 7 . We can assume that E ω ≠ 0 , and hence also E ω - 1 ≠ 0 .

Let g ¯ ∈ C G ⁢ ( f ¯ ) , with preimage g ∈ G ^ . Then f g = ± f . If f g = f , then 𝑔 stabilizes every eigenspace E ω j ; for j ≠ 0 , 4 , these are all totally singular. And if f g = - f , then 𝑔 swaps E ω and E - ω , hence stabilizes E ω + E - ω , which is also totally singular. We conclude that C G ⁢ ( f ¯ ) stabilizes a totally singular subspace of 𝑉, hence is a reducible subgroup of 𝐺, a contradiction.

Hence F ^ has exponent dividing 4, as claimed. Since F ^ is contained in the normalizer of a maximal torus by [7, II, 5.16], we have

where W ⁢ ( G ) is the Weyl group and 𝑟 is the rank of 𝐺. Since | W ⁢ ( G ) | = 2 r - δ ⁢ r ! with δ ∈ { 0 , 1 } , it follows that | W ⁢ ( G ) | 2 ≤ 4 r . Hence | F ^ | ≤ 4 2 ⁢ r , as required. ∎

Suppose 𝐺 is of exceptional type, of rank 𝑟. Then | C G ⁢ ( H ) | ≤ c r , where 𝑐 is as in the table below.

Proof

First assume that G ≠ E 8 . We claim that the non-identity elements of prime-power order in 𝐹 can only have the following possible orders:

To see this, let 1 ≠ f ∈ F have prime-power order. If C G ⁢ ( f ) is connected, then as it is irreducible, it is semisimple. Then the order o ⁢ ( f ) is equal to a coefficient of the expression for the highest root in terms of simple roots (see for example [6, (4.5)]), which gives the conclusion in these cases. This deals with G = G 2 , or F 4 , as these are simply connected; hence all their semisimple element centralizers are connected by [7, II, 3.9]. Now consider G = E 7 with C = C G ⁢ ( f ) disconnected. Then | C / C 0 | = 2 by [7, II, 4.4]. In the simply connected cover G ^ , a preimage f ^ of 𝑓 satisfies f ^ g = z ⁢ f ^ for some g ∈ G ^ , where Z ⁢ ( G ^ ) = ⟨ z ⟩ of order 2, and hence 𝑓 is a 2-element. Let C 0 = D ⁢ T , where 𝐷 is a semisimple group and 𝑇 a central torus, and write f = d ⁢ t with d ∈ Z ⁢ ( D ) , t ∈ T . Let g ∈ C ∖ C 0 . Then C T ⁢ ( g ) 0 = 1 as 𝐶 is irreducible in 𝐺, and hence 𝑔 acts on 𝑇 as an involution. It follows that t g = t - 1 , and hence t 2 = d g ⁢ d - 1 ∈ Z ⁢ ( D ) . If t 2 has order greater than 2, then it must have order 4 and be contained in a factor A 3 , A 3 ⁢ A 3 or A 7 of 𝐷. In the first case, d g ⁢ d - 1 has order 1 or 2 (being a product of two elements of the same order in Z ⁢ ( A 3 ) ), a contradiction. In the second case, t ∈ C G ⁢ ( A 3 ⁢ A 3 ) = A 1 , so t 2 ∈ Z ⁢ ( A 1 ) again has order at most 2. Finally, if D = A 7 , then T = 1 and f = d ∈ Z ⁢ ( D ) , which has order 4. This establishes the claim for G = E 7 , and the argument for G = E 6 is similar.

From the previous paragraph, we can list the possible centralizers of the elements of 𝐹 of orders specified in the following table; the table also gives the traces of their actions on the adjoint module L ⁢ ( G ) :

From Table 1, we see that 𝐹 is a { 2 , 3 } -group, hence is solvable, and so there exist Sylow 2- and 3-subgroups P 2 , P 3 of 𝐹 such that F = P 2 ⁢ P 3 . We can bound the orders of P 2 and P 3 as in the previous proof. Since P 2 has exponent dividing 4 and is contained in the normalizer of a maximal torus, we have

where W ⁢ ( G ) is the Weyl group of 𝐺. Similarly,

We can use the trace values given above to reduce these bounds for some cases. For example, consider P 3 < E 7 . If | P 3 | = 3 a , then since the trace of every non-identity element of P 3 is −2, we have

The right-hand side can only be a non-negative integer if a ≤ 3 , and hence we have | P 3 | ≤ 3 3 . Similar calculations give the following bounds for other cases:

The bounds in the conclusion now follow from these together with (2.2): for example, consider G = E 7 . Here we have

and this is less than c 7 for c = 17.3 .

Now consider G = E 8 . For 1 ≠ f ∈ F , C G ⁢ ( f ) is irreducible and connected, so as above, 𝑓 has order equal to a coefficient of the highest root, hence to 2 , 3 , 4 , 5 or 6. Moreover, any element of order 5 in 𝑓 has centralizer A 4 ⁢ A 4 and trace −2 on L ⁢ ( G ) .

Observe that 𝐹 is a { 2 , 3 , 5 } -group. Suppose first that 𝐹 is solvable so that F = P 2 ⁢ P 3 ⁢ P 5 , where each P i is a Sylow 𝑖-subgroup. If | P 5 | = 5 a , then

which forces a ≤ 3 . Hence, as above,

giving the conclusion.

Now suppose that 𝐹 is non-solvable. Any non-abelian composition factor of 𝐹 is a simple { 2 , 3 , 5 } -group, and inspection of the simple groups shows that the only possibilities are A 5 , A 6 and U 4 ⁢ ( 2 ) . However, U 4 ⁢ ( 2 ) is excluded, as it has an element of order 12 which is not in the list of possible orders of elements of 𝐹. Let 𝑅 be the solvable radical of 𝐹 (i.e. the largest solvable normal subgroup). Then F / R has socle F 1 × ⋯ × F t , a direct product of non-abelian simple groups F i , each isomorphic to A 5 or A 6 . If t ≥ 2 , then F / R has an element of order 15, which is not possible. Therefore, t = 1 and F / R has socle A 5 or A 6 . Then 𝐹 has a solvable subgroup 𝐽 of index 5 or 10. As above, we have | J | < 147 8 , and hence

This completes the proof of the lemma. ∎