# FINITE GROUP THEORY: SOLUTIONS FALL MORNING 5. Stab G (l) =.

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1 FINITE GROUP THEORY: SOLUTIONS TONY FENG These are hints/solutions/commentary on the problems. They are not a model for what to actually write on the quals FALL MORNING 5 (i) Note that G acts transitively on the set of l. Picking l = [1 : 0], we see that the stabilizer of l is the upper-triangular Borel Stab G (l) =. Since the stabilizer groups of all the l are conjugate (by transitivity), it suffices to prove that this particular one has a unique p -sylow. By counting its size, the p - Sylow has order q = p n. By inspection, the unipotent radical 1 N = 1 is a p -Sylow in Stab G (l). Since all p -Sylows are conjugate, the statement that it is unique is equivalent to it being normal, which we check explicitly. You should know how to do the computation #G (F q ) = (q 2 1)(q 2 q ). In particular, the biggest power of p dividing this is q = p n, so N is also a p -Sylow of G. Since all p -Sylows are conjugate, to count the total number of p -Sylows we just have to count the number of N. Alternatively, show that they are in bijection with l. To recover l from N, take the span of the fixed vector of N. (ii) First assume l 1 = [0, 1] and l 3 = [1, 0]. Then we are asking for g fixing l 1,l 3 and taking any l 2 to [1, 1]. Note that we can express l 2 = [a, 1] with a 0. Then take a g =. 1 Now for the general case. Argue that we can find any g taking l 1 to [0, 1] and l 3 to [1, 0]. Then g might not send l 2 to [1, 1], but by the first special case there is an h sending g (l 2 ) to [1, 1] and fixing the other lines, so h g does the trick. (iii) We proved in (i) that P i is the unique p -Sylow subgroup of Stab G (l i ) for some l i, and Q i is the unique p -Sylow subgroup of Stab G (l i ). By (b) there exists g such that g (l i ) = l i. Therefore g Stab G (l i )g 1 = Stab G (g l i ), and so the unique p -Sylow subgroups match. 1

2 2 TONY FENG SPRING MORNING 1 (a) Proceed by contradiction. We assume that G = g H g 1. (2.0.1) g G Let s count how many distinct g H g 1 appear above. By orbit-stabilizer, it is G /N G (H ). Now note that H N G (H ), so G /N G (H ) G /H. Each conjugate of H has H elements, and each contains the identity of G. So the total number of elements of G accounted for by the right side of (2.0.1) is if H < G. 1 + G /N G (H ) ( H 1) 1 + G /H ( H 1) < G (b) The stabilizers are all all conjugate. By (a), there is some g G not in any stabilizer FALL AFTERNOON 1 (a) In this case g H g 1 H are elements in G fixing both x and g x. If G is Frobenius then this has only the identity. Conversely, if G is not Frobenius then some g G lies in Stab G (k x ) and Stab G (k x ). Then kh k 1 (k )H (k ) 1 is non-trivial, so (k ) 1 kh k 1 (k ) H is non-trivial. (b) Take S = F q, with G by affine transformations. Then H is the stabilizer of FALL AFTERNOON 5 Note that F p 3 as a 3-dimensional vector space over F p = Z/p. Picking a basis for it, we can identify GL 3 (Z/p) with the group of F p -linear automorphisms on F p 3. The subgroup of F p 3-linear automorphisms gives an inclusion (i) First compute the size of SL 3 (F p ): F p 3 GL 3 (F p ). # SL 3 (F p ) = (p 3 1)(p 3 p)(p 3 p 2 ) (p 1) = (p 2 + p + 1) 2 (p 1) 2 (p + 1). Check that if l p 2 + p + 1 then l p(p 1)(p + 1). The l-sylow then comes from (F p 3 ) Nm=1 SL 3 (F p ). (ii) In this case the 3-Sylow is the semidirect product of the 3-Sylow in (F p 3 ) Nm=1 with Gal(F p 3/F p ), which is not even commutative SPRING MORNING 3 (i) Note that x y x 1 y 1 lies in P 2, since it can be written as x (y x 1 y 1 ) with both factors in P 2 by normality. Similarly, it lies in P 7. But any element in the intersection of P 2 and P 7 has order simultaneously a power of 2 and of 7, so the

3 FINITE GROUP THEORY: SOLUTIONS 3 order must be 1. (ii) Let n 2 denote the number of 2-Sylows and n 7 denote the number of 7-sylows. By the Sylow theorems, we know: n 2 1 (mod 2) and n 2 7. n 7 1 (mod 7) and n 7 8. We want to show that n 2 = 1 or n 7 = 1. If not, then by inspection we must have n 2 = 7 and n 7 = 8. We ll show that there are not enough elements in the group to allow this to happen. Any two Sylow 7-subgroups can intersect in only the identity element, since they are cyclic. So each Sylow 7-subgroup contributes 6 new elements of order 7, for a total of 8 6 = 48 distinct non-identity elements in G of order 7. Any two Sylow 2-subgroups can intersect in a group of size at most 4. Therefore, two distinct 2-Sylow subgroups contribute at least 8+4 = 12 elements not already accounted for by the above count of elements of order 7. But = 60 already exceeds the size of G. (iii) Make a non-split semi-direct product Z/7 Z/8 by having Z/8 act through the nontrivial homomorphism Z/8 (Z/7) = Aut(Z/7). (The non-normality follows from the fact that the two subgroups don t commute.) Make a non-split semi-direct product (Z/2) 3 Z/7 by having Z/7 act through Z/7 F 8 GL 3(Z/2) SPRING A2 (i) By assumption, we can find g G such that g x g 1 = y. We want to try to get g N, the normalizer of P. In other words, we want to choose g so that g P g 1 = P. We are free to translate g on the left by C (y ) and on the right by C (x ). If we can get g P g 1 to be in C (y ), then by the Sylow theorems applied to C (y ) we can left-translate by something in C (y ) so that g P g 1 = P. Is it the case that g P g 1 C (y )? This is asking if conjugation by y induces the identity on g P g 1. In other words, does conjugation by g 1 y g induce the identity on P? But g 1 y g = x, which lies in C (P ) by assumption! (ii) The normalizer of N is the group of upper-triangular matrices. The matrices a b and b a and conjugate [why?] in G, but not in N FALL MORNING 1 (i) Arbitrary (entry-wise) choices of lift will lie in GL 3 (Z/p 5 ), because the determinant commutes with reduction. The kernel is in bijection with 3 3 matrices with entries modulo p 4, hence has size p 36.

4 4 TONY FENG (ii) #G = p 36 (p 3 1)(p 3 p)(p 3 p 2 ). An explicit p -Sylow is the pre-image of the unipotent radical FALL MORNING 2 (i) We claim that the unipotent group is a Sylow subgroup. This has size p n(n 1)/2. To check that it works, we compute the size of SL n (F p ): (p n 1)(p n p)...(p n p n 1 ). p 1 The power of p is n 1 = n(n 1) 2. (ii) Lots of possibilities here, e.g. let P i be the subgroup of matrices supported above the i th superdiagonal. Alternatively, you could take matrices supported after the i th column. (i) Use the exact sequence SPRING MORNING 3 0 K GL 2 (Z/9) GL 2 (Z/3) 0. You should be know how to compute # GL 2 (Z/p) for any prime p ; for p = 3 it s (3 2 1)(3 2 3). It remains to compute #K. This kernel is the group of matrices with entries in Z/9, congruent to 1 modulo 3. Show that any such matrix is of the form Id+3M. Note that this only depends on the entries of M modulo 3, and any M is possible. Therefore, K is in bijection with Mat 3 3 (Z/3), which has size 3 4. (ii) If g has 3-power order in G, then its image in GL 2 (Z/3) does as well. If the converse is true, then some 3-power exponent of g lies in K. Argue that the bijection K Mat 3 3 (Z/3) is in fact a group homomorphism, so that K has 3-power order. (Alternatively, this can be seen by pure counting.) (iii) Any Sylow 2-subgroup of G maps isomorphically onto its image in GL 2 (Z/3), because the kernel has to have a power of 3, since it lies in K. Therefore, it sufficesto show the same result for G replaced by GL 2 (Z/3). By counting sizes, check that a Sylow 2-subgroup has size 16. We can view GL 2 (Z/3) = GL 2 (F 3 ) as the group of linear automorphisms of F 9 viewed as a 2-dimensional F 3 -vector space (picking a basis for F 9 over F 3 ). Then the inclusion of the subgroup of automorphisms which are moreover F 9 -linear corresponds to an embedding F 9 GL 2(Z/3).

5 FINITE GROUP THEORY: SOLUTIONS 5 Additionally, the Galois action of Gal(F 9 /F 3 ) is F 3 -linear and gives an embedding Z/2 GL 2 (Z/3), which preserves the subgroup F 9, with the generator acting as x x 3 by the Galois theory of finite fields FALL AFTERNOON 1 (i) Applying the inductive hypothesis to G /G N H /N, we find that G /G N = H /N. This implies that G N = H. (ii) First, use the usual argument that Z 0 (orbit-stabilizer for the conjugation action of H on itself). If G Z = H and G Z = 0, then H = G Z, but then G would not surject onto H /[H,H ].

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