MATH 451 Mid-Term Questions: Group Theory, Exams of Algebra

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NAME: John Q. Public

Question Marks 1 20 2 20 3 20 4 20 5 20 1

Question 1. Throughout this question, let p be a prime. (a) Suppose that G is a finite p-group and that X is a nonempty G-set such that |X| 6 ≡ 0 mod p. Prove that there exists a point x ∈ X such that g x = x for all g ∈ G. (b) Suppose that G is a p-group and that H E G. Prove that if H 6 = { 1 }, then H ∩ Z(G) 6 = { 1 }. (Hint: Consider the action of G on H r { 1 } by conjugation.) (a) Let |G| = pn. Let X = Ω 1 t · · · t Ωt be the decomposition of X into G-orbits. If α ∈ Ωi, then |Ωi| = [G : Gα] = pmi^ for some 0 ≤ mi ≤ n. Since |X| = ∑ti=1 |Ωi| and p does not divide |X|, there exists 1 ≤ i 0 ≤ t such that mi 0 = 0. Hence, letting Ωi 0 = {x}, it follows that g x = x for all g ∈ G. (b) Let |G| = pn. If g ∈ G, then gHg−^1 = H and so g( H r { 1 } )g−^1 = H r { 1 }. Thus G acts by conjugation on H r { 1 }. Since |H| = pm^ for some 1 ≤ m ≤ n, it follows that p does not divide |H r { 1 }|. Hence there exists h ∈ H r { 1 } such that ghg−^1 = h for all g ∈ G. Clearly h ∈ H ∩ Z(G).

Question 3. Suppose that G be a simple group of order 168. Let P be a Sylow 7-subgroup of G and let H = NG(P ).

(a) Prove that |H| = 21. (b) Prove that NG(H) = H. (Hint: Notice that H 6 NG(H) 6 G.) (c) Prove that there exists an element g ∈ G such that gHg−^1 6 = H and gHg−^1 ∩ H 6 = { 1 }. (a) If s is the number of Sylow 7-subgroups of G, then s divides 24 and s ≡ 1 mod 7. Since G is simple, s 6 = 1 and so s = 8. By considering the transitive action of G by conjugation on the set of its Sylow 7-subgroups, we see that [G : NG(P )] = 8 and hence |H| = |NG(P )| = 21. (b) Since H 6 NG(H) 6 G, it follows that d = [G : NG(H)] divides [G : H] = 8. Also by considering the transitive action of G on the coset space G/NG(H), we see that there is an embedding of G into Sd. Thus 7 divides |Sd| and so d = 8. It follows that NG(H) = H. (c) Suppose that gHg−^1 ∩ H = 1 whenever gHg−^1 6 = H. Then the 8 distinct conjugates of H intersect pairwise in 1. Hence

|(

g∈G

Hg^ ) r { 1 }| = 8 × 20 = 160.

But this means that G has a unique Sylow 2-subgroup, which is a contradiction.

Question 4. Prove that 〈 x, y | x^2 = 1, y^2 = 1, (xy)^3 = 1 〉 is a presentation of S 3.

Let X = {x, y} and let N be the normal closure of {x^2 , y^2 , (xy)^3 } in F (X). For each w ∈ F (X), let w¯ = wN ∈ F (X)/N. By von Dyck’s Theorem, there exists a surjective homomorphism ϕ : F (X)/N → S 3 such that ϕ(¯x) = (1 2) and ϕ(¯y) = (2 3). In particular, |F (X)/N | ≥ 6. On the other hand, let

w¯ = ¯xn^1 ¯ym^1 · · · x¯nt^ ¯ymt^ ∈ F (X)/N,

where each ni, mi ∈ Z. Since ¯x^2 = 1 and ¯y^2 = 1, we can suppose that each 0 ≤ ni, mi ≤ 1. Using the relations ¯xy¯x¯y¯x¯y¯ = 1 and ¯x = ¯x−^1 and ¯y = ¯y−^1 , we can now reduce ¯w to one of the following words:

1 , ¯x, y,¯ ¯xy,¯ y¯x,¯ x¯y¯x.¯

Thus |F (X)/N | ≤ 6 and it follows that ϕ : F (X)/N → S 3 is an isomorphism.