Group Theory: Math30038, Sheet 8

GCS Solutions

1. Let P be a Sylow p-subgroup of the finite group G. Suppose that N ⊴G.

(a) Show that P ∩ N ∈ Syl_{p}(N). Solution P ∩ N ≤ P so |P ∩ N| is a power of p. Also P N/N ≃ P/P ∩ N so |P N : N| = |P : P ∩ N| but

|P N : N| · |N : P ∩ N| = |P N : P ∩ N| = |P N : P | · |P : P ∩ N|

and therefore |N : P ∩ N| = |P N : P |. Now |P N : N| is a divisor of |G : P | and so is coprime to p. Thus P ∩ N is a Sylow p-subgroup of N.

(b) Show that P N/N ∈ Syl_{p}(G/N). Solution P N/N has p-power order by the 2nd isomorphism theorem. Now |G : P N| divides |G : P | an so is coprime to p. Now |G/N : P N/N| = |G : P N| since if T is a left transveral for P N in G, then {tN | t ∈ T } is easily seen to be a left transversal for P N/N in G/N.

2. Show that every group of order 15 must be abelian.

Solution Sylow’s theorem tells us that there are unique Sylow 5subgroups and 3-subgroups P and Q respectively. These subgroups must be normal in G since conjugation must leave each one invariant. Now for all a ∈ P and b ∈ Q we have a^{−1}b^{−1}ab ∈ P ∩ Q = 1 so a and b commute. Now if a ∈ A, then A ≤ C_{G}(a) since A is cyclic (of prime order). We have just shown that B ≤ C_{G}(A) so ⟨A, B⟩ ≤ C_{G}(a). Now ⟨A, B⟩ has order divisible by both 3 and 5 and so G = ⟨A, B⟩. Thus a ∈ Z(G) so A ≤ Z(G). Similarly B ≤ Z(G) so G = ⟨A, B⟩ ≤ Z(G) and therefore G is abelian.

3. Show that every group of order 35 must be abelian.

Solution The solution is a copy of the previous solution.

4. Show that there is no non-abelian finite simple group of order less than 60. Solution A group of p-power order has a non-trivial centre. This will prevent G from being simple unless G = Z(G) is abelian (in which case G will be simple if and only if |G| = p but this is not needed). Thus there is no non-abelian finite simple group of prime power order. Now using the results of the next question we may focus on groups of the following orders: 24, 40, 48, 54, 30 and 56.

(a) A simple group of order 24 would have 3 Sylow 2-subgroups, and so there would (see Poincaré’s theorem) be a non-trivial homomorphism from G to S_{3} which will be injective by simplicity and therefore 24 would divide 6 which is absurd.

(b) A group G of order 40 must have a unique Sylow 5-subgroup and so can not be simple.

(c) A simple group of order 48 must have 3 sylow 2-subgroups, and so (see —G— = 24) we can deduce that 48 divides 6 which is absurd.

(d) A group of order 54 must have a Sylow 3-subgroup H of index 2 in G. Now any subgroup of index 2 is normal so G can not be simple.

(e) A simple group of order 30 would have 6 Sylow 5-subgroups. Since any pair of these groups intersects in the identity, it follows that there are 24 elements of order 5. Similarly this this group must have 10 Sylow 3-subgroups and so 20 elements of order 3. However, 24 + 20 > 30 so this is absurd.

(f) A simple group of order 56 would have 8 Sylow 7-subgroups, any pair of which would intersect in the trivial group. Thus there would be 48 elements of order 7. This leaves 8 elements remaining, which must all belong to a Sylow 2-subgroup. This Sylow 2-subgroup is unique and therefore invariant under conjugation and thus is normal in G. This is absurd.

5. Let p and q be distinct prime numbers.

(a) Show that a group of order pq can not be simple. Solution Suppose w.l.o.g. that p < q. By Sylow’s theorem there is a unique Sylow q-subgroup Q. Now Q is invariant under conjugation by elements of G, and so must be a normal subgroup of G.

(b) Show that a group of order p^{2}q can not be simple. Solution Suppose (for contradiction) that G is simple. By Sylow’s theorem the number of Sylow q-subgroups is p or p^{2}, and this number must be congruent to 1 modulo q. Therefore q divides p − 1 or p^{2} − 1 = (p − 1)(p + 1). In the first event q < p. In the second event q < p or (p, q) = (2, 3). Now, if (p, q) = (2, 3) it follows that |G| = 12. In a non-simple group of order 12 there must be 4 Sylow 3-subgroups, any pair of which intersect in the identity. There are therefore 8 elements of order 3. The number of elements of order dividing 4 is therefore at most 12 − 8 = 4. Let P be a Sylow 2 subgroup of G, a group of size 4, all elements of P will have order dividing 4. Thus P must consist of all the elements of G which are not of order 3. Thus there is a unique Sylow 2-subgroup which violates simplicity. Thus we may assume that q < p. Now (see Poincaré’s theorem) we have a non-trivial homomorphism G → S_{q} which must be injective by the simplicity of G. Now by Lagrange’s theorem |G| divides q! and so p divides q! which is false.

(c) Show that a group of order p^{2}q^{2} can not be simple. Solution Assume (for contradiction) that G is simple. The number of Sylow q subgroups q and must be p or p^{2} (by simplicity) so q divides p − 1 or divides p^{2} − 1 = (p − 1)(p + 1). Thus either q < p or q = 3, q = 2 (so |G| = 36). Reversing the roles of p and q we obtain that either p < q or |G| = 36. Thus we are done unless |G| = 36, but then a Sylow 3-subgroup has index 4 and so (see Poincaré’s theorem) there is a non-trivial homomorphism from G to S_{4} which must be injective by simplicity. Now by Lagrange’s theorem 36 divides 24 which is absurd.