MATH 251 Homework 6 Solutions 1. Let G be a finite group, N ¢ G ...

4. Let G be a finite group, p a prime dividing |G| and N ✂ G.(a) Prove that Syl p (N) = {S ∩ N | S ∈ Syl p (G)}.(b) Prove that Syl p (G/N) = {SN/N | S ∈ Syl p (G)}.Proof : (a) Let T ∈ Syl p (N). Then T is a p-subgroup of G. By the Third Sylow Theorem, there existsS ∈ Syl p (G) with T ⊆ S. Hence T ≤ S ∩ N. Note that S ∩ N is a p-subgroup of N since S ∩ N ≤ N,S ∩ N ≤ S and so |S ∩ N| divides |S|, which is a power of p. But T is a maximal p-subgroup of N (sinceT ∈ Syl p (N)). Hence T = S ∩ N. So Syl p (N) ⊆ {S ∩ N | S ∈ Syl p (G)}.Let S ∈ Syl p (G). We proved that there exists S 1 ∈ Syl p (G) such that S 1 ∩ N ∈ Syl p (N) (indeed, letT ∈ Syl p (N); then T = S 1 ∩ N for some S 1 ∈ Syl p (G)). By the Second Sylow Theorem, S = S g 1 for someg ∈ G. Hence (S 1 ∩ N) g = S g 1 ∩ N g = S ∩ N since N ✂ G. In particular, |S ∩ N| = |(S 1 ∩ N) g | = |S 1 ∩ N|.Since S 1 ∩N ∈ Syl p (N) and S∩N is a p-subgroup of N with |S∩N| = |S 1 ∩N|, we get that S∩N ∈ Syl P (N).So {S ∩ N | S ∈ Syl p (G)} ⊆ Syl p (N).(b) Let S ∈ Syl p (G). By the Second Isomorphism Theorem, SN/N ∼ = S/(S ∩ N), which is a p-group since|S/(S ∩ N)| divides |S|. Soord p (|SN/N|) = ord p (|S/(S ∩ N)|) = ord p (|S|/|S ∩ N|) = ord p (|S|) − ord p (|S ∩ N|)Since S ∈ Syl p (G), ord p (|S|) = ord p (|G|). By (a), S ∩ N ∈ Syl p (N) and so ord p (|S ∩ N|) = ord p (|N|).Henceord p (|SN/N| = ord p (|S|) − ord p (|S ∩ N|) = ord p (|G|) − ord p (|N|) = ord p (|G|/|N|) = ord p (|G/N|)Since SN/N is a p-subgroup of G/N and |SN/N| is the size of a Sylow p-subgroup of G/N, we get thatSN/N ∈ Syl p (G/N). So {SN/N | S ∈ Syl p (G)} ⊆ Syl p (G/N).Let X ∈ Syl p (G/N). Let S ∈ Syl p (G). We proved that SN/N ∈ Syl p (G/N). By the Second SylowTheorem, X = (SN/N) x for some x ∈ G/N. Then x = gN for some g ∈ G andX = (SN/N) gN = {(gN) −1 (sN)(gN) | s ∈ S} = {g −1 sgN | s ∈ S} = S g N/NNote that S g ∈ Syl p (G). So Syl p (G/N) ⊆ {SN/N | S ∈ Syl p (G)}.✷5. Let G be a group of order 700.(a) Prove that G is not simple.(b) Prove that G has a subgroup of order 14.⎧⎨ n 2 ∈ {1, 5, 7, 25, 35, 175}Proof : Note that 700 = 2 2 · 5 2 · 7. We easily get that n 5 = 1⎩n 7 ∈ {1, 50}(a) Since n 5 = 1, G has exactly one Sylow 5-subgroup, which has to be normal in G. So G is not simple.(b) Let P ∈ Syl 7 (G). Suppose first that n 7 = 1. Then P ✂ G. By Proposition 5.6, there exists H ≤ G|P ||H|with |H| = 2. Then P H ≤ G and |P H| =|P ∩H|= 7·21= 14 (note that |P ∩ H| = 1 since it divides|P | = 7 and |H| = 2). Suppose next that n 7 = 50. Then |N G (P )| = 70050 = 14. ✷2