# Write up and turn in solutions to the highlighted problems from the ```HOMEWORK 11 FOR MATH-UA.0343-1: ALGEBRA 1 - FALL 2013
Write up and turn in solutions to the highlighted problems from the list below. However, solve all
other problems as well. The quizzes will cover the complete list of homework problems.
Due date: Friday, November 22, in the beginning of the recitation. NO LATE HOMEWORK
WILL BE ACCEPTED.
Problems.
1. Problems 22 and 23 on page 103 of Herstein.
2. Let G be a finite group and p a prime. Fix a subgroup H ≤ G such that p divides |H|.
(a) Let Q be Sylow p-subgroup of H. Prove that there exists a Sylow p-subgroup P of G such that
Q = P ∩ H.
Hint: use problem 3 from Homework 10.
(b) Assume that H is normal in G and that S ≤ G is a Sylow p-subgroup of G. Prove that S ∩ H
is a Sylow p-subgroup of H.
Hint: Fix an arbitrary Sylow p-subgroup Q of H and let P ≤ G be the Sylow p-subgroup from
part (a) of this question. Use Sylow’s second theorem for P and S.
(c) Assume that H is normal in G and that S ≤ G is a Sylow p-subgroup of G. Prove that (SH)/H
is a Sylow p-subgroup of G/H.
(d) Let P ≤ G be a Sylow p-subgroup of G and assume that N (P ) ⊆ H ⊆ G. Prove that H = N (H)
(here, as usual, N (H) is the normalizer of H in G).
Hint: fix x ∈ N (H). Show that P and xP x−1 are both Sylow p-subgroups of H. Use Sylow’s
second theorem to deduce that they must be conjugate in H.
3. The following problems from page 102 in Herstein’s book: 7, 13 .
Hint for problem 13: Let G be a group of order p2 q. If p = q show that G has a non-trivial center.
If p &gt; q then show that a Sylow p-subgroup of G must be normal. If q &gt; p show that if G has
a subgroup of order pq then we are done (what is the index of such a subgroup? Recall problem
1 on Homework 10). Deduce therefore that if G is simple then it must have p2 Sylow q-subgroups
and q Sylow p-subgroups. Show that this yields a contradiction by counting the the total number of
elements in these subgroups.
4. In this problem we will show that there is no simple group of order 120. The proof will be by
contradiction, so assume that G is a simple group of order 120.
(a) Let P be a Sylow 5-subgroup of G. Show that |G : N (P )| = 6.
Hint: Deduce from Sylow’s third theorem, in combination with the assumption that G is simple,
that G has exactly 6 Sylow 5-subgroups.
(b) Deduce from problem 4 in Homework 7 that there is a homomorphism ψ : G → S6 whose kernel
is contained in N (P ). Consider the subgroup H = ψ(G) of S6 . Show that the assumption that
G is simple implies that H is isomorphic to G, and hence H is also a simple group of order 120.
(c) Show that H ∩ A6 is a normal subgroup of H, and therefore it equals either H or {e}. Use the
counting principle for products of subgroups to show that |H : H ∩ A6 | is either 2 or 1, and use
this information to deduce that H ⊆ A6 .
(d) Show that |A6 : H| = 3 (what is the order of A6 ?). Use part (b) of problem 1 from Homework
10 to deduce that A6 is not a simple group, in contradiction to what we proved in class.
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