Problem 1. [5 points.]
Consider R the ring of all rational numbers of the form a/b where a, b are intergers such that b
is not divisible by 5. What are the prime ideals of R? What are the maximal ideals of R?
R is the localization of Z at the prime ideal (5).
The prime ideals in R are prime ideals in Z which don’t intersect Z \ (5) i.e. which are contained
in the ideal (5).The correspondence is
p → S −1 p, for S = Z \ (5).
The prime ideals in Z are either (0) or (p) for p prime, and only (0) and (5) are contained in (5).
Hence the prime ideals in R are either (0) or the maximal ideal
a
m = { : a ≡ 0 mod 5, b 6≡ 0 mod 5}.
b
Problem 2. [5 points]
Let p > q be distinct primes. Show that any group with pq elements is solvable.
Hint: Such a group cannot be simple.
Let |G| = pq. Consider a Sylow p-subgroup H. We must have
np |q, np ≡ 1
mod p =⇒ np = 1( since p > q).
Thus H must be normal in G.
Since H has p elements, it is cyclic. Similarly, G/H has q elements, hence it must be cyclic. Then,
1⊂H⊂G
is the desired series with successive abelian quotients, showing G is solvable.
Problem 3. [6 points]
Let n ≥ m ≥ 7. Describe all group homomorphisms φ : Sn → Sm . (You are allowed to use any
homework problems without proof.)
Let K be the kernel of φ. Then, K must be normal in Sn . We saw in the homework that this
implies either K = {1}, K = An or K = Sn .
When K = Sn , the homomorphism φ is trivial.
When K = {1}, the homomorphism φ must be injective, hence
|Sn | ≤ |Sm | =⇒ n! ≤ m! =⇒ n ≤ m =⇒ n = m.
Therefore, φ must be bijective, hence an automorphism, hence an inner automorphism since n 6= 6:
φ(σ) = τ στ −1 ,
for some τ ∈ Sn .
When K = An , we must have
Im φ ∼
= Sn /An ∼
= Z/2Z.
Thus, the image of φ is {1, τ } where τ is a transposition. We must have
φ(σ) = 1 if σ even, φ(σ) = τ for σ odd.
Problem 4. [9 points]
(i) [3] How many abelian groups with 72 · 25 are there?
We have G = A × B with |A| = 72 , |B| = 25 .
The groups A correspond to the partitions of the exponent 2: 2 or 1 + 1. There are two
options for A.
The group B correspond to partitions of the exponent 5: 5, 4 + 1, 3 + 1 + 1, 3 + 2, 2 + 2 + 1,
2 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1. There are 7 options for B.
Therefore there are 2 × 7 = 14 abelian groups of order 72 · 25 .
(ii) [6] Show (by an explicit construction) that there exists at least one non-abelian group with
72 · 25 elements.
Consider H = Z/7Z × Z/7Z. We have Aut(H) = GL2 (F7 ) which has (72 − 1)(72 − 7) =
7 · 25 · 32 . elements. This number is divisible by 25 , so there must be a Sylow subgroup of
order 25 , say K. We have
φ : K → Aut(H)
the inclusion map. The semidirect product H oφ K is the desired group of order 72 · 25 .
The dihedral group is another possible answer.
Problem 5. [6 points.]
Consider a simple group G with 60 elements. Find the number of Sylow 3-subgroups of G.
We have
n3 |20, n3 ≡ 1
mod 3 =⇒ n3 = 1, 4 or 10.
If n3 = 1, then the 3-Sylow is normal, contradicting that G is simple.
If n3 = 4, we have 4 Sylow subgroups, all conjugate. Consider the action of G on the Sylow
subgroups by conjugation. Each element g ∈ G gives a permutation of the 4 subgroups, giving a
homomorphism
φ : G → S4 .
Since G is simple, the kernel of φ must be trivial, hence φ is injective. But |G| = 60 > |S4 | contradiction.
Therefore n3 = 10.