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MAT301 Assignment 6 (the last one!)

Due: 23:59 EDT, Monday, August 14, 2023

Justify all claims in your solutions and state the results that you use. You may only use results that have

been covered in Weeks 1–12. Notice that the deadline is after the one in the Syllabus.

Exercise 1. Some exercises on the Week 10 exercise sheet will give you all the tools you need for this

problem. You should work through those exercises, but you can use them without including proofs of them.

Let b1 = (2, 2) and b2 = (?4, 4).

1. Is β0 = {b1, b2} a basis of Z2?

2. Let B = spanZ(β0). Prove that B is a free abelian group of rank 2.

3. Find a basis β = {b′1, b′2} of B, a basis γ = {c1, c2} of Z2, and positive integers d1 | d2 such that

b′1 = d1c1 and b

′

2 = d2c2.

4. Express Z2/B as a direct product of cyclic groups.

Exercise 2. Let A be an abelian group and let B be a free abelian group of finite rank. Prove that for

every surjective homomorphism π : A→ B there exists a homomorphism ι : B → A such that π ? ι = idB .

Exercise 3. Use the Sylow’s theorems to prove that every group of order 45 is abelian.

Exercise 4. Let p be a prime and let Fp = Z/pZ. Note that SL2(Fp) acts on F2p via matrix multiplication.

Let P 1(Fp) be the set of all lines ? ? F2p through the origin of F2p. Recall that a line l in F2p is a subset

of form {at+ b : t ∈ Fp} for some a, b ∈ F2p.

1. For each A ∈ SL2(Fp) and ? ∈ P 1(Fp), define A · ? = {Av : v ∈ ?}. Prove that this defines an action of

SL2(Fp) on P 1(Fp).

2. Prove that |P 1(Fp)| = p+ 1.

(Hint 1: find p+1 vectors v1, . . . , vp+1 ∈ F2p such that the elements of P 1(Fp) are spanFp(v1), . . . , spanFp(vp+1).)

(Hint 2: what can you say about a, b if the respective subset has to contain (0, 0)?)

3. Define

PSL2(Fp) := SL2(Fp)/Z

where

Z := {kI2 : k ∈ Fp} ∩ SL2(Fp) = {kI2 : k ∈ Fp, k2 = 1}.

Using the action of SL2(Fp) on P 1(Fp) from Part 1, construct an injective homomorphism from

PSL2(Fp) to the symmetric group on P 1(Fp).

4. Using the injective homomorphism from Part 3, prove that PSL2(F2) ～= S3, PSL2(F3) ～= A4. (You can

use the fact that S4 has a unique subgroup of index 2 without proving it.)