GRE Math Subject Preparation Seminar
ORGANIZER: SCOT ADAMS
(seminar diary)
NOTE: This seminar has ended.
- MEETING 11 (Friday 23 October 2015)
- Worked problems 44, 54, 61 from practice test 2
- these three problems touch on: differential equations, probablity and measure theory
- Described how to use the Sylow Theorems to prove that, up to isomorphism,
there's only one group of order 15.
- This depends on:
- Every group of order 3 is cyclic.
- Every group of order 5 is cyclic.
- If A and B are normal subgroups of a group G, and if A∩B={1G}, then
- AB is a normal subgroup of G,
- ∀a∈A, ∀b∈B, ab=ba and
- (a,b) ↦ ab : A × B → AB is a group isomorphism.
- By Sylow, any group of order 15 has a unique subgroup of order 3.
- This is because, by Sylow, the number k of such subgroups satisfies
- k is congruent to 1 mod 3 and
- k is a divisor of 15.
- NOTE: {divisors of 15} = {1,3,5,15}. Then k ∈ {1,4,7,10,...} ∩ {divisors of 15} = {1}.
- By Sylow, any group of order 15 has a unique subgroup of order 5.
- This is because, by Sylow, the number ℓ of such subgroups satisfies
- ℓ is congruent to 1 mod 5 and
- ℓ is a divisor of 15.
- NOTE: {divisors of 15} = {1,3,5,15}. Then ℓ ∈ {1,6,11,16,...} ∩ {divisors of 15} = {1}.
- Let G be a group and let n be a positive integer.
If G has a unique subgroup H of order n, then H is a normal subgroup of G.
- Proof: Given g∈G. We wish to prove that gHg-1 = H.
Since |gHg-1| = |H| = n, it follows, by uniqueness, that gHg-1 = H. QED
- If a group G has subgroups A and B, if |A|=3 and if |B|=5, then |A∩B|=1.
- This is because |A∩B| is a divisor of |A| and |B|, and the only common divisor of 3 and 5 is 1.
- Proof that if G is a group and |G|=15, then G is isomorphic to the product of (the cyclic group of order 3)
with (the cyclic group of order 5):
- By Sylow, let A be the unique subgroup of G such that |A|=3.
- Then A is cyclic and is a normal subgroup of G.
- By Sylow, let B be the unique subgroup of G such that |B|=5.
- Then B is cyclic and is a normal subgroup of G.
- It suffices to show that G is isomorphic to A×B.
- |A∩B|=1, so A∩B={1G}.
- [ AB is a normal subgroup of G ] and [ AB is isomorphic to A×B ].
- Because AB is isomorphic to A×B, it suffices to show that AB=G.
- So, because AB⊆G and |G|=15, it suffices to show that |AB|=15.
- AB is isomorphic to A×B, so |AB|=|A×B|=(|A|)(|B|)=(3)(5)=15. QED
- MEETING 10 (Friday 16 October 2015)
- worked problems 42, 46, 49, 59 from practice test 1
- these four problems touch on: complex analysis, probability theory and elementary group theory
- MEETING 09 (Friday 9 October 2015)
- worked problems 29-34 from practice test 1
- requests from students, and proposals for addressing those requests:
- complex analysis and complex algebra
- probability theory
- measure theory
- problem 54 from practice test 2
this is phrased as a probability problem, but could be looked at as a measure theory problem
- differential equations
- elementary group theory, including Sylow theorems
- MEETING 08 (Friday 2 October 2015)
- MEETING 08 (Friday 25 September 2015)
- MEETING 07 (Friday 18 September 2015)
- MEETING 06 (Monday 14 September 2015)
- discussed when to hold the seminar
- discussed the format of the seminar
- worked problems 13 and 14 from practice test 1
NO SEMINAR on Monday 7 September 2015.
Click here for more information about the seminar schedule.
- MEETING 05 (Monday 31 August 2015)
- MEETING 04 (Monday 24 August 2015)
NO SEMINAR on Monday 17 August 2015
- MEETING 03 (Monday 10 August 2015)
- MEETING 02 (Monday 3 August 2015)
- worked problems 4, 5 and 6 from practice test 1
- 7 question quiz: problems 52-58 from practice test 4
- small group work: discussion of solutions
- lecture style: discussion of solutions
- only did Problem 52 and part of Problem 53
- will continue with Problem 53
- MEETING 01 (Monday 27 July 2015)
- worked problems 2 and 3 from practice test 1
- 8 question quiz: problems 1-8 from practice test 3
- small group work: discussion of solutions
- lecture style: discussion of solutions
- discussion: effectiveness of 8 question quiz
- ok, but do more advanced problems
- MEETING 00 (Monday 20 July 2015): Organizational meeting
- introductions (why math?, favorite math course?)
- set a time for the seminar in the summer (Mondays 1pm-2pm, until M 31 Aug)
- discussed ideas for how to run the seminar; some ideas:
- work problems in small groups
- solve problems and create analogous problems
- organize problems by topic
- organize problems by required background
- study group that meets outside of seminar?
- worked problem 1 from practice test 1