MATH 5a: Solutions to Homework Set 8
December 2013
1. We want to nd all Sylow 2-subgroups and 3-subgroups of D12 and S3 S3 .
The Sylow 2-subgroups of D12 have order 4. By Sylows theorem, the number n2 of Sylow 2-subgroups
is odd and divides 3; hence n2 cf
MATH 5a: Solutions to Midterm
2013
1. (5 points) Let H G. We will show that the map x x1 sends each left coset of H in G onto a
right coset of H and gives a bijection between the set of left cosets and the set of right cosets of H in G.
Let aH be a left c
MATH 5a: Solutions to Homework Set 7
November 2013
1. (a) Want to prove that GB = cfw_ G| (B ) = B is a subgroup of G containing Ga . First note
that GB is not empty since id(B ) = B . Now let , GB . Note that 1 (B ) = 1 (B ) = B so
1 (B ) = B . Thus 1
MATH 5a: Solutions to Homework Set 6
November 2013
1. (a) Z2 has every element of order 1 or 2.
i=1
(b) Q/Z = cfw_ p + Z | p < q integers. We can also take Zn .
n=1
q
(c) Z Z2
(d) Sn
i=2
(e) Z2
i=1
2. We want to determine the proper subgroups of Zp Zp2 .
MATH 5a: Solutions to Final
2013
1. We have that |S3p | = (3p)!. The Sylow p-subgroup of S3p has order p3 . Consider
H = (1 2 p), (p + 1 p + 2 2p), (2p + 1 2p + 2 3p)
be a subgroup of S3p generated by a = (1 2 p), b = (p + 1 p + 2 2p), c = (2p + 1 2p + 2
MATH 5a: Solutions to Homework Set 5
November 2013
1. If h, g G(r) then as G is abelian, (gh)r = g r hr = 1 1 = 1. Hence gh G(r). Similarily,
= (g r )1 = 1, so g 1 G(r). Thus, G(r) G.
If g G(m) G(k ) then g m = g k = 1. So |g | divides (m, k ) = 1, and he
MATH 5a: Midterm Review
November 2012
1
Finite Groups and their Subgroups
Given a positive integer n, it is in general very hard to determine how many isomorphism types of groups
of order n there are. There are however some cases where we can say exactly
MATH 5a: Final Review
December 2012
1
Group action
1.1
Short overview of concepts
Let G act on a set X .
The action is transitive if X = and if for any x, y X , there exists g G such that g x = y .
The action is faithful if for any g, h G, there exists