M2P2 Algebra II
Solutions to Sheet 1
1. FTFFTFFTFTTTT if I got them all right. . .
2. (a) is easily checked using the permutations representing the group elements given in lectures.
(b) If H = then as |D8 : H | = 2, the right coset H = cfw_,
KMB, 5th November 2013
M2PM2 Algebra II, Solutions to Problem Sheet 5.
a) Yes image C4 , kernel C3 .
b) No (does not satisfy (xy ) = (x)(y ) for example if x = (1 2 3) and
y = (2 3 4), then x and y have order 3 but xy = (2 1)(3 4) has order 2).
KMB, 14th November 2013
M2P2 Algebra II, Solutions to Sheet 4.
1. (a) D20 has an element of order 10, and S5 doesnt (check cycle types for
example). So S5 has no subgroup isomorphic to D20 .
(b) Let x = (1 2 3) (4 5) and y = (1 2). Then x has order 6, y h
M2P2 Algebra II: Solutions to Sheet 3.
1. (i) We saw in lectures (Prop 3.2) that an n-cycle can be written as a product of n 1 2-cycles,
so the answer is yes, and from the proof we can even see in this case that g = (15)(14)(13)(12).
M2PM2 Algebra II
Solutions to Sheet 2
1. (a) Say g is an isometry that xes two points v and w = v . Say x is a general point in R2 .
Then d(g (x), v ) = d(x, v ) = a say (as g (v ) = v ), and similarly d(g (x), w) = d(x, w) = b.
Let denote t
KMB, 12th November 2013
M2PM2 Algebra II Solutions to Problem Sheet 6
1. (a) G is cyclic, hence abelian, so N G automatically. If G = x , then every
coset in the factor group G/N is of the form N xr = (N x)r , so G/N is generated
by N x, so is cyclic.
KMB, 15th November 2013
M2PM2 Algebra II, Solutions to Problem Sheet 7
1. (a) The only non-zero term in the sum dening the determinant is the one
mentioning a13 a24 a32 a41 a55 , which corresponds to the 4-cycle = (1324), which
has signature 1. Hence the
KMB, 12th December 2013
M2PM2 Algebra II, Solutions to Problem Sheet 9.
1. P =
0 (many other P s work).
2. As the only eigenvalue is 0, the char poly must be xn . So by CayleyHamilton,
An = 0.
3. By induction
KMB, 21st November 2013
M2PM2 Algebra II: Solutions to Problem Sheet 8
1. Dene A B if P invertible such that B = P 1 AP .
Then A A as A = I 1 AI .
And A B B = P 1 AP A = P BP 1 B A.
Finally A B, B C B = P 1 AP, C = Q1 BQ C = Q1 P 1 AP Q =
(P Q)1 A(P Q) A
M2PM2 Algebra II, Progress test 1, 15/10/2013, solutions.
i. This is false a counterexample is the dihedral group D12 , the symmetries of a regular
hexagon. If it were cyclic then it would have to have an element of order 12. But what could
M2PM2 test 2, 19/11/13, solutions and mark scheme.
i. The group C2 C4 C12 has 2 4 12 = 8 12 = 96 elements. One peasy mark. If
(a, b, c) is a general element, then it has order 6 i (a, b, c)6 = (1, 1, 1) i a6 = b6 = c6 = 1.
Lets gure out the solutions