M2PM2 test 2, 19/11/13, solutions and mark scheme.
Q1.
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,
Business
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This power point is made by mike.everything in this template is a
RS#01: Background on the Guatemalan Coup of 1954
Read the background on the Guatemalan coup, and complete the chart and questions at the end of the reading.
In the late 19th and early 20th centuries,
KMB, 3/10/13
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) I
KMB, 5th November 2013
M2PM2 Algebra II, Solutions to Problem Sheet 5.
1.
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
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
KMB, 25/10/13
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 e
KMB, 15/10/13
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 )
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 =
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
KMB, 12th December 2013
M2PM2 Algebra II, Solutions to Problem Sheet 9.
1. P =
1
1
0
1
1
1
,
0
0
1
0
0
0
, 1
0
1
1
1
1
0
0
0 (many other P s work).
1
2. As the only eigenvalue is 0, the char poly must
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
M2PM2 Algebra II, Progress test 1, 15/10/2013, solutions.
Q1.
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
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