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11 Nov 2014: Evaluate 1000! (mod 10249). 7. Without using a calculator, evaluate 20!2120 (mod 23) and 1710000 (mod 30).

15 Jan 2014: Hence find the complete character table of S5. Repeat, replacing S4 by the subgroup 〈(12345), (2354)〉 of order 20 in S5.

27 Oct 2014: 5. Find the eigenvalues and give bases for the eigenspaces of the following complex matrices: 1 1 00 3 20 1 0.

18 Sep 2014: 20. Application: SL2(Z) = Z/4 Z/2 Z/6. Recall that Isom(H2) = PSL2(R), so SL2(Z)acts on H2 in a natural way, with kernel {1}.G =

19 May 2014: ivKaye, I, pp11-13, p34; Kaye, III, pp119-20. v Kaye, I, pp336-7, p387. ... ix Bruce, pp47-50. x Cunningham, p44, pp76-81. xi M’Gregor, II, pp14-20.

18 Sep 2014: 20. 1. 1 Introduction. Let M be a compact surface with a geometric structure. ... 20. 2. If χ(M) = 0 then M is a torus and admits a Euclidean structure.

17 Mar 2014: 314 32 185 44 91 21 52 17. 556 24 158 38 140 20 72 15. ... NA NA 150 20 59 5 90 8. NA NA 148 19 87 4 50 5.

18 Jul 2014: of degrees 20 and 30. By Theorem 2.1 it only remains to show they are scaled as. ... The rest is[15, Lemma 5.2.31 and Theorem 5.3.20]. 2. Theorem 4.5.

27 May 2014: 20 TOM FISHER. By the substitution (8) this corresponds toz11 z12 z13z21 z22 z23. ... 2, 333–377. [20] J. Harris, Algebraic geometry, A first course, Graduate Texts in Mathematics, 133, Springer-.

15 Apr 2014: 2. Equations for X(n). We derive equations of Klein [19], [20], [21] for the modular curves X(n). ... It is shown in [2, Corollary 23.28] that X(11) P4 has degree 20.