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19 Jan 2015: 1 21 42 56 24 24α 14 2 0 1 0 0β 15 1 1 0 1 1γ 16 0 0 2 2 2. ... 24 = g5.]. 11 Let a finite group G act on itself by conjugation.

21 Oct 2015: 4. (a) Show that the symmetric group S4 has a subgroup of order d for each divisor d of 24,and find two non-isomorphic subgroups of order 4.

25 Feb 2015: Show by direct calculation that K = K{0}is a cyclic group and deduce that K is finite field with 24 elements.

19 Feb 2015: Show by direct calculation that K = K{0}is a cyclic group and deduce that K is finite field with 24 elements.

26 Feb 2015: X4 2X 2, X4 18X2 24, X3 9, X3 X2 X 1, X4 1, X4 4.

18 Feb 2015: Hint 'f. V rfm(f)ΓQ. ,. where Vf is the two-dimensional -adic representation of ΓF attached [18, 24, 22]to f. ... 24] A. Wiles: On ordinary λ-adic representations associated to modular forms.

22 Sep 2015: so b = 1. 24. Exercise 1.5.1. There are many ways of explaining this. ... By inspection m = 1 is not asolution, but m = 2 is a solution (24 = 42).

7 Aug 2015: It is known [24] that Ū(g) is a Fréchet-Stein algebra. We view Ū(g) as aquantisation of the algebra of rigid analytic functions on g in much the same

3 Apr 2015: Automorphy of some residually dihedral Galois representations. Jack A. Thorne. April 3, 2015. Abstract. We establish the automorphy of some families of 2-dimensional representations of the absolute Galoisgroup of a totally real field, which do not

20 May 2015: 24 SIMON WADSLEY. Proof. (a). tr AB =. ni=1.  mj=1.