## Theory of Groups of Finite Order by William Burnside

By William Burnside

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The contents of this ebook were used in classes given by means of the writer. the 1st was once a one-semester path for seniors on the college of British Columbia; it used to be transparent that sturdy undergraduates have been completely able to dealing with straightforward workforce thought and its program to basic quantum chemical difficulties.

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1 1 + e1 γk y3 x + aβ x(x − 1)(2x − 1)γ 2 y 2 − x2 γy 2 3! + x 2 φ2 = e1 xy + e1 k (a + k) dy y 2 + ad x 2 y 2 + δy + βγ x 3 pm−5 , y pm−5 . Placing x = 1 and y = p in (23) and by (16) p Q−p 1 R Q1 = R, and by (19) a≡0 (mod p). A continued multiplication, with (11), (22), and (23), gives 2 2 2 2 (Q1 P x )p = Qp1 P xp = P (x+l)p . 35 Let x be so chosen that (mod pm−5 ), (x + l) ≡ 0 then Q = Q1 P x is an operator of order p2 which will be used in place of Q1 , 2 Qp = 1 and h ≡ 0 (mod pm−5 ). From (21), (10) and (16) 2p 2 ≡ kpm−4 , α2 p 2 ≡ 0 e1 p2 ≡ 0 and (mod pm−3 ).

B5 . The congruences for this case are −e y z ≡ kx (mod p), γz + δz ≡ 0 (mod p), δ xz ≡ δy (mod p), (II) (III) (IV) (V) (VI) (VII) 1 xz +δ ex z 2 + e yz ≡ e1 x + γx + δx cz ≡ 0 (mod p), e y z ≡ ex (mod p). (mod p), (II), and (III) being linear in z and z give k ≡ 0 or ≡ 0, and γ ≡ 0 or ≡ 0 (mod p), (IV) gives δ ≡ 0, (V) being linear in x gives 1 ≡ 0 or ≡ 0 (mod p), (VI) gives c ≡ 0 and (VII) e ≡ 0. 30 Elimination of x and y from (IV) and (VII) gives δez 2 ≡ δe (mod p). δe is a quadratic residue or non-residue (mod p) according as δ e is a residue or non-residue.

Therefore, b1 = 1. Substituting 1 for b1 and 1 + α2 pm−5 for α1 in congruence (6) we find (1 + α2 pm−5 )p ≡ 1 (mod pm−3 ), α2 ≡ 0 or (mod p). Let α2 = αp and equations (7) and (5) are replaced by m−4 β 1+αp R−1 1 P R1 = Q P (8) m−4 αp R−1 1 Q R1 = QP (9) , . From (8), (9) and (3) (10) [−y, 0, x, y] = 0, βxy, x + αxy + aβx (11) [−y, x, 0, y] = [0, x, axypm−4 ]. y 2 + βky x 2 pm−4 , From (2), (10), and (11) [z, y, x]s = [sz, sy + Us , sx + Vs pm−4 ], (12) where Us = β Vs = s 2 s 2 xz, αxz + kxy + ayz + βk + βk s 3 1 x2 z + aβ 2 s 2 x 2 z 1 (2s − 1)z − 1 xz.