Theory and applications of finite groups, by G.A. Miller, H. F. Blichfeldt [and] L. E. Dickson.

~1901 ACTUAL GROUP FOR THE BITANGENTS 369
Hence, if k1 is a new constant,
(17) 24+ 4 hX2= lX,
X2 X4 hi
k1X4 = X13+12 Al(2 +alhl)x23.
a2 a3
Permuting 1, 2, 3 cyclically, we get
k2x4 = X12+23 h2(2 +a2h2)x13,
a3 al
k3x4 = X23 + -h3(2+a3h3)x12.
al a2
Since the three expressions for x4 must be identical, the three
k's are equal and will be designated by k. Also,
0 = 1 -+ahi(2+aoh)) = (1 +aoh)2 (i = 1, 2, 3),
so that = - l/ai. Thus
~(18) kX ~ =X23 X13 X12
(18) kX 4= --— ~-.
al a2 a3
Since Xi is derived from X3 by permuting 1, 2, 3 cyclically, we
have
- a2 - a3 - al
X1= —, X2, — X3 -a3 al a2
Permuting 1, 2, 3 cyclically in (17), we get
X24 X34 X23 kaixi
(19') ~+- - - ka~ l
X2 X3 al
X34 X14 -X13 X14 X24 -X12
— + —= — -k2x, + xkaX.
X3 X1 a2 a 1 x2 a3
Adding and employing (18) and (16), we get
X14 X24 X34
-+ +-=-k(alxl +a2x2+aax3).
X1 X2 X3
Hence
9 x14 X23 (a2X2 + a3) x24= 13 (a, +a33
(19) x - k(a2x2.+.aaxa), = -k(alxl +aaxa),
" Xi a1 X2 a2
X34 X12
x -- k(0alxl+a2x2).
A3 a3