190 THEORY OF COLLINEATIONS. 
The function (iii) becomes 
1— (A, +82) + (a;-+6, —1) (4;+-8:—1). 
Substituting for 4,, 8., their values from equation (10) 
Art. 172, after putting c,=c.=0, y,=y.=0, and ¢,=y.=1, 
we get 
(a, rae 8) (2, a 1) (0; ar 1) (4, a ih) . (2/3, = Uae ie (iii) 
This quantity can not vanish because of the presence of the 
terms a,3, and b,a,, which do not occur in (i) or(ii). Hence 
T, is not of type II. 
(2). Let us next suppose that 7 is not a double root of 
A(p)=0, but that (39) has equal roots. The condition that 
(39) has equal roots is 
4 (a,b, — a,b,) — (a, + 6)? = 0. (ay) 
Writing down the same conditions for T, and T, we have the 
problem: Do the first two conditions 
4 (a,b, — a,b,) —(a,+6,)* =9, . (j) 
4 (2 — 4231) — (a, +82)? = 0, (i) 
UAC Be NAc On) alain Bs) (JJ) 
cause the third expression to vanish? Since A, = AA,, we 
may write the third in the form 
(a,-++ bas (a,+ Be)? — 4(,a, + 4,8, + bya, Ale b.0,)*. (3333) 
This function does not vanish because of the presence of the 
two terms a,3, and b,«,; hence again 7, is not of type II. 
Both cases (1) and (2) lead to the same result, viz.: the 
conditions R, and D do not define a five-parameter group of 
type II. 
In like manner we may take the two conditions R, and D 
and write M and A(¢) in the reduced forms, thus: 
ay b, «ch | a,—P by Cy 
M= Qa bo C5 and AN (0) = a» bs—P eo = (i) (38’) 
0 i) 1 0 0 1-/ 
The problem is now identical with the one just solved for F, 
and D and leads to the same results. 
We have thus proved that there is no five-parameter group 
of type II defined by D and a set of linear relations on M. It 
follows that there can be no five-parameter group of type II 
