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

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NUMBER OF INVARIANTS

259

2~. Let us now suppose that G has just k linearly independent absolute invariants of the first degree. We may
assume that the n variables were chosen originally such that
xl, x2,..., Xk are the invariants in question. Then G
is intransitive, breaking up into k+1 sets of intransitivity,
containing respectively 1, 1,..., 1; n-k, variables: xi,
X2,..., Xk; (Xk+l,.~ ~, xn), by Theorem 6.
Thus, if n=2, k= 1, and if xi is the absolute invariant, the matrix of
any transformation of G will be of the form
I 0
c d *
The reducible group may now be transformed into an intransitive group of
the following special form
1 0
o d J
The characteristic x(St) is accordingly equal to k+x(S't),
where S't is the transformation corresponding to St in that component of G which involves the set (X.+1,..., Xn). Now,
X (S't) =0, or we would have a new invariant by 1~. Hence,
x(St) =kg,
t=l
which completes the proof.
COROLLARY 1. The number of linearly independent absolute invariants of degree m in xi,..., Xn is
- l X(St(m)),
t t=1
where x(St(m) represents the sum of the homogeneous products
of degree m in the multipliers of St, namely
alm+a2m+... +1lm-lag2+... +a1m-20a2a3+..
Proof. For brevity we take n=m=2. When the variables
xl, X2 are subjected to a linear transformation
a b'
S=
c d