304 
MATHEMATICS: E. J. WILCZYNSKI 
of F in the sense, that when F is given, the coefficients of <E> are determined 
uniquely. We may express this symbolically by the equation 
* = U(p), (2) 
where U (p) is the symbol for a one-valued function of the coefficients p. 
Now let T be any transformation of G, and let 
F = T (F) 
be any form of [F] equivalent to F under G. Since we assumed that F was 
generic, and since 
F = T~\F), 
we conclude that F is also generic. In fact we find 
S(F) = ST~ 1 (F) = 
so that ST~ l is an operation of G which transforms F into a canonical form 
of the sub-class [#]. 
Let 5 be the most general operation of G which transforms F into a canoni- 
cal form of the sub-class [<£], so that 
5 (?) = *, 
and let ir denote the coefficients of i. These coefficients will depend on the 
coefficients p of F, in exactly the same way as the coefficients ir of <£ depend 
upon the coefficients p of F. That is, we shall have, in a manner analogous 
to (2), 
7T = U (P). (3) 
We also know that 
i = 
for we have seen already that 5T _1 will transform F into <£, and we are 
assuming that every generic form F of [F] has a unique canonical form of the 
sub-class [$>]. 
But, if two forms of a class are equal, their corresponding coefficients are 
equal. Therefore we have ir = t or, according to (2) and (3). 
U(p)=U (p). 
In other words the coefficients of $ are indeed absolute invariants of F under 
the group G, and therefore our theorem is demonstrated. 
If the forms of the class [F], which are generic forms from the point of view 
of the canonical form chosen, do not constitute the whole of [F] there will 
remain in [F] certain exceptional forms constituting a sub-class [H] of [F], 
But the theorem may be applied to these exceptional forms as well, whenever 
a unique canonical form exists for a generic one of the exceptional forms; but 
of course, the canonical form in this case will not belong to the class [<£], but 
