COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

116 J. S. Frame
3. The decomposition of Kronecker powers. If the cycle symbols E in a
class symbol are replaced by A~-~(2k), then the first, second, third, fourth,
. . . powers of an element in class 1” 28 3” 46. . . have the character values
a, a+2/$ a+3y, a+2fl+46, . . . (3.1)
in the S-dimensional orthogonal representation 8, of F. Using the formulas
worked out by Murnaghan [9], the character 8:” of the mth Kronecker
powers of 8, may be split into characters [;I], one for each partition (2.)
of m, which are irreducible for the full S-dimensional orthogonal group 0,.
This is a substantial refinement of the Schur decomposition which yields
characters irreducible for the full linear group. Constituents of odd powers
of 8,, including the first power 8, itself, will be classified as type z (pairs
of associated characters) or type w (self-associated characters which vanish
in the odd coset of A + or F ‘). For m = 0, 1, 2, 3, 4, all but one of these
Murnaghan characters are irreducible for the subgroup F of 0,. We denote
them by their degrees with subscripts, and express their values in
terms of a, p: y, 6,. . . as follows
[O] = 1, = I
[l] = 8, = a
[12] = 28, = ; -/3
0
[2] = 35, = (a+2)(a-1)/2+/I
[13] = 56, = “; -ap+y
0
[2 l]= 160, = (a+2) &-2)/3--y
f3] = 112, = (a+4) a(a- 1)/6+a@+y
P"l = 70, = (i)-(i)B+(!)+ay---b
[2 l*]= 350, = (af2) a(a-1)(x-3)/8-(af2)(a-1)/I/2- ; +6
0
[22] = 300, = (a+2)(a+l) a(a-3)/12$-b@- 1)-xyfap’
[3 I] = 567, = (a+4)(a+ l)(a- I)(a-2)/S +(;p-(p:i)-b
[4] = 210,+84, = (a+6)(a+ 1) a(a-1)/24+ (O.:‘)P+(~)+ay+d
(3.2)
The first eleven of these characters are found to be irreducible for F
by summing their squares over the group. It is clear that [4j cannot be irreducible
for F, since its degree 294 = 2.3.72 does not divide the group order
The characters of the Weyl group Es 117
IFI. We shall see presently how to split this character by using permutation
characters induced from the subgroups H and M.
It is probably simpler not to use all these formulas explicitly, but to calculate
8,, 28, and 160, by these formulas, and then express the rest by Kronecker
products as follows.
In [2]: 35, = s;-l,-28, (3.3)
In [13]: 56, = 8,(28,- lx)- 160,
In [3]: 112, = 8,(35x- I,)- 160,
In [2 12]: 350, = 28yl--28,
In [l”]: 70, = 8,(56,) - 350, - 28,
In [22]: 300, = 8,(160,)-28,(35-J
In [3 I]: 567, = 35,(28,-I.)-28,-350,
In [4]: 210,+84, = 8,(112,)-35,-567,
Several irreducible components of the Kronecker 5th power of 8, can
be split off in like manner as follows. Here the irreducible character 56:
is the associate of 56,, with values of opposite sign in the odd classes of
types c and d.
In [15] : 56: is associate of 56, (3.4)
In [2 I31 : 448, = 8,(70,)- 56,- 562
In [221]: 840, = 56,(2&f l,)-8,(28,+70,)
In [3 12]: 1296,= 160,(28,)-8,(35,+300,+70,)+56~
In [3 21: 1400, = 8,(300,)- 160,- 840,
In [4 11: 1008, = 8,(210,-84,)
In [5]: 560, = S&34,) - 112,
Before the characters [4], [4 11, and [5] of OS can be split in F, it is necessary
to obtain the character 84,. Note first that the difference of the symmetrized
Kronecker squares of the characters 56, and 28, splits into two
characters, one the associate 350: of the known character 350,, and the
other a new character of degree 840:
840, = 56L21-28yl -350;. (3.5)
4. Induced permutation characters. Three permutation characters of
A, denoted 120,, 135,, and 960, respectively, are induced by the subgroups
of A index 120, 135, and 960, called H, M, and S above. Both H and M
have exactly three double cosets in A, containing 1 f 56+63 cosets of
H, or 1 + 64+ 70 cosets of M respectively. Hence by a theorem of Frame
[3] these permutation characters each split into the l-character 1, and a