134 4. EquationsThus we can write (123). (1243) = (1432). Unlike a product of numbers,
this product is not commutative; sometimes, taking the factors in a differ-
ent order gives a different result. For example, (1243). (123) = (1324) #
(123). (1243).
Verify the following: (12)(34). (132) = (234); (13). (234) = (1423);
(12). (1432) = (243).
The set of 24 permutations along with this operation of “multiplication”
constitutes the group of all permutations of four numbers. Of these 24, we
select a certain subset in the following manner to make up the grozlp of
the polynomial. Consider all possible polynomial equations with rational
coefficients of the form P(tl, t2, t3, t4) = 0 which are satisfied by the ti. For
example, verify that the following hold:
(a) tl + t2 + t3 + t4 = 4(b) tltzt& = -1(c) t: + t; + t; + t: = 4(d) tl + tz = 2(e) t3 + t4 = 2(f) (tl - t2)4 = 32Cd (t2 - t3)4 = -8
(h) (t3 - t4)4 = 32(i) (tl - Q4 = -8(j) (tl - t2)4 + 4(t2 - t3)4 = 0.Equations (a), (b), and (c) are symmetrical in the ti, and remain valid no
matter how we permute the variables. However, for the others, there are
some permutations of the roots which will render them false. For example,
(d) remains valid under the permutation (13)(24) (which converts it to (e)),
but not under the permutation (123), since t2 + t3 is not equal to 2.
The group of the polynomial t4 - 4t3 + 6t2 - 4t - 1 over the field Q
(the smallest field which contains the coefficients) is the set of all permu-
tations which preserve the validity of any equation over Q of the form
P(tl, t2, t3, t4) = 0. Denote this group by G. Show that G always contains
E, does not contain (123), and contains along with any two permutations,
their product (in either order).
It can be shown that in fact there are eight permutations in G. With this
information, it is not too hard to see what they are. Consider the equation
(d). Argue that any permutation which replaces tl by t2 must also replace
t2 by tl, so that (123), (124), (1234) and (1243) do not belong to G. In a