Abel's theorem in problems and solutions - School of Mathematics

100 Chapter 2
342. Suppose the monodromy group F of the function be soluble.
Prove that the monodromy group H of the function is also
soluble.
The functions and are continuous single-valued
functions on the entire plane. Their Riemann surfaces thus consist of a
single sheet and therefore the corresponding monodromy groups consist of
a single element and are therefore soluble. As a consequence, taking into
account the definition of the functions representable by radicals (§2.11)
and the results of Problems 338, 339, 342, one obtains the proposition
of Theorem 11.
REMARK 1. This remark is for that reader who knows the theory
of analytic functions. Theorem 11 holds for a wider class of functions.
For example, to define a function one can be allowed to use, besides
of constant functions, the identity function, the functions expressed
by arithmetic operations and radicals, also all analytic single-valued functions
(for example, etc.), the multi-valued function ln and some
others. In this case the monodromy group of the function will be
soluble, though it is not necessarily finite.
2.14 The Abel theorem
Consider the equation
We consider as a parameter and for every complex value of we
look for all complex roots of this equation. By virtue of the result
of Problem 269 the given equation for every has 5 roots (taking into
account the multiplicities).
343. Which values of can be multiple roots (of order higher than
1, cf., §2.8) of the equation
For which values of are these roots multiples?
It follows from the solution of the preceding problem that for
and equation (2.8) has four distinct roots, and for the other
values of it has 5 distinct roots. Let us study the function