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

208 Problems of Chapter 2
350. From the result of Problem 349 we obtain that the monodromy
group of the function is the group of all permutations of degree
5, which is not soluble (cf., Theorem 5, 1.15). On the other hand if
the function were representable by radicals, then the corresponding
monodromy group should be soluble (cf., Theorem 11, 2.13). From the
contradiction so obtained it follows that the function is not representable
by radicals.
351. Hint. If such a formula existed then, taking in it the values
and we would obtain that
the function (cf., 350) is representable by radicals.
352. The function expressing the roots of equation (2.9) in
terms of the parameter possesses a Riemann surface which consists of
a separated sheet, on which and of 5 sheets which represent
the scheme of the function expressing the roots of the equation
in terms of the parameter Hence the monodromy group corresponding
to the function coincides with the monodromy group of the scheme
of the function i.e., with group of all degree 5 permutations,
which is not soluble (cf., 349). On the other hand if the function
were represented by radicals then the corresponding monodromy group
would be soluble (cf., Theorem 11, 2.13). From the contradiction so
obtained it follows that the function is not representable by radicals
and that the general equation of degree for is not solvable by
radicals.