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

Solutions 207
moving along a curve which does not pass through the points
and Moreover, every passage from one sheet to another, whilst
crossing a cut, coincides with the passage indicated on the scheme at that
branch point from which this cut starts (cf., note 1 §2.10). Consequently
the connection of the sheets at the branch points must be made so as
to obtain a connected scheme. Since points different from and
are not branch points (cf., 345), and at each one of the points
and only two sheets can join (cf., 346), in order to
obtain a connected scheme we must put one arrow in correspondence with
each one of the points and i.e., all of these four points
are branch points. All the distinct connected schemes are shown in Figure
115. Any connected scheme of the Riemann surface of the function
matches one of these three schemes after a permutation of the sheets and
of the branch points (here we do not claim that all these three schemes
can be realized).
FIGURE 115
349. We prove that the permutation group of the scheme for all the
three schemes shown in Figure 115 contains all the elementary transpositions
(cf., §1.15), i.e., the transposition (1, 2), (2, 3), (3, 4), (4, 5).
For the first scheme this is evident because these transpositions correspond
to the branch points. In the second and in the third scheme
one of the branch points corresponds to the transposition (1, 2). The
transpositions (2, 3) and (3, 4) are obtained in both cases as the products
(2, 3) = (1, 2) · (1, 3) · (1, 2) and (3, 4) = (1, 3) · (1, 4) · (1, 3). The
transposition (4, 5) is obtained for the second scheme as the product
(4, 5) = (1, 4) · (1, 5) · (1, 4) and for the third scheme it simply coincides
with one of the branch points.
Consequently the required group contains, in all cases, all elementary
transpositions, and consequently (Theorem 4, §1.15) it coincides with the
entire group of the permutations of degree 5,