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

Solutions 203
340. To every sheet of the scheme of the Riemann surface of the
function there corresponds a pack of sheets in the scheme of the
Riemann surface of the function (Proposition (a) of Theorem
10, §2.11). The permutations of the scheme of the function
which correspond to the turns around the branch points of the function
permute the packs without destroying them (Proposition (b) of Theorem
10). But thus all permutations of the group H also permute the
packs, without destroying them. Therefore to every permutation of the
group H there corresponds a permutation of the packs. Moreover, if
to the permutation there corresponds the packs permutation and
to the permutation the packs permutation then to the permutation
there corresponds the packs permutation We obtain a
homomorphism of the group H into the permutation group of the packs.
The permutation of the packs, obtained by a turn around an arbitrary
point corresponds to the permutation of the sheets by a turn around
the point in the scheme of the Riemann surface of the function
(Proposition (c) of Theorem 10). Consequently the group of the permutations
of the packs, generated by the group H, coincides with the group
F (more precisely, it is isomorphic to F).
The abovedefined homomorphism is thus a surjective homomorphism
of the group H onto the group F.
341. The kernel of the homomorphism, built in the solution of Problem
340, consists in those permutations of the group H which transform
each pack into itself. Let and be two permutations of such a type.
If the sheets of the packs are numbered in this way:
then both permutations and permute cyclically the sheets of every
pack (cf., Proposition (d) of Theorem 10). Consider an arbitrary pack. If
cyclically displaces the sheets of this pack by sheets, and displaces
them by sheets, then both permutations and displace the
sheets of the given pack by sheets. In this way the permutations
and permute identically the sheets in every pack, i.e.,
342. If is the homomorphism defined in the solution of Problem
340 and is its kernel, then the quotient group H/ is isomorphic
to the group F (Theorem 3, §1.13). Since the group is commutative
(cf., 341) and the group F is soluble by hypothesis, the group H is soluble
as well (cf., 166).
343. Let If is a multiple root
of the equation then is a root of the equation