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

Solutions 145
FIGURE 43
190. The permutations of degree 5 may be represented as product of
independent cycles only in the following ways: a) b)
c) d) e) f) Applying the results
of Problems 182 and 183 we obtain that in the cases (a)‚ (b)‚ and
(c) the permutations are even‚ whereas in the cases (d)‚ (e) and (f) the
permutations are odd.
191. Suppose that the normal subgroup N in the group contains
a permutation of the type (a) (see 190). Without loss of generality
we may suppose that We prove that any
permutation of the type (a) belongs to N. If in the row
there is an even number of inversions then the permuta-
tion is even. Hence by the definition of a
normal subgroup N contains the permutation If
in the row there is an odd number of inversions then in
the row there is an even number of inversions (because
the order of elements is reversed in three pairs). In this case the per-
mutation is even. Hence N contains the
reject the four tetrahedra having a vertex in common with the chosen tetrahedron.
The remaining tetrahedra are five. Amongst them four tetrahedra have disjoint sets
of vertices‚ whereas the remaining tetrahedron has one vertex in common with each of
the four disjoint tetrahedra. The choice of the first tetrahedron thus forces the choice
of the others‚ so there are in all only two choices. (Translator’s note)