RePoSS #11: The Mathematics of Niels Henrik Abel: Continuation ...

6.6. Combination into an impossibility proof 115
The first case, in which µ = 5, was eliminated, ABEL said, since that assumption
would require all the values (6.12) to be different. Considering transpositions of the
form ( 1k
k1 ), ABEL found that xm 1 v would take on another 20 different values, which
would also be distinct from those in (6.12). 36 Thus in total, xm 1 v would take on 25 different
values, and since 25 did not divide 5! = 120 a contradiction had been obtained.
Secondly, ABEL assumed µ = 1 and found that the function v would only take on
one value under all permutations of x2, . . . , x5 and thus the case had been reduced to
the one above, giving v in the form (6.11).
Thirdly, for µ = 4, the function x m 1 v would take on the different values xm 1 v1, . . . , x m 1 v4,
and the function v would take on the values v1, . . . , v4 under permutations of x2, . . . , x5.
Thus, the function
v1 + v2 + v3 + v4
was a symmetric function of x2, . . . , x5, and therefore of the form (6.11). Writing v5 as
v5 = (v1 + · · · + v5) − (v1 + · · · + v4) ,
ABEL concluded that the symmetric function v1 + · · · + v5 could be incorporated in
the constant term of (6.11), and therefore v5 itself was of the form (6.11).
These first three cases were not very difficult to follow. However, the remaining
two cases were subjected to much criticism from his contemporaries (see section 6.9).
In a letter to the Swiss mathematician E. J. KÜLP (⋆1801), 37 who in a private corre-
spondence had asked for clarifications, ABEL described a refined argument, which I
have incorporated in the present description.
The fourth case, in which µ = 2, reduced to the well known situation of a function
having only two values under permutations. ABEL concluded that since x m 1
v took
on the two values x m 1 v1 and x m 1 v2 under all permutations of x1, . . . , x5, the function v
would take on only two values, say v1 and v2, when only x2, . . . , x5 were permuted.
Letting
φ (x1) = v1 + v2, (6.13)
ABEL found that φ (x1) was symmetric under permutations of x2, . . . , x5 and thus
of the form (6.11). The expression φ (x1) had to take on the five different values
φ (x1) , . . . , φ (x5) under all permutations of x1, . . . , x5 since only transpositions of the
form ( 1k
k1 ) effected the value of φ.
36 To see this, it suffices to realize that any permutation σ of five quantities can be written as a product
of a permutation ˜σ fixing the symbol 1 and a transposition τ of the form (1k). Then, if an application
of σ to v gives vu, it follows that
(x m 1 v) ◦ σ = (xm 1 ◦ ˜σ ◦ τ) (v ◦ σ) = xm ˜σ(k) vu.
37 (Abel→Külp, Paris, 1826/11/01. In Hensel, 1903, 237–240).