E, Solvable Equations of Prime Degree.
The group of an equation can always be considered as a permutation group. H f(x) is a polynomial in a field F, let alt a2,., ., an be
the roots of f(x) in the splitting field E = F( an). Then each
automorphism of E over F maps each root of f(x) into a root of f(x), that is, permutes the roots. Since E is generated by the roots of f(x), different automorphisms must effect distinct permutations. Thus, the group of E over F is a permutation group acting on the roots ror an irreducible equation this group is always transitive. For let a and a' be any two roots of f(x), where f(x) is assumed irreducible. F(a ) and F(a 1) are isomorphic where the isomorphism is the identity on F, and this isomorphism can be extended to an automorphism of E (Theorem 10). Thus, there is an automorphism sending any given root into any other root, which establishes the "transitivity" of the group.
A permutation g of the numbers 1, 2, , • • , q is called a linear substitution modulo q if there exists a number b 4 0 modulo q such that o(i) e bi + c(mod q), i = 1,2, . • • ,q.
THEOREM 7. Let f( x ) be an irreducible equation of prime de- gree q in a field F. The group G of f( x) (which is a permutation group of the roots, or the numbers 1,2, ■ ■ ■ , q) is solvable if and only if, after a suitable change in the numbering of the roots, G is a group of linear substitutions modulo q, and in the group G all the substitutions withb = 1, or(i) = c + l(c = 1, 2 , .. ., q) occur.
Let G be a transitive substitution group on the numbers 1, 2, ... , q and let Gj be a normal subgroup of G. Let 1, 2, . . . , k be the images of 1 under the permutations of G1; we say: 1,2, ■ ■ ■ , k is a domain of transitivity of G,. If i < q is a number not belonging to this domain of transitivity, there is a a e G which maps 1 oil i, Then a(l, 2,. . ., k) is a domain of transitivity of cK^a"1. Since Gj is a normal subgroup of G, we have G L - aGjO^1. Thus, a( 1? 2, . • • , k) is again a domain of transitivity of Gj which contains the integer i and has k elements. Since i was arbitrary, the domains of transitivity of Gt all contain k elements. Thus, the numbers 1,2, . . . , q are divided into a collection of mutually exclusive sets, each containing k elements, SO that k is a divisor of q. Thus, in case q is a prime, either k = 1 (and then Gj consists of the unit alone) or k - q and Gt is also transitive.
To prove the theorem, we consider the case in which G is solvable. Let G - Go 1 Gj 1 . . . ^ - 1 be a sequence exhibiting the solvability. Since Gs is abelian, choosing a cyclic subgroup of it would permit us to assume the term before the last to be cyclic, i.e., Gg is cyclic. If a is a generator of Gs, a must consist of a cycle containing all q of the numbers 1,2,,, ., q since in any other case Gs would not be transitive [ if a = (11J .. . m)( n , ,, p), ,, then the powers of o would map 1 only into I, i, j , . ..m, contradicting the transitivity of Gs ]. By a change in the number of the permutation letters, we may assume
Now let r be any element of Gs j. Since Gg is a normal subgroup of Gs J , гот 1 is an element of Ga, say rcrr'1 = a b. Let r(i) = j or r_1(j) = i, then гот"1^ j) — j) = j + b (mod q). Therefore, ra(i) = r(i) -1- b (mod q) or r(i-il) = r(i) + b for each i. Thus, setting г(0) = C, we have г(1) = c + b, r( 2) = r( 1) + b = c 2b and in general г(і) = c ib (mod q). Thus, each substitution in G s l is a linear substitution. Moreover, the only elements of Gs_1 which leave no element fixed belong to Ga, since for each a ^ 1, there is an і such that ai + b = і (mod q) [ take і such that (a-1) і e - b],
We prove by an induction that the elements of G are all linear substitutions, and that the only cycles of q letters belong to Gs- Suppose the assertion true of G^. Let т і Gs n l and let a be a cycle which belongs to Gs (hence also to Gs_n). Since the transform of a cycle is a cycle, T lOT is a cycle in G and hence belongs to G .
J J ' S-n 11
Thus тлат ~ аь for some b. By the argument in the preceding paragraph, 7 is a linear substitution bi + c and if r itself does not belong to G then r leaves one integer fixed and hence is not a cycle of q elements.We now prove the second half of the theorem. Suppose G is a group of linear substitutions which contains a subgroup N of the form c(i) - i c. Since the only linear substitutions which do not leave an integer fixed belong to N, and since the transform of a cycle of q elements is again a cycle of q elements, N is a normal subgroup of G. In each coset N . j where r(i) = bi + c the substitution a~lr occurs, where ff = i + c. But a~1r( i) = (bi + c) - c = bi. Moreover, if r(i) - bi and t" (i) = b'i thenrr'(i) = bb'i. Thus, thef actorgroup (G/N) is isomorphic to a multiplicative subgroup of the numbers 1,2, ... , q-1 mod q and is therefore abelian. Since (G/N) and N are both abelian, G is solvable.
Corollary 1. If G is a solvable transitive substitution group on q ' ( letters (q prime), then the only substitution of G which leaves two or more letters fixed is the identity.
This follows from the fact that each substitution is linear modulo q and bi + c = i (mod q) has either no solution (b = 1, c / 0) or exactly one solution(b / 1) unless b e 1, c e0 in which case the substitution is the identity.
Corollary 2. A solvable, irreducible equation of prime degree in a field which is a subset of the real numbers has either one real root or all its roots are real.
The group of the equation is a solvable transitive substitution group on q (prime) letters. In the splitting field (contained in the field of complex numbers) the automorphism which maps a number into its complex conjugate would leave fixed all the real numbers. By Corollary
1, if two roots are left fixed, then all the roots are left fixed, SO that if the equation has two real roots all its roots are real.