Galois Group

Let L be an extension field of K, denoted L/K, and let G be the set of automorphisms of L/K, that is, the set of automorphisms sigma of L such that sigma(x)=x for every x in K, so that K is fixed. Then G is a group of transformations of L, called the Galois group of L/K. The Galois group of L/K is denoted Gal(L/K) or Aut(L/K).

Let f(x) be a rational polynomial of degree n and let K be the splitting field of f(x) over Q, i.e., the smallest subfield of C containing all the roots of f. Then each element of the Galois group G=Gal(K/Q) permutes the roots of f in a unique way. Thus G can be identified with a subgroup of the symmetric group S_n, the group of permutations of the roots of f. If f is irreducible, then G is a transitive subgroup of S_n, i.e., given two roots alpha and beta of f, there exists an element sigma of G such that sigma(alpha)=beta.

The roots of f are solvable by radicals iff G is a solvable group. Since all subgroups of S_n with n<=4 are solvable, the roots of all polynomials of degree up to 4 are solvable by radicals. However, polynomials of degree 5 or greater are generally not solvable by radicals since S_n (and the alternating group A_n) are not solvable for n>=5.

The inverse Galois problem asks whether every finite group is isomorphic to a Galois group Gal(K/Q) for some number field K.

The Galois group of C/R consists of the identity element and complex conjugation. These functions both take a given real to the same real.

See also

Abhyankar's Conjecture, Finite Group, Fundamental Theorem of Galois Theory, Galois's Theorem, Galois Theory, Group, Solvable Group, Symmetric Group

Portions of this entry contributed by David Terr

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Birkhoff, G. and Mac Lane, S. "The Galois Group." §15.2 in A Survey of Modern Algebra, 5th ed. New York: Macmillan, pp. 397-401, 1996.Jacobson, N. Basic Algebra I, 2nd ed. New York: W. H. Freeman, p. 234, 1985.

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Galois Group

Cite this as:

Terr, David and Weisstein, Eric W. "Galois Group." From MathWorld--A Wolfram Web Resource.

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