One possible approach
Hi, happyg1,
Let me go out on a limb here and try to guess the answer to the question implicit in the reply by StatusX. My guess is that you are trying to use Theorem 4.16 in Jacobson, Basic Algebra to find a polynomial over Q which has Galois group S_7, as a homework problem for your course in modern algebra. This theorem states that if f is a polynomial of degree p, where p is prime, which is irreducible over Q and has exactly two non-real roots over C (aha!), then the Galois group of f is S_p.
So your problem is reduced to finding an irreducible polynomial of degree seven which has precisely five real roots. But this is fairly elementary given that trial and error is likely to succeed. On scrap paper, you can use analytic geometry to concoct a family of polynomials with five real roots (i.e. plot x^7 and add terms to pull the graph up/down in just the right way to give five roots), then you can probably use your favorite sufficient condition for irreducibility to find a rational coefficient polynomial in this family which is irreducible, and in your writeup you can simply present your polynomial, prove it is irreducible over Q, use Sturm's theorem to carefully verify that it has precisely five real roots, and then use Theorem 4.16 to show that it must have Galois group S_7.
(I think I see a clever approach which uses a "coincidence of small groups", but never mind.)
"Trial and error": actually, in a sense most irreducible polynomials of degree p have Galois group S_p. The hard part is finding ones with smaller Galois groups, since they get rather rare rather quickly! OTH, every finite solvable group (maybe even every finite group) arises as the Galois group of some polynomial over Q. These topics are discussed in various algebra books.
