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**Definition 8.6.1.**
Let G be a group acting on a set S. We say that G acts
** transitively**
on S if for each pair of elements x,y in S
there exist an element g in G such that y = gx.

If G is a subgroup of the symmetric group S_{n}, then G is called a
** transitive**
group if it acts transitively on the set { 1, 2, ... , n }.

**Proposition 8.6.2.**
Let f(x) be a separable polynomial over the field K, with roots
r_{1} , ... , r_{n} in its splitting field F.
Then f(x) is irreducible over K if and only if
Gal(F/K) acts transitively on the roots of f(x).

**Lemma 8.6.3.**
Let p be a prime number, and let G be a transitive subgroup of S_{p}.
Then any nontrivial normal subgroup of G is also transitive.

**Lemma 8.6.4.**
Let p be a prime number,
and let G be a solvable, transitive subgroup of S_{p}.
Then G contains a cycle of length p.

**Proposition 8.6.5.**
Let p be a prime number,
and let G be a solvable, transitive subgroup of S_{p}.
Then G is a subgroup of the normalizer in S_{p}
of a cyclic subgroup of order p.

Let f(x) be a polynomial of degree n over the field K,
and assume that f(x) has roots
r_{1},
r_{2},
... ,
r_{n}
in its splitting field F.
The element
of F defined by

=
(r_{i} -
r_{j})^{2},

It can be shown that the discriminant of any polynomial f(x) can be expressed as a polynomial in the coefficients of f(x), with integer coefficients. This requires use of elementary symmetric functions, and lies beyond the scope of what we have chosen to cover in the book.

We have the following properties of the discriminant:

**(i)**0 if and only if f(x) has distinct roots;**(ii)**belongs to K;**(iii)**If 0, then a permutation in S_{n}is even if and only if it leaves unchanged the sign of(r

_{i}-r_{j}) .

**Proposition 8.6.6.**
Let f(x) be a separable polynomial over the field K,
with discriminant
,
and let F be its splitting field over K.
Then every permutation in Gal(F/K) is even
if and only if
is the square of some element in K.

We now restrict our attention to polynomials with rational coefficients.
The next lemma shows that in computing Galois groups
it is enough to consider polynomials with integer coefficients.
Then a powerful technique is to reduce the integer coefficients
modulo a prime and consider the Galois group of the
reduced equation over the field GF(p).

**Lemma 8.6.7.**
Let f(x) = x^{n} + a_{n-1} x^{n-1}
+ **· · ·** +
a_{1} x + a_{0}
be a polynomial in **Q**[x], and assume that

a_{i}
= b_{i} / d
for d, b_{0},
b_{1}, ... ,
b_{n-1} in **Z**.

If p is a prime number,
we have the natural mapping
:
**Z**[x] -> **Z**_{p}[x]
which reduces each coefficient modulo p.
We will use the notation
( f(x) ) = f_{p}(x).

**Theorem [Dedekind]**
Let f(x) be a monic polynomial of degree n,
with integer coefficients and Galois group G over **Q**,
and let p be a prime such that f_{p}(x) has distinct roots.
If f_{p}(x) factors in **Z**_{p}[x]
as a product of irreducible factors of degrees
n_{1}, n_{2}, ... , n_{k},
then G contains a permutation with the cycle decomposition

(1,2, ... ,n_{1})
(n_{1}+1,
n_{1}+2,
... ,
n_{1}+n_{2})
**· · ·**
(n-n_{k}+1, ... ,n),

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