Simple example of a Galois Group:

Let $Q$ be a field of rationals. Let $x^2-2$ be a polynomial with rational coefficients. That is, let $x^2-2 \in Q[x]$.

Since $\sqrt{2}$ is not a rational, the roots of equation $x^2-2$ will not be in $Q$. However, if we extend this field by adjoining $\sqrt{2}$, then we have a new field over which the polynomial $x^2-2$ has roots. This new field $Q(\sqrt{2})$ is called splitting field of the polynomial. Elements of this field have form $\{a+b\sqrt{2}|a,b \in Q\}$.

Let $K=Q,L=Q(\sqrt{2})$. 

We ask for all automorphisms of $L$  that fix $K$, namely $\phi:L \rightarrow L$ that fix $K$. These automorphisms must take a root of above polynomial to a root. Thus, we have two automorphisms.

$Q(\sqrt{2})$ should map to $\sqrt{2}$. Identity automorphism $Id$ meets this criteria. 

Consider automorphism $\phi_1(\sqrt{2})\rightarrow -\sqrt{2}$. Clearly, this will carry elements of $Q(a+b\sqrt{2}) \rightarrow a - b \sqrt{2}$. 

If we collect all these automorphisms into a set, we have a set consisting of $\{Id,\phi_1\}$

Clearly $\phi_1\circ\phi_1 = Id$. Hence, above set forms  a group of order $2$.  This Automorphism group is called Galois group denoted as $Gal(L/K)$.