Question: Given,

$U\_t-U\_{xx}-2U\_x=0$

Using method of separation of variables, find ALL possible solution.

My answer :

$U(x,t)=X(x)T(t)$

$T'(t)/T(t)$=$[X''(x)+2X'(x)]/X(x)$=$\lambda$

From here I'm stuck. Could someone show me some working steps for me to proceed further.

Answer: 
A) The Galois group of a degree $n$ polynomial need not be a transitive subgroup of $S\_n$. E.g. what is the Galois group of a biquadratic extension? What you have in mind is that the Galois group of an *irreducible polynomial* is a transitive subgroup of $S\_n$, where $n$ is the degree of the polynomial.

B) Why does a Galois group of order 8 have to be a subgroup of $S\_4$? What tells you that you cannot have, say, a cyclic group of order 8 as a Galois group (which would then be a transitive subgroup of $S\_8$)?

So with these remarks in mind, perhaps you should ask the precise question that you meant to ask.

Anticipating what your question *might* be: MAGMA tells me that the group generated by
$$
(1, 4)(2, 5)(3, 8)(6, 7),\\ (1, 5, 6, 8)(2, 7, 3, 4),\\ \text{and } (1, 6)(2, 3)(4, 7)(5, 8)
$$ is a transitive subgroup of $S\_8$ that is isomorphic to $C\_2\times C\_4$. It also tells me that there are no such subgroups in any $S\_n$ for $n$ smaller than 8.
Actually, the simplest example is $\Bbb Q(\zeta\_{15})$, since cyclotomic extensions are well-known.

A quick calculation shows $\operatorname{Gal}(\Bbb Q(\zeta\_{15})/\Bbb Q) = (\Bbb Z/15\Bbb Z)^\times = (\Bbb Z/3\Bbb Z)^\times \times (\Bbb Z/5\Bbb Z)^\times = C\_2 \times C\_4$.