Let $F$ be any field. The compositum of separable extensions of $F$ contained in the algebraic closure $\overline{F}$ of $F$ will itself be separable, and so there is a largest separable extension of $F$ contained in $\overline{F}$ (namely, the compositum of all separable extensions). This is called the separable closure of $F$ in $\overline{F}$. (See for example Lang's Algebra, revised 3rd Edition, Theorem 4.5 and discussion following, pp. 241f).
Now start with a non-perfect field; for example, take $\mathbb{F}_p(x)$, the field of rational functions with coefficients in the field of $p$ elements.
Let $K$ be the separable closure of $F$ as above; because $F$ is not perfect, $K$ cannot equal $\overline{F}$. In particular, $K$ is not algebraically closed.
However, every nontrivial algebraic extension of $K$ is not separable (in fact, it will be purely inseparable): because if $L$ is an algebraic separable extension of $K$, then $L$ is also an algebraic separable extension of $F$, hence must be contained in $K$, so $L=K$.
Thus, no nontrivial algebraic extension of $K$ is separable, so no nontrivial algebraic extension of $K$ is Galois over $K$; and yet there are nontrivial algebraic extensions of $K$, since $K$ is not algebraically closed.
No I don't think this is right. For example the extension $\Bbb{Q}(\sqrt{3},\sqrt{2})$ is a Galois extension of $\Bbb{Q}$ because it is the splitting field of $(x^2 - 3)(x^2-2)$. This polynomial has exactly 4 distinct roots in $\Bbb{Q}(\sqrt{3},\sqrt{2})$ and it can easily be shown that its Galois group is the Klein 4-group $V_4$. But $V_4$ has order 4 that is not equal to $4!=24$.
Now you are thinking that the Galois group has to be $S_4$. Let me tell you why this is not possible. Let us call $\sqrt{3}$ root #1, $\sqrt{2}$ root #2, $-\sqrt{3}$ root #3, $-\sqrt{2}$ root #4. Given a cycle in $S_4$ let that cycle act on the roots simply by permuting the numbers $1,2,3$ and $4$. For example, the cycle $(12)$ exchanges $\sqrt{3}$ and $\sqrt{2}$ and keeps the negative guys fixed. But then this cannot possibly be a valid element of the Galois group because:
Elements of the Galois group must send for example $\sqrt{2}$ to another root of the minimal polynomial of $\sqrt{2}$ over $\Bbb{Q}$. The minimal polynomial of $\sqrt{2}$ when viewed as an element of $\Bbb{Q}(\sqrt{2},\sqrt{3})$ is $x^2 -2$. Therefore the only possibility for where $\sqrt{2}$ can be sent to is $-\sqrt{2}$, because this is the only other root of this polynomial in any splitting field. So therefore the cycle $(12)$ above cannot be a valid element of the Galois group (if we view the Galois group as sitting inside of $S_4$).
It follows that the Galois group in this case can only be viewed as a proper subgroup of $S_4$ and hence cannot have order $4!=24$.
Edit: Since you seem to be having some trouble understanding the Galois group, let me explain a bit more here. Now I assume that you know what an $F$ - algebra is (otherwise how would you understand field extensions?)
The following is the start of how one describes the Galois group:
Let $A = F[x]$ where $F$ is a field. Let $\iota_A$ denote inclusion of $F$ into $A$. Then any $F$ - algebra homomorphism from $F[x]$ to some other $F$ - algebra $B$ (where the homomorphism in question for $B$ is just the inclusion map $\iota_B : F \to B$) is completely determined by specifying the image of $x$ in $B$. This is because if we have an $F$ - algebra homomorphism $\varphi$ from $F[x] \rightarrow B$, we must have that
$$\iota_B = \varphi \circ \iota_A.$$
In particular this means that $\varphi$ must be the identity on the coefficients of a polynomial in $F[x]$. This explains why $\varphi$ is completely determined by its action on $x$. Now we claim that we have a bijection of sets
$$\Big\{\operatorname{Hom}_{\text{$F$ -algebra}} \big(F[x],B\big)\Big\} \longleftrightarrow B $$
where the bijection is given by $f$ that maps $\varphi$ on the left to $\varphi(x)$ with inverse $g$ that maps an element in $b \in B$ to the homomorphism $\varphi_b$ which is evaluation at $b$. Viz. $\varphi_b$ is just the homomorphism that sends $x$ to $b$. You can check that $f$ and $g$ are mutual inverses.
Now a corollary of this is that we have a bijection
$$\Big\{\operatorname{Hom}_{\text{$F$ -algebra}} \big(F[x]/(f(x)),B\big)\Big\} \longleftrightarrow \Big\{b\in B : \varphi_b(f(x)) = 0 \Big\}. $$
I will leave you to work out the details of how this comes from the fact I stated before. Essentially it is due to the universal property of quotient rings that says given a unique ring homomorphism $\varphi$ from $F[x]$ to $B$ we have unique ring homomorphism from the quotient of $F[x]/(\ker \varphi)$to $B$ such that $\varphi$ factorises through the quotient. I can edit my post to explain this more if you wish.
This is the start of how one gets a description of the Galois group because giving a homomorphism from some field say $F(\alpha)$ to itself (which is automatically an automorphism by the Rank - Nullity Theorem) is by my description above equivalent equivalent to specifying a root of the minimal polynomial of $\alpha$ over $F$ in $B$. But then our $B$ here is exactly what we started with, that is $F(\alpha)$ so that $\alpha$ must be sent to another root of its minimal polynomial over $F$.
Does this help to explain more to you?
Best Answer
Since you know that $|G(E/\mathbb Q)|=6$ and since the Galois group permutes the roots, of which there are three, it follows that the Galois group is some group with 6 elements that can be identified with a subgroup of $S_3$. That narrows it down to precisely $S_3$ (up to isomorphism).