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?
The Galois group of $F/\mathbb Q$ is cyclic of order $6$ as you described above, and if $\sigma$ is a generator for this group, then since the unique subgroup of order $2$ is $\{1, \sigma^2, \sigma^4 \}$, we can assume without loss of generality that $\sigma^2 = u$. If you compute $\sigma$, you can compute $\sigma^2$ and explicit $E$ as the fixed field of $\sigma^2$.
Now the extension $F/\mathbb Q$ is simple, so $\sigma$ is characterized entirely by where it maps $f$ (let me write $f = \zeta_7$ for clarity). It suffices to find a primitive root of $(\mathbb Z/7\mathbb Z)^{\times}$ and we got it. After some testing you realize that $3^6 = 1$ but $3^2, 3^3 \neq 1 \pmod 7$, so $3$ is a primitive root. This means that $\sigma(\zeta_7) = \zeta_7^3$ is a possibility for $\sigma$ as a generator of the Galois group.
Now $\sigma^2(\zeta_7) = \zeta_7^9 = \zeta_7^2$ (as promised by $\sigma^2 = u$ and $u(f) = f^2$), so we're trying to find the fixed field corresponding to the group $\{ 1,\sigma^2,\sigma^4\}$.
Because of the simple relation $\zeta_7^7 = 1$, it is not hard to solve the equation $\sigma^2(x)-x = 0$ for $x \in F$ ; write
$$
x = a_0 + a_1 \zeta_7 + \cdots + a_5 \zeta_7^5
$$
and apply $\sigma^2$ to it ; the equation $\sigma^2(x) = x$ will give you linear conditions on the coefficients. I'm afraid finding generators involves linear algebra. If you have the patience of writing down the equation by hand, you'll see the computations do not require any smart-ideas and you get
$$
E = \mathbb Q(\zeta_7 + \zeta_7^2 + \zeta_7^4).
$$
The minimal polynomial can be computed, it is
$$
(X - (\zeta_7 + \zeta_7^2 + \zeta_7^4))(X - (\zeta_7^3 + \zeta_7^5 + \zeta_7^6)) = X^2 + X + 2.
$$
The roots of this polynomial are $\frac{-1 \pm \sqrt{-7}}2$, so you can write
$$
E = \mathbb Q(\sqrt{-7}).
$$
Note : The magical reason why I get only one canonical element (i.e. $\zeta_7 + \zeta_7^2 + \zeta_7^4$) as a generator for $E/\mathbb Q$ is because $[E:\mathbb Q] = 2$, so I know the extension can only be generated by one element and that a basis for $E/\mathbb Q$ will be given by $\{1,x\}$ for some $x \in E$. This means when I tried to compute the fixed field of $\sigma^2$, I know the term for $1$ would be irrelevant and the rest would only give me one term.
Interesting remark : You know that $\mathbb Z / 6 \mathbb Z$ has only one subgroup of index $2$ and that the extension $F/\mathbb Q$ is Galois, so by the Galois correspondence $F$ has only one subfield of order $2$ over $\mathbb Q$. The element $\zeta_7 + \sigma^2(\zeta_7) + \sigma^4(\zeta_7)$ is invariant by $\sigma^2$ and is not in $\mathbb Q$, so you could've skipped the linear algebra computations and know already that $E = \mathbb Q(\zeta_7 + \zeta_7^2 + \zeta_7^4)$. We don't escape the computation of the minimal polynomial to simplify this though. (These ideas of averaging are often used in invariant theory ; I averaged the orbit of $\zeta_7$ over the subgroup $\{1,\sigma^2,\sigma^4\}$ to find the invariant substructure I wanted.)
Hope that helps,
Best Answer
There is a slight divergence of nomenclature. Everyone agrees on what $\mathrm{Aut}(E/F)$ is. The question is what to call it.
Some books (e.g., Hungerford, Rotman's Galois Theory), always refer to $\mathrm{Aut}(E/F)$ as the "Galois group" of $E$ over $F$ (or of the extension), whether or not the extension is a Galois extension.
Other books (e.g., Lang), use the generic term "automorphism group" to refer to $\mathrm{Aut}(E/F)$ in the general case, and reserve the term Galois group exclusively for the situation in which $E$ is a Galois extension of $F$.
So, in Lang, even just saying "Galois group" already implies that the extension must be a Galois extension, that is, normal and separable. In Hungerford, just saying "Galois group" does not imply anything beyond the fact that we are looking at the automorphism of the extension.
Wikipedia is following Convention 2; your book is following convention 1.
There is also the question of whether to admit infinite extensions or not. A lot of introductory books only consider only finite extensions when dealing with Galois Theory, and define an extension to be Galois if and only if $|\mathrm{Aut}(E/F)| = [E:F]$. This definition does not extend to the infinite extension, so the definitions are restricted to finite (algebraic) extensions, with infinite extensions not considered at all. Other characterizations of an extension being Galois (e.g., normal and separable) generalize naturally to infinite extensions, so no restriction is placed. Likewise, some books explicitly restrict to algebraic extensions, others do not; but note that most define "normal" to require algebraicity, because it is defined in terms of embeddings into the algebraic closure of the base field, so even if you don't explicitly require the extension to be algebraic in order to be Galois, in reality this restriction is (almost) always in place.
This is not such a big deal as it might appear, because one can show that an arbitrary (possibly infinite) Galois extension $E/F$ is completely characterized in a very precise sense by the finite Galois extensions $K/F$ with $F\subseteq K\subset E$ with $[K:F]\lt\infty$, as the automorphism group $\mathrm{Aut}(E/F)$ is the inverse limit of the corresponding finite automorphism groups.