A question regarding the definition of Galois group In my book, Galois group is defined to mean the set of automorphisms on $E/F$ that "leave alone" the elements in $F$.
On Wikipedia it says:
"If $E/F$ is a Galois extension, then $Aut(E/F)$ is called the Galois group of (the extension) $E$ over $F$, $\dots$"
And Wikipedia's definition of Galois:
"An algebraic field extension $E/F$ is Galois if it is normal and separable. Equivalently, the extension $E/F$ is Galois if and only if it is algebraic, and the field fixed by the automorphism group $Aut(E/F)$ is precisely the base field $F$." 
So in one case, Wikipedia, the extension is restricted to be algebraic. So the set of automorphisms on $\mathbb{Q}(\pi) / \mathbb{Q}$ is not a Galois group. 
My question: How is it possible to have two different definitions of what a Galois group is? Do these not conflict? Or what am I missing here?
Many thanks for your help. 
Edit:
I'm using J. Gallian, Contemporary Abstract Algebra and Allan Clark, Elements of Abstract Algebra. Both use the same terminology, not the same as Wikipedia. 
 A: 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.
A: I don't know which book you are using or what the precise statement therein is, but here is how the terminology is used (at least within the English speaking mathematical world) [Edit: This may be too categorical a statement; see Arturo Magidin's answer]:
If $E/F$ is any field extension, then $Aut(E/F)$ denotes the field automorphisms of $E$ that leave each element of $F$ fixed.
If $E/F$ is a Galois extension (i.e. finite, separable, and normal) then one
writes $Gal(E/F)$ for $Aut(E/F)$ and calls it the Galois group of $E$ over $F$.
So Galois groups are a special case of automorphism groups. The reason for introducing this new terminology (i.e. $Gal(E/F)$) for an existing more general
concept (i.e. $Aut(E/F)$) is in part historical, and also to serve as a reminder that
in the particular case when $E/F$ is Galois, there is a rich theory (the Galois correspondence) relating $Gal(E/F)$ and the field theoretic structure of the extension $E/F$, which does not hold in more general contexts.
Note that there is also a notion of an infinite degree Galois extension $E/F$
(if $E/F$ is an infinite extension of $F$, one says that it is Galois if
$E$ is the union of finite subextensions of $F$; in particular, $E$ is still necessarily algebraic over $F$), and in this context one also uses the
notation $Gal(E/F)$ for $Aut(E/F)$.  But in this case one also equips $Gal(E/F)$ 
with extra structure (one makes it not just a group, but a topological group).
Also, this case is typically not treated in undergraduate textbooks and courses (at least in the U.S.), so if you are learning this material for the first time, you won't want to worry about it.
Finally, I want to say that studying $Aut(\mathbb Q(\pi)/\mathbb Q)$ (just to give one non-Galois theoretic example) is interesting and important; it just isn't part of Galois theory.
A: It is supposed to be proved in your Galois theory book as an important theorem, if not, firstly, consider the degree of L|F, and secondly, consider the fixed field, or to take a look at Van Der Wareden's book Algebra is to be of some help.
Edit:Here I mean that the fixed field of Aut(E|F) is exactly equal to F when E|F is
(1)Finite
(2)Separable
(3)Normal
and the converse is also true, if this isn't your question, I will delete my post immediately, thanks. 
