Meaning of $Gal(L/L)$ for some field $L$? In my notes it says $Gal(L/L)=1$ and I am confused on the notation clearly there is only one automorphism of $L$ that map all elements of the base field $L$ to itself namely the identity map. But what does  $Gal(L/L)=1$ mean? Shouldn't it be  $|Gal(L/L)|=1$ or is the $1$ some kind of notation to refer explicitly to the identity map?
 A: Personally, I've always seen $\text{Gal}(E/F)$ represent the Galois group while $\lvert \text{Gal}(E/F) \rvert$ represent the order. If someone writes $\text{Gal}(E/F)=1$, I would assume they meant $\lvert \text{Gal}(E/F) \rvert=1$, but I am pretty sure $id$ is the notation for the identity permutation, not $1$.
A: What is meant to be expressed is that the Galois group is the trivial group, containing only the identity map. 
The notation is sloppy. Even if $1$ were to stand for the identity, which is reasonable but perhaps a bit uncommon in that context, a correct way to write it would be $Gal(L/L) = \{1\}$. 
You are right that a correct way to express the intended meaning would be $|Gal(L/L)|= 1$.  
A: Since the galois group is indeed a group, (informally of permutations of roots), when the group has only one element, it is a trivial group. The group contains only one element, the identity element guaranteed by the axioms.
All order $1$ groups are isomorphic, and it's reasonably common that people write a trivial group as $1$. This has similarity to the identity element of $\Bbb R^*$ or $\Bbb Q^*$ or $\Bbb C^*$ under multiplication, and so forms a trivial subgroup. 
Also $1$ is commonly used to denote an identity matrix $1_{n\times n}$, when the dimension is understood from context, and so $n\times n$ is omitted.
$1$ is also used denote the identity map $1_A:A\to A$ of some set $A$, and again the when the set is understood from context the subscript is omitted, although $\operatorname{id}_A$ is also common. Proof Wiki mentions these as common symbols and has references there also.
Anyway given a wealth of examples where the number $1$ can be thought of as an identity element, it seems fitting to just call the trivial group $1=\{1\}$ with the trivial group operation inferred from context. This is probably an abuse of notation, but it's kind of convenient also.
A: If $G$ is a group, it is common practice to denote the identity of $G$ by $1_G$ or just $1$.  Writing $G = 1$ is an abuse of notation for $$G = \{1\}$$ which is intended to mean that there are no elements of $G$ except for the identity element.  
Another way to interpret this is to choose once and for all a group consisting of a single element (there is up to isomorphism exactly one such group), and denote that group by the symbol $1$.  Writing $G = 1$ is an abuse of notation for $G \cong 1$.  This is the same as saying that $G$ is a group with one element.  
I am not sure which abuse of notation is intended by the author of your notes, but the precise formulations of these abuses are logically equivalent.   
Similarly, if $M$ is a module over some ring, the zero element of $M$ is usually written as $0$.  If this is the only element of $M$, one commonly writes $M = 0$ rather than $M = \{0\}$.
