How do I proof that Aut(E/F) is a group? (E is an extension field of F)

Question

I need help proving this. Should I use the basics that define a group or can I use the fact that it is a set of automorphisms?

Answer 1

Basically yes, you can use only the fact that it is a set of automorphisms, closed under composition and inverse (and of course contains the unit element, which will be the identity now).

Answer 2

The set of automorphisms of some object, i.e. maps of the object (viewed as a set) to itself that respect certain properties of the object and have an inverse map that also respects these properties, is always a group (under composition of maps as group operation):

• The composiion of maps that respect certain properties of the object aso respects theese properties: If neither of two steops breaks something, the steps together don't break it either. Therefore the compostiotion of automorphisms is an automorphism.
• Composition of maps is always associative, as $((f\circ g)\circ h)(x)=f(g(h(x)))=(f\circ (g\circ h))(x)$ for all $x$
• The identity map $\text{id}$ of the object does nothing, hence repects all properties of the object, and is its own inverse, hence is an automorphism
• Composing the identity with any map yields the given map again, i.e. $\text{id}\circ f=f\circ \text{id}=f$ for all $f$, because $\text{id}(f(x))=f(x)=f(\text{id}(x))$ for all $x$
• By definition of automorphism, there exists an inverse map, which is also an automorphism, and the composition with it produces the identity.