Question on Automorphism group as subgroup

[NeoInTheMatrix]

The problem is as follows:-

Statement of the problem: Given group G, show that the automorphism group of G is a subgroup of the permutation group of G.

I can show that Auto(G) is a subset of Perm(G) easily. So I have to show that subgroup conditions hold: (1) for each x in Auto(G), x inverse is also in Auto(G), (2) the identity is in Auto(G), and (3) Auto(G) is closed under composition.

2&3 are very easy. For part 1, I have to show that given f(x)^-1, it is bijective and homomorphic. Showing that the inverse is bijective is easy. The only problem I have is showing that it's a homomorphism. I start with the following relation on the inverse:-

Relevant Equation:-
f(x)^-1 = f(x^-1)

Attempt at solution:-
So given x,y in G, f(xy)^-1 = f((xy)^-1) = f(y^-1 x^-1) = f(y^-1)f(x^-1) = f(y)^-1f(x)^-1.

But the multiplicands are reversed from what I want to show a homomorphism, namely:-
f(xy) = f(x)f(y).

any ideas? Am I missing another obvious way to approach this?

 

[micromass]

f^-1 (x) = f(x^-1)

Why do you think this relation is true? It's not what the definition of f^-1 tells you...

 

[NeoInTheMatrix]

Why do you think this relation is true? It's not what the definition of f^-1 tells you...

f is a homomorphism since it is part of the automorphism group of G. So therefore:-

e' = f(e) = f(x x^-1) = f(x)f(x^-1) .. where e' is the identity in the image of f. But this means that f(x^-1) is f^-1(x) since we can also have e' = f(x^-1 x) = f(x^-1)f(x) by similar logic.

 

[micromass]

No, this only shows that f(x^{-1})=f(x)^{-1}. The definition of f^{-1} is something completely different and has nothing to do with inverses in G.
The inverse is defined to satisfy f(f^{-1}(x))=x=f^{-1}(f(x)).

 

[NeoInTheMatrix]

No, this only shows that f(x^{-1})=f(x)^{-1}. The definition of f^{-1} is something completely different and has nothing to do with inverses in G.
The inverse is defined to satisfy f(f^{-1}(x))=x=f^{-1}(f(x)).

I'll edit the question to move the -1 to the outside. I've never heard of f^-1 as something different from the inverse.

 

[Kreizhn]

In this case f is a group element, so f^{-1} needs to be the element that takes f to the identity automorphism. You're considering how the homomorphism maps inverse elements, rather than the inverse homomorphism.

 

[micromass]

I'll edit the question to move the -1 to the outside. I've never heard of f^-1 as something different from the inverse.

But f^-1 IS the inverse of f. But what does inverse mean? In this case, it means that f\circ f^{-1}=id_G=f^{-1}\circ f. This is the definition of the inverse.

So f^-1(x) and f(x)^-1 have completely different meanings:
f^-1(x) is the element in G such that f(f^-1(x))=x, while f(x)^-1 is the element in G such that f(x)f^-1(x)=e. These are two different notions...

 

[NeoInTheMatrix]

In this case f is a group element, so f^{-1} needs to be the element that takes f to the identity automorphism. You're considering how the homomorphism maps inverse elements, rather than the inverse homomorphism.

Now it makes sense. Thanks.

 

[NeoInTheMatrix]

With that misunderstanding cleared up, the proof seems pretty straightforward:-

Consider f in Auto(G). f is surjective and injective as well as homomorphic. The inverse f^-1 is surjective as well. Given any x in G, x = f^-1(f(x)) such that x is in the image of f^-1 since the image of f(x) is G.

The inverse f^-1 is injective as well since the kernel is simply {e}. Since f is onto, every y in the image of f^-1 can be written as y = f^-1(f(y)). But f^-1(f(y)) = e only for y = e and hence only for f(y) = e. So the kernel of f^-1 is only e.

f^-1 is homomorphic. Since f is onto, every value y = f(x) for some x in g. So every value in the image of f^-1 can be written in the form f^-1(f(x)). Therefore

f^-1(f(xy)) = f^-1(f(x)f(y)) = xy = f^-1(f(x)) f^-1(f(y)) using f^-1(f(x)) = x on each part.

Given the surjectiveness of f, we can rewrite the variables as f(x) = a, f(y) = b for arbitrary a,b in G, to reduce it to the form:-

f^-1(ab) = f^-1(a)f^-1(b).