Application in permutations group

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The discussion focuses on a problem involving automorphisms and permutation groups in the context of group theory. The key point is to show that the function fg defined as fg(x) = a1(g)*x*a2(g^-1) is a bijection, thus belonging to the symmetric group Sg. It is established that fg has an inverse, h_g(x) = a1(g^-1)*x*a2(g), confirming that fg is well-defined and bijective. Additionally, the condition f_g = id(G) leading to g = e indicates a unique identity element, while the property f_g * f_h = f_(g*h) demonstrates that this function forms a group monomorphism from G into Sym(G). This establishes an embedding of G into the symmetric group that preserves the group operation.
Dassinia
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Hello
I am studying for my exam and there's a question that i don't know how to solve, I have some difficulties with symmetric/permutations groups

1. Homework Statement

Consider a finite group of order > 2.
We write Aut(G) for the group of automorphisms of G and Sg for the permutations group of the set G.
Consider a1 and a2 ∈ Aut(G) two automorphisms.
We suppose that the g=1 is the only element g ∈ G such as ∀ x ∈ G we have a1(g)*x=x*a2(g)
For g,x ∈ G we have
fg(x):=a1(g)*x*a2(g-1)
Show that for each g ∈ G the application fg: G→G is in Sg

Homework Equations

The Attempt at a Solution


I really don't know how to do that..
 
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I don't get the problem here. Since all f_g are well-defined functions on the finite set G with an obvious inverse
h_g (x) := a1(g^-1)*x*a2(g) = f_g^-1 (x), i.e. a bijection, they have to be in Sg, if Sg is Sym(G) the set of all bijections on G.

The additional condition says that f_g = id(G) implies g = e. It is also true that f_g * f_h = f_ (g*h).
Therefore you have a group monomorphism f: G → Sym(G) with f(g) := f_g, i.e. an embedding of G into Sym(G) that respects the group operation.
 
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