Given a group action from G to G/H, show that N(H)/H is isomorphic to G/H

[potmobius]

Homework Statement

G is a group, H is a subgroup of G, G acts on G/H in the standard manner, and N(H) is the normalizer of H in G. Show that there is an isomorphism between Aut(G/H) and N(H)/H, where Aut(G/H) is the set of G-equivariant bijections f:G/H -> G/H

Homework Equations

The Attempt at a Solution

I know from a previous theorem that there exists an isomorphism from Aut(X) to X, so here its only necessary to prove that G/H and N(H)/H are isomorphic. Since G is the entire group in this case, it is also a normal group that contains H. So it can potentially be N(H). But I don't how to to guarantee an isomorphism.

 

[mighty2000]

Thank you for your post. Here is a possible solution to your problem:

First, let's define the map f: Aut(G/H) -> N(H)/H as follows: for any f in Aut(G/H), we define f': G/H -> G/H as f'(gH) = f(g)H. In other words, f' is just the restriction of f to the cosets of H in G. We claim that f' is well-defined and an isomorphism.

To see that f' is well-defined, note that for any gH in G/H, we have (gh)H = g(HhH) = gH, since H is a subgroup of G. This means that the value of f' on gH only depends on the coset gH and not on the representative g.

Next, we show that f' is a bijection. To see this, let f' be an element of Aut(G/H). Then for any gH in G/H, we have f'(gH) = gH, since f' is an automorphism. This means that f' is surjective. To see that f' is injective, suppose that f'(gH) = f'(g'H) for some gH, g'H in G/H. Then f(g)H = f(g')H, which means that f(g') = f(g)h for some h in H. But since f is an automorphism, this means that g' = gh for some h in H, so g'H = (gh)H = gH. Therefore, f' is injective.

Finally, we show that f' is a homomorphism. To see this, let f', g' be elements of Aut(G/H). Then for any gH in G/H, we have (f'g')(gH) = f'(gH) = f(g)H = f'(gH), since gH is a coset of H and f' is an automorphism. Therefore, f'g' = f'g, which shows that f' is a homomorphism.

Thus, we have shown that f' is a well-defined, bijective, and homomorphism from Aut(G/H) to N(H)/H. Therefore, f' is an isomorphism, as desired.

I hope this helps. Let me know