I finally can answer your question. I was a little confused about the details before; but it's definitely a semi-direct product.
Let's first of all, ignore the function \( f \). Let's write a product,
\(
(xy,\beta \alpha) = (\varphi(\alpha)(x),\beta)(y, \alpha)\\
x,y \in N\,\,\alpha,\beta \in G\\
\)
And call this group,
\(
N \propto_\varphi G\\
\)
Where \( \varphi(\alpha) : G \to \text{Aut}(N) \). Where \( \text{Aut}(N) \) is the group of isomorphisms of \( N \).
Now; there are many choices of \( \varphi(\alpha) \) (something to do with Euler's phi function will be involved in the estimate of how many). Now when you write \( g^{\alpha} = \alpha g \alpha^{-1} \in \text{Aut}(G) \), you are choosing an isomorphism of \( \text{Aut}(G) \to \text{Aut}(N) \); let's call this \( \psi : \text{Aut}(G) \to \text{Aut}(N) \). We can do this because we're only going to care about \( f \) in the final result. And this is just considering \( f(g,\alpha,x) \) at an implicit level in the preimage and considering it equivalent.
Now, here is where I wasn't making any sense before. You wrote your equation backwards from the usual semi-direct product. The right way I should've written; which I apologize for saying. Is that it's,
\(
N \propto_{\psi(g)} G = N \propto_{\varphi} G\\
\)
You are now looking for projections,
\(
f(N \propto_{\psi(g)} G) \to G\\
\)
I fucking knew it was semi-direct products! Took me a while to think about it though...
Regards, James. I hope this helps.
Long story short; you have a lot of group theory at your disposal, Mphlee. May I recommend dummit & foote.
Let's first of all, ignore the function \( f \). Let's write a product,
\(
(xy,\beta \alpha) = (\varphi(\alpha)(x),\beta)(y, \alpha)\\
x,y \in N\,\,\alpha,\beta \in G\\
\)
And call this group,
\(
N \propto_\varphi G\\
\)
Where \( \varphi(\alpha) : G \to \text{Aut}(N) \). Where \( \text{Aut}(N) \) is the group of isomorphisms of \( N \).
Now; there are many choices of \( \varphi(\alpha) \) (something to do with Euler's phi function will be involved in the estimate of how many). Now when you write \( g^{\alpha} = \alpha g \alpha^{-1} \in \text{Aut}(G) \), you are choosing an isomorphism of \( \text{Aut}(G) \to \text{Aut}(N) \); let's call this \( \psi : \text{Aut}(G) \to \text{Aut}(N) \). We can do this because we're only going to care about \( f \) in the final result. And this is just considering \( f(g,\alpha,x) \) at an implicit level in the preimage and considering it equivalent.
Now, here is where I wasn't making any sense before. You wrote your equation backwards from the usual semi-direct product. The right way I should've written; which I apologize for saying. Is that it's,
\(
N \propto_{\psi(g)} G = N \propto_{\varphi} G\\
\)
You are now looking for projections,
\(
f(N \propto_{\psi(g)} G) \to G\\
\)
I fucking knew it was semi-direct products! Took me a while to think about it though...
Regards, James. I hope this helps.
Long story short; you have a lot of group theory at your disposal, Mphlee. May I recommend dummit & foote.
