The problem is: Prove that the function \( \theta_x : g \mapsto g^x = x^{-1}gx \) on a group G is an isomorphism from G to G. Here's my attempt so far: Let \( g_1,g_2 \in G \) such that \( \forall x \in G \): \( {g_1}^x{g_2}^x \;=\; \( x^{-1}g_1x\)\(x^{-1}g_2x) \;=\; x^{-1}g_1\(xx^{-1}\)g_2x \;=\; x^{-1}g_1g_2x \;=\; \(g_1g_2\)^x \), then g[sup]x[/sup] is a homomorphism from G to G. Suppose \( {g_1}^x \;=\; {g_2}^x \), then \( x^{-1}g_1x \;=\; x^{-1}g_2x \) so g[sub]1[/sub] = g[sub]2[/sub] by cancellation. So g[sup]x[/sup] is 1:1. Now to show g[sup]x[/sup] is onto, I need to show that each (or some arbitrary) element in the codomain of g[sup]x[/sup] has at least one element in its domain. Is the codomain the image of g[sup]x[/sup], and is x the element in the domain of g[sup]x[/sup]? Just, you know, making sure.

I think what I need to show to prove g[sup]x[/sup] is surjective is that its image is the whole of G. If it maps all x in G to G, maybe it's "clearly" onto. The question doesn't specifically say "for all x". But I guess I can assume it means "some arbitrary x".

You seem to sometimes confuse the parameter \(x\) of \(\theta_x\) with its argument \(g\). The parameter \(x\) is just a fixed element of \(G\), and we have to show that \(\theta_x:G\to G\) is an isomorphism. You have done most of the work, what remains is to show surjectivity, which means that for any \(h\in G\) there exists \(g\in G\) such that \(g^x=h\). This equation is easily solved: \(g=xhx^{-1}\equiv h^{x^{-1}}\). In other words, \(\theta_x^{-1}=\theta_{x^{-1}}\).

Ok, thanks. So \( \theta_x \) is a function of g, not a function of x. Or rather it maps some g in G to the ccnjugate of x?

No that's inaccurate; if \(g^x=h\), then g is conjugate to h. So x conjugates g. We haven't covered conjugation as such, but when we did G-sets, we got this example: For H ≤ G, G is an H-set if we define h·g = hgh[sup]-1[/sup], ∀h ∈ H, ∀g ∈ G. Then we have · : G x H → G; e·h = h; (g[sub]1[/sub]g[sub]2[/sub])h = g[sub]1[/sub](g[sub]2[/sub]h).