# Group homomorphisms

Discussion in 'Physics & Math' started by arfa brane, Jun 18, 2012.

1. ### arfa branecall me arfValued Senior Member

Messages:
6,844
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.

3. ### arfa branecall me arfValued Senior Member

Messages:
6,844
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".

5. ### temurman of no wordsRegistered Senior Member

Messages:
1,330
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, $\theta_x^{-1}=\theta_{x^{-1}}$.

7. ### arfa branecall me arfValued Senior Member

Messages:
6,844
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?

8. ### arfa branecall me arfValued Senior Member

Messages:
6,844
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).