Shhowing a map is well defined and bijective

[bjogae]

Homework Statement

The group $$G$$ acts transitively from the left on the set $$X$$. Let $$G_x$$ be the little group of the element $$x \epsilon X$$. Show that the map $$i:G/G_x$$, $$i(gG_x)=gx$$ is well defined and bijective.

Homework Equations

transitive action:for any two x, y in X there exists a g in G such that g·x = y

The Attempt at a Solution

Transitive action shows that $$x \epsilon X$$, $$g \epsilon G$$ $$->$$ $$g·x \epsilon X$$. This shows that the mapping is a surjection. Now how do i show that it's an injection? And obviously the bijection thing shows that the function is well defined, right? Even the injection would suffice for this?

 

[rasmhop]

And obviously the bijection thing shows that the function is well defined, right? Even the injection would suffice for this?

No. In your proof of injectiveness you will assume that i is a function, but before you have shown that it's well-defined we only know that it's a relation. Of course I can't use the example you gave as it's a function and bijective, but there do exists bijective relations that are not functions. Consider for instance f(0) = 0, f(0) = 1 defined from $\{0\}$ to $\{0,1\}$. As a relation f is both injective since only one element map to 0 and only one element map to 1, and it's surjective since there is an element that map to 0 and an element that map to 1, but it's not a function.

To show that it's well-defined you suppose $gG_x = hG_x$ and then show gx=hx, but this is clear since gx=hx iff $h^{-1}gx=x$ iff $h^{-1}g \in G_x$ iff $h^{-1}gG_x = G_x$ iff $gG_x=hG_x$. This also shows injectiveness since we have bi-implications all the way so if gx =hx, then $gG_x = hG_x$. So in this case you could do injectiveness and that i is well-defined together, but this is not true in general.

Transitive action shows that $$x \epsilon X$$, $$g \epsilon G$$ $$->$$ $$g·x \epsilon X$$. This shows that the mapping is a surjection.

You may have the right idea (as this resembles the right way), but as written this is flawed (and you use the symbol x for two distinct objects which can be confusing). You first consider an arbitrary element $y \in X$ and then you want to show that there exists some $gG_x \in G/G_x$ which the property $i(gG_x)=y$. Since the action is transitive we can find an element $g \in G$ such that $gx = y$. [Stating that $gx \in X$ doesn't really give us any information or use transitivity since that is true for all group actions.] Now $i(gG_x) = gx = y$ so i is surjective.


As a small aside: when you type math into LaTeX, then use \in instead of \epsilon for inclusion and use \to to make the arrow.

 

[bjogae]

Thanks for the help.