# Shhowing a map is well defined and bijective

1. Oct 26, 2009

### bjogae

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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?

Last edited: Oct 26, 2009
2. Oct 26, 2009

### rasmhop

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.

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.

3. Oct 28, 2009

### bjogae

Thanks for the help.