My problem is in proving the following:

The text suggests that I construct the functionLet H be a subgroup of a group G. Then |Hg| = |H| = |gH| for all g in G.

$\displaystyle f: Hx \to H : hx \mapsto h$

for all x where $\displaystyle h \in H$ and $\displaystyle x \in G$. All the text says is that f is "easily shown to be a bijection."

First, this looks suspiciously like a projection map of some kind. I'll tackle that later...I should be able to show that f is a surjection.

The problem I am having is this: f maps a set of cosets onto a single element of H. Okay, but looking at a couple of examples of the function acting explicitly on a given group:

Let

$\displaystyle H = \{e, a, a^2, a^3 \}$ and H < G then

$\displaystyle f(e \cdot a) = e \text{ and } f(a \cdot e) = a $ This implies that f is multiple valued, ie f(a) is equal to both e and a. Which can't be right.

Should I be taking the function of a coset, $\displaystyle Hx = \{e, a, a^2, a^3 \}x$ , in which case hx is a representative of that coset? I'm obviously looking at something wrong, but I don't know what.

Thanks!

-Dan