# Thread: Projection function of some kind?

1. ## Projection function of some kind?

My problem is in proving the following:
Let H be a subgroup of a group G. Then |Hg| = |H| = |gH| for all g in G.
The text suggests that I construct the function
$f: Hx \to H : hx \mapsto h$

for all x where $h \in H$ and $x \in G$. 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
$H = \{e, a, a^2, a^3 \}$ and H < G then
$f(e \cdot a) = e \text{ and } f(a \cdot e) = a$ 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, $Hx = \{e, a, a^2, a^3 \}x$ , 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

2. Originally Posted by topsquark
My problem is in proving the following:
The text suggests that I construct the function
$f: Hx \to H : hx \mapsto h$
for all x where $h \in H$ and $x \in G$. All the text says is that f is "easily shown to be a bijection."
Look $f(h_1g)=f(h_2g)$ be definition means $h_1=h_2$.
Operate by $g$ and get $h_1g=h_2g$, injective.

If $h\in H$ then $f(hg)=h$, surjective.

3. Originally Posted by Plato
Look $f(h_1g)=f(h_2g)$ be definition means $h_1=h_2$.
Operate by $g$ and get $h_1g=h_2g$, injective.

If $h\in H$ then $f(hg)=h$, surjective.
Thank you. I see what I was doing wrong earlier. And thanks for the bijection proof as well.

-Dan