1. ## Group theory

I have a feeling that this question should not be hard but somehow I don't really understand it.. Please help with some detail explanation if you can... (I don't just need an answer but really want to understand it...)

Let f(x,y) = (-x,y)
h(x,y) = (-1/2 -sqrt3/2) (x)
(sqrt3/2 -1/2 ) (y)
(i.e. rotate counterclockwise and angle of 2pi/3

Define a group G = {f^k.h^j|k = 0,1 ; j = 0,1,2}
Find a formula that expresses (f^i.h^j)*(f^s.h^t) = f^a.h^b
Show that G is nonabelian with order 6

Thank you so much!!!

2. 1) $f^2 = \text{id}$
2) $h^2 = \text{id}$
3) $hf = f^2h$

That should be sufficient (I hope I did not make any mistake for I did that in my head).

3. er.. i'm sorry but i dont understand your post.. do you mind elaborate a little more?

4. h(x,y) = (-1/2 -sqrt3/2) (x)
(sqrt3/2 -1/2 ) (y)
(i.e. rotate counterclockwise and angle of 2pi/3
That's not what your formula says. I presume you mean h(x,y)= ((-1/2- sqrt(3))x+ (sqrt(3)/2-1/2)y, (-1/2- sqrt(3))x+ (-1/2- sqrt(3)/2)y).

What part do you not understand?

Do you understand that $f^2= id$? What is f(x,y)? What is f(f(x,y))?

Do you understand that $h^2= id$? What is h(x,y)? What is h(h(x,y))?

Do you understand that $hf= h^2f$? What is h(f(x,y))? what is h(h(f(x,y)))?

The problem asked you to "Find a formula that expresses (f^i.h^j)*(f^s.h^t) = f^a.h^b". Well, you have it now don't you?

Now, write out all the different members of G.

5. Originally Posted by HallsofIvy
That's not what your formula says. I presume you mean h(x,y)= ((-1/2- sqrt(3))x+ (sqrt(3)/2-1/2)y, (-1/2- sqrt(3))x+ (-1/2- sqrt(3)/2)y).
You see, that's the part i dont understand... I'm sorry i was try to type in a matrix but failed. h(x,y) is the formula for rotating (x,y) and angle of 2pi/3.
It is a 2*2 matrix, The first line has entires (-1/2) and (-sqrt3)/2. The second line has entries (sqrt3)/2 and (-1/2). and of course that times (x,y) ....

6. never mind, i know how to solve this already... thank you tho...
everyone else who is interested, look up for dihedral group!!!!

7. Originally Posted by pc31
never mind, i know how to solve this already... thank you tho...
everyone else who is interested, look up for dihedral group!!!!
Let $A$ be the rotation matrix of angle $2\pi/n$ counterclockwise.
Let $B$ be the reflection matrix through y-axis.
Then $G = \{ A^iB^j | 0\leq i < n, 0\leq j < 2 \}$ is (isomorphic) to the dihedral group $D_{n}$.