1)
2)
3)
That should be sufficient (I hope I did not make any mistake for I did that in my head).
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!!!
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).h(x,y) = (-1/2 -sqrt3/2) (x)
(sqrt3/2 -1/2 ) (y)
(i.e. rotate counterclockwise and angle of 2pi/3
What part do you not understand?
Do you understand that ? What is f(x,y)? What is f(f(x,y))?
Do you understand that ? What is h(x,y)? What is h(h(x,y))?
Do you understand that ? 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.
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) ....
Any help please?