1. ## Elements of Prism

In the attachment I have given the picture that goes with the question.

The positions of the square faces are 1 and 6 and the positions of the rectangular faces are 2,3,4 and 5 for top, right, bottom and left respectivley. Let G denote the group of rotational symmetries of P

a) Using the given numbering, write down all the elements of G as permutations of the face positions 1,2,3,,5 and 6.

So far I have

(1)(2)(3)(4)(5)(6) =e

(1)(2345)(6) (rotation through horizontal axis by pi/2)

(1)(24)(35)(6) (rotation throught horiz. axis by pi)

(1)(2543)(6) (rotation through horiz. axis by 3pi/2)

(16)(2)(35)(4) (rotation through vertical axis by pi)

My question here is have I got them all. I wasn't sure if there should be a rotation through a diagonal axis say, starting from the bottom left of face 1 and ending at top right corner of face 6 and then rotating, or would I just end up with the same permutaions I have now.

Hopefully you can just let me know if I have the right amount and if not a clue as to where the rest come from

Thanx

2. Originally Posted by bex23
In the attachment I have given the picture that goes with the question.

The positions of the square faces are 1 and 6 and the positions of the rectangular faces are 2,3,4 and 5 for top, right, bottom and left respectivley. Let G denote the group of rotational symmetries of P

a) Using the given numbering, write down all the elements of G as permutations of the face positions 1,2,3,,5 and 6.

So far I have

(1)(2)(3)(4)(5)(6) =e

(1)(2345)(6) (rotation through horizontal axis by pi/2)

(1)(24)(35)(6) (rotation throught horiz. axis by pi)

(1)(2543)(6) (rotation through horiz. axis by 3pi/2)

(16)(2)(35)(4) (rotation through vertical axis by pi)

My question here is have I got them all. ...

The green axis yield 4 rotations.
The red axis yield 2 rotations.
The blue axis yield 2 rotations.

That means you must come out with 8 rotations in total.

3. Originally Posted by earboth

The green axis yield 4 rotations.
The red axis yield 2 rotations.
The blue axis yield 2 rotations.

That means you must come out with 8 rotations in total.
For the green axis I get the identity plus 3 rotations before coming back to where I started totaling 4 as required.

For the red axis I get (16)(2)(35)(4) but when I rotate again I end up with the identity.

For the blue axis I get (16)(24)(3)(5). If I rotate again I end up with the identity.

So I am a little confused as to how the red and blue axes can have two rotations each whereas I get only one. So at the moment I have 6 rotations in total including the identity.

Hopefully you can clarify the axes problem I seem to be having.

Thanks again

Bex