Cayley table: symmetries (group theory)

I have an octagon (irregular) whose elements of symmetry I have found are;

s(oct) = {e, r(pi), q(pi/2), q(3*pi/4)}

I have to construct a Cayley table but have come to a halt part way through.

__o.............. ...____ e ____............____.r(pi)....... ____q(pi/2)____ .......____q(3*pi/4)__

e.............. .....e .............r(pi)....... q(pi/2) .......q(3*pi/4)

**r(pi)**............ r(pi)............ e........... q(7*pi/4) .......q ?

**q(pi/2)**....... q(pi/2)........ q?........... r ? ............r ?

**q(3*pi/4.)** ..q(3*pi/4)..... q?........... r ? ............r ?

The quarters should have all the same type (r or q)

I'm trying to visualise it as a fixed point on a disc, but am still having trouble. Also can I use the "constant diagonal" to help & if so where?

Any helpful explanation as to whether I'm going about this the right way would be great

Re: Cayley table: symmetries (group theory)

first of all, you'll have to explain your notation. i'm guessing r(pi) is a rotation through pi radians, but what is q(pi/2) (i suspect it is a reflection of some sort)?

the second thing i notice is you have r(pi)*q(pi/2) = q(7pi/4). if this is not a typo on your part, then you haven't found "all" the symmetries of your octagon (groups satisfy closure under the group operation). but without actually seeing the octogon, merely stating it is "irregular" is NOT enough information.

also, cayley tables are silly. if you have an assignment to construct one, i understand, but the tables themselves don't often yield any real insight into the geometric situation.

i suspect that r(pi)*q(pi/2) is actually q(pi), and that your octogon has the same symmetry group as a rectangle (you have one rotational symmetry, and two orthogonal reflective symmetries). without actually seeing it, however, there's no way for me to know. if you have an axis of symmetry at the line at an angle of 3pi/4 radians (which is the SAME line as one at angle of 7pi/4 radians), then "flipping" about that line and rotating pi radians, should yield a second axis of symmetry at an angle of pi/4 radians, which doesn't seem to be accounted for in your table.

my advice is to post a picture, so it is clear what you are talking about.

Re: Cayley table: symmetries (group theory)

Yes the r is a rotation & q is a reflection.

Yes, this particular irregular octagon did have the same symmetry group as a rectangle.

Yes, rules are important. I hadn't seen, or hadn't read all the way up to, the table showing the rule regarding Cayley tables. (Headbang)

o r(theta) q(theta)

r(phi) r(phi+theta) q(phi/2 + theta)

q(phi) q(phi-theta/2) r(2*phi - 2*theta)

So now I have a table which I can calculate the values for.

Phew (Bow)