# Homework Help: 3 different multiplication tables?

1. Oct 13, 2006

### pivoxa15

Are there 3 different multiplication tables for groups with 4 elements?

The reason is that an element 'a' could have 3 different inverses, a,b,c.

Each potential inverse will make a different table. Hence 3 different tables.

2. Oct 13, 2006

### 0rthodontist

I happen to know this one since it was a HW problem a while ago: yes, there are 3 tables for 4 elements, up to the naming of the elements other than unity.

3. Oct 13, 2006

### pivoxa15

Thanks for that. It means there could be a table where aa=e, bb=e, cc=e where e is the identity element. This means a,b and c are their own inverses which seems strange to me. I wonder what group has elements with this property.

I think the group of reflections in a square is one. There are 4 elements in this group and when each one is applied twice, one is back where one started from.

Last edited: Oct 14, 2006
4. Oct 16, 2006

### pivoxa15

I asked my prof and he said 2 tables only for 4 elements. Because the order of each element must divide the order of the group (Lagrange's theorem), which is 4. So the orders of the elements are either 1,2 or 4. We allow one group to be cyclic. Since there is only one unique cyclic group, the second group must not contain any elements of order 4. So it must contain one element of order 1 (the identity element) and the rest are of order 2 (since it can't be 1 or 4). That is it, there are no other groups with 4 elements.

5. Oct 16, 2006

### matt grime

Stating the orders of the elements in now way fixes the group. It so happens that there are only two groups of order 4, but the explanation you gave does not demonstrate this. You can simply do it by brute force.

6. Oct 16, 2006

### pivoxa15

I thought my explanation was logically sound and good. Just out of interest, how would you do it?

Also I've got a small unrelated question which is how many different reflections (about a plane) are there in a cube? I can see 9 standard ones. But what about cutting the cube from vertex to opposite vertex, in which case we would have 4 more. So a total of 13 reflections. Correct?

Last edited: Oct 17, 2006
7. Oct 17, 2006

### matt grime

As I said before, just writing out orders of elements does not specifiy a group.

It so happens that using certain other observations writing them down will do it in some cases. In particular if you have the result that if G is a group and every non-identity element has order 2, then G is abelian, and you have the structure theorem for abelian groups, then you can invoke your argument with these facts.

And as I also said above, you can prove it properly simply by brute force and working out what all the possible mulitplication tables are by hand. It won't take more than one minute to do this for groups of order 4 since the only case you need to worry about are the Gs where all elements have order 2 (or 1). This means of the 16 entries in a multiplication table 10 of them are filled in already, so you can just play with the other 6 to see what works (there will be only one choice).

8. Oct 17, 2006

### matt grime

In what sense are the 9 reflections you see standard? Am I supposed to instantly think 'oh those 9!'? The symmetry group of the cube has 48 elements (just count them by thinking about where vertices can map to). 24 are orientation preserving (rotations) and 24 are orientation reversing (reflections). Note, that in this classification, the identity is a rotation, though it is entirely moot as to what you might call it I suppose. It has determinant one, anyway, as a linear map when we consider the cube centred on the origin on R^3.

9. Oct 17, 2006

### pivoxa15

The 9 obvious ones for me are 3 cuts across the faces of the cube (4 pairs of vertices reflected). And 6 diagonal cuts across the edges of the cube (2 pairs of vertices reflected). They look symmetrical so definitely correct.

I think there are 4 other ones which are cuts across opposite vertices of the cube. These are reflection where only one pair of vertices are reflected. But this site http://www.maths.uwa.edu.au/~schultz/3P5.2000/3P5.20N-Cube.html [Broken]
claims only 3^2=9 reflections for a cube.

You are claiming 24 different reflections?

Last edited by a moderator: May 2, 2017
10. Oct 17, 2006

### matt grime

Yes, there are 24: the symmetry group has 48 elements, and 24 are orientation preserving.

That link does not say there are 9 reflections. It says there are 9 reflection hyperplanes. That is not the same thing.

http://www.ams.org/featurecolumn/archive/cubes7.html

(which is the first hit on googling for 'symmetry group of the cube', by the way).

11. Oct 17, 2006

### pivoxa15

What do you mean by orientation preserving?

Could the 24 extra symmetries arise from a combination of rotation and reflection? So there are 24 different rotations, 13 different reflections and 11 rotation&reflections that produce unique configurations not achieved by single rotation or reflection or their powers.

If not than I have already shown 13 different reflections. Could you describe the other 11 different reflections?

12. Oct 17, 2006

### matt grime

Orientation preserving, etc, is to do with determinants of the matrices representing these linear maps. In R^3 rotations have determinant 1, reflections determinant -1.

Get a cube, big one. Label the corners 1-8 and work what the rotations are - these are the ones that have to fix orientation. Now try to work out the remaining ones. I presume you tried this rather than just trying to do it in your head....? A rotation followed by a reflection is still a reflection in some plane - just think about the symmetries of an n-gon if need be.

Last edited: Oct 17, 2006
13. Oct 17, 2006

### pivoxa15

So orientation preserving if and only if det=1 ?

Should I first think about different symmetrical planes for which the remaining 11 reflections could take place?

14. Oct 17, 2006

### matt grime

You should either make a cube or draw a picture.

15. Oct 18, 2006

### pivoxa15

I have a die plus I have already drawn many pictures. For any reflection, I need a plane in which the reflection can take place. I can think of planes which cut half way into the cube, so the reflections would be on one side of the cube only but I have a feeling that is not allowed as it is not allowed with square reflections. Plus it would lead to more than 11 reflections. Apart from that I can't think of different planes of reflection which look symmetrical enough for the remaining 11 reflections in a cube. At the moment I have 3 full planes across faces of cube, 6 full planes across edges and 4 full across vertices. What other symmetrical planes are there?

16. Oct 18, 2006

### matt grime

So you've drawn a picture of all 24 non-orientiation preserving maps? Seriously, I am asking you if you've listed all of the non-rotations. It is easy to do that systematically, so do it: label the vertices (not the faces: a die won't help), start by labeling one vertex 1, and the 3 vertices joined to it as 2,3,4. 1 can map to any of eight vertices, 2 can map to any of the three neighbouring vertices. 3 can now map to anyone of 2 vertices. One choice is orientation preserving and the other is not. There are theus 48 symmetries and 24 of them are reflections. So draw them all and look at what happens.

And, if you don't want to do that, then think back to the square case. Start with reflection in some line. Now apply some rotation by an angle of T. What is the result? A reflection in the line of the original reflection but rotated by T/2. So if you have 24 rotations and some reflection why not compose them and see what happens to the plane of reflection?

(I really hope you aren't spending much time on this: it is not important, and there is always the chance that I'm leading you down a cul-de-sac).

Last edited: Oct 18, 2006