# Groups and symmetries

1. Oct 25, 2009

### mariab89

1. The problem statement, all variables and given/known data

(a) How many distinct cyclic subgroups of D6 are there? Write them all down explicitly. (Here, D6 is the dihedral group of order 12, i.e. it is the group of symmetries of the regular hexagon.)

(b) Exhibit a proper subgroup of D6 which is not cyclic.

2. Relevant equations

3. The attempt at a solution

so far i know that..

D6 = {I, R1, R2, R3, R4, R5, S1, S2, S3, S4, S5, S6} where, I is the identity, R1-R5 are rotations (60, 120, 180, 240, 300 degrees respectively) and S1-S6 are the 6 reflections across the 6 different reflective axes of the hexagon.

I'm not sure where to go from here, any help at all would be greatly appreciated!

thanks!:)

2. Oct 25, 2009

### aPhilosopher

Well, you know that a cyclic subgroup is generated by a single element. You could figure out which subgroup each element generates.

3. Oct 25, 2009

### mariab89

So, then would I itself be a subgroup?
also, {I, R1, R2, R3, R4, R5} and {I, S1, S2, S3, S4, S5, S6}?

4. Oct 25, 2009

### aPhilosopher

I isn't a subgroup but {I} is.

{I, R1, R2, R3, R4, R5} is a cyclic subgroup. What is it's generator? Does it have any subgroups? Are you sure that {I, S1, S2, S3, S4, S5, S6} is a subgroup? What is S1*S2?

A useful notation is (x) for the subgroup generated by x. So what is (R2)? What is (S5)?

What you will end up with is something like this.

(R1) = (R5) = {I, R1, R2, R3, R4, R5, R6}
(R2) = (R4) = ?
(S1) = ?
......

Does that make sense?

5. Oct 26, 2009

### mariab89

ohh ok

so then D6 has 4 distinct cyclic subgroups...
(I) - generator is the identity
(R1)=(R5)= {I, R1, R2, R3, R4, R5} - generator is R1 or R5
(R2)= (R4) = {I, R2, R4} - generator is R2 or R4
(R3) = {I, R3}

I'm just wondering what about the reflections?

would they be cyclic subgroups as well?.. for example..
(S1) = {I, S1}
(S2) = {I, S2} etc...

6. Oct 26, 2009

### aPhilosopher

If it's generated by a single element, then it's cyclic. Any ideas on the non-cyclic subgroup?

7. Oct 26, 2009

### mariab89

Well.. I was thinking maybe the group of reflections. But then that wouldn't form a group since (reflection * reflection = rotation)

any hints?

8. Oct 27, 2009

### aPhilosopher

A rotation times a reflection generates a reflection as well, correct?

If you took Si as one generator and Si*R3 as another, I bet you'd get a subgroup with two generators.