Question

question on groups and symmetries

1) The dihedral group of order 12 is D6 = {e, r, r^2, r^3, r^4, r^5, s, rs, (r^2)s, (r^3)s, (r^4)s, (r^5)s } where |r| = 6, |s| = 2 and sr = (r^5)s
a) Let H = and show that H is normal to D6( the sideway triangle symbol). b) Construct the Cayley table of D6/H. Is this group abelian? Is it cyclic?

Answer 1

@nbouscal

Answer 2

to show that H is normal to D6 we have to show that aH = Ha right?

Answer 3

@satellite73

Answer 4

to show it is normal this is what i did

Answer 5

(e)H = {e, r^2, r^4} = H(e)
(r)H = {r, r^3, r^5} = H(r)
(r^2)H = {r^2, r^4, e } = H(r^2)
(r^3)H = {r^3, e} = H(r^3)
(r^4)H = {r^4, r^2} = H(r^4)
(r^5)H = {r^5, r^4, r^2} = H(r^5)

and i did this for the s, rs, ....

but could yoi please tell me if i did the above right and how on earth do i make a cayley table out of this

Answer 6

@KingGeorge

Answer 7

The above looks correct to me. That successfully shows that H is normal in \(D_6\).

To construct the Cayley table, start with two things. First, write all your group elements across the top and the side (in the same order, and e is usually first). It should look something like the following.

e r r^2 r^3 r^4 ................
e
r
r^2
r^3
r^4
r^5
s
sr
sr^2
sr^3
sr^4
sr^5

Make sure to continue the top line.

Answer 8

Once you've drawn all that out, the entries in the table will be \(gh\), where \(g\) is the row label, and \(h\) is the column label.

Answer 9

alright perfect i shall follow

Answer 10

but is D6/H a significance ?

Answer 11

@KingGeorge

Answer 12

While making the Cayley table, it's irrelevant that H is normal to D6

Answer 13

ok i can clearly see that this is abelian but how would i know its cyclic i see a pattern but how exactly would i word that this D6 group is cyclic?

Answer 14

D6 should not be abelian.

Answer 15

wth that is weird i am getting it is abelian :s

Answer 16

alright i am going to be posting a picture

Answer 17

@KingGeorge

Answer 18

i stopped right there but whatever i pick i get it being abelian or did i draw the cayley table wrong

Answer 19

I think you're drawing it a bit wrong. You seem to be switching between \(gh\) and \(hg\). It's important that you remain consistent throughout the entire table.

Answer 20

hmm what do you mean by that :s wont it always be the same like

Answer 21

like say i do (r^4)o(r^3) = r^7 = r right

Answer 22

and by the table i get (r^3)o(r^4) = r^7 = r also :s

Answer 23

That's all fine. However, when you start multiplying elements with s, things get a bit weird.

For example, in the eighth row and second column, you have \(rsr=r^2s\), when it should be \(rsr=rr^5s=s\).

Answer 24

o really :$ and thats by the condition they gave us whereever we see sr we replace it with r^5*s?

Answer 25

Correct. \(sr=r^5s\).

Answer 26

and how would i knwo it is cyclic?

Answer 27

You would know it's cyclic if you can find some element \(g\) such that \(|g|=12\).

Answer 28

well clearly e is no?

Answer 29

e is definitely not.

Answer 30

I think an easy way is to look at the diagonal of the table you've constructed. If every element in the diagonal is different, then your group is cyclic.

Answer 31

Let me know what you've got. After you post something I can post the table I've made.

Answer 32

alright i am still drawing the table i made it again ill finish half of it and just tell me if it is right

Answer 33

Sounds good. Just post it, and I'll let you know.

Answer 34

@KingGeorge

Answer 35

That looks good so far. I would also recommend always writing elements in the form \(r^ns\) if you have both r's and s's in some element. For example, instead of writing \(sr^4\), write \(r^2s\).

You don't have to edit what you've already written, or start over, just follow this for the rest of the matris.

Answer 36

hmm i see what you mean

Answer 37

and just to clarify s*s is just s right

Answer 38

Also, I don't think what I said earlier with unique entries on the diagonal was correct. Just ignore that comment.

\(s*s=s^2=e\)

Answer 39

ooo yes truee truee and yeah i was going to ask you about that i was getting some similar terms

Answer 40

I'll post the table I made now. It's a little bit different format, but you should be able to check against it.

Answer 41

hmm alright thanks a lot @KingGeorge and we can see this is cyclic because

Answer 42

ok you said they are suppose to have order of 12 right

Answer 43

each row has a different element though

Answer 44

There must be an element with order 12 for \(D_6\) to be cyclic. If you look though, you should find that \(D_6\) is not cyclic.

Answer 45

hmm because there is no cycle occuring

Answer 46

Precisely.