# Write all permutations in the cyclic group $C_{12}$ ​of order 12 in cycle notation.

I am tasked to find out the permutations of the cyclic group $$C_{12}$$ of order 12 and am having a little bit of trouble on where to start. I understand the Dihedral groups, but fail to see the difference between the two and how I can attempt this question.

Do I just need to find all the possible rotations and reflections that can occur on $$C_{12}$$?

So for example...

Do nothing ($$e$$)

$$\begin{pmatrix}1 & 2 & 3 & 4& 5& 6&7&8&9&10&11&12\end{pmatrix}$$

Rotate 30 degrees

$$\begin{pmatrix}12 & 1 & 2 & 3& 4& 5&6&7&8&9&10&11\end{pmatrix}$$

Rotate 60 degrees

...

However, we overcount by 12 in this way. Since order of dihedral group $$D_n$$ = $$2n$$

• I don't understand how permutations on 12 "objects" can take profit of the fact that besides these objects belong to an algebraic structure for example here $C_{12}$... Nov 16, 2021 at 5:14
• This does not make sense as stated: "the permutations of the cyclic group $C_{12}$". Do you mean that you are being asked to give a permutation representation of $C_{12}$? Nov 16, 2021 at 6:01

$$C_{12}$$ is the cyclic group of order 12. So it's not $$D_{12}$$ (aka $$D_{24}$$) which has order 24.

So, looking at the two elements you've identified:

• $$e$$, or "do nothing". That doesn't have order 12.
• "rotate $$30^\circ$$", which you've written as $$(12,1,2,3,4,5,6,7,8,9,10,11)$$.

Let's call that one $$a$$. "rotate by $$30^\circ$$" has order 12, and the group it generates is therefore $$C_{12}$$, the cyclic group of order $$12$$.

You could identify all the elements of $$\langle a\rangle$$ which have order $$12$$. The elements themselves are easy to identify, they are $$e=a^0$$, $$a=a^1$$, $$a^2$$, $$a^3$$, $$a^4$$, etc up to $$a^{11}$$ (since $$a^{12}=e=a^0$$).

Note that I also want to point out that when you wrote $$e$$ and $$a$$ above, you didn't write them in "cycle notation". Cycle notation might look like this:

• $$p=(1,4,6,2)(3,7)$$ which means $$p$$ maps $$1\rightarrow4$$, $$2\rightarrow1$$, $$3\rightarrow7$$, $$4\rightarrow6$$, $$5\rightarrow5$$, $$6\rightarrow2$$, $$7\rightarrow3$$, and for $$n\geq 8$$, $$n\rightarrow n$$.

So $$e$$, in cycle notation, is just $$()$$. Your $$a$$, which I think you have mapping $$1$$ to $$12$$, would actually be written $$(1,12,11,10,9,8,7,6,5,4,3,2)$$, or (equivalently), $$(6,5,4,3,2,1,12,11,10,9,8,7)$$ or a few other possibilities, depending what you choose to "start" the cycle with.

Then, $$a^2$$ would map $$1\rightarrow11$$, $$11\rightarrow9$$, $$9\rightarrow7$$, etc, so you'd write it as $$(1,11,9,7,5,3)(2,12,10,8,6,4)$$. That has order $$6$$. I can tell, because $$6$$ is the lowest common multiple of the lengths of all the cycles in $$a^2$$.

Write $$a$$ in cycle notation, then work out $$a^2$$, $$a^3$$, $$a^4$$ etc up to $$a^{11}$$, and identify which ones have order $$12$$. There should be four of them.