Using the hints provided in Adams solution, I have calculated the solutions to be:

1) 6 2) 120 3) 4 5) 12, 360.

as per the title it is my task to find the order of the group elements in the following 4 cases:

  1. ($\begin{smallmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 4 & 2 & 6 & 8 & 9 & 1 & 3 & 7 & 5\end{smallmatrix}$)
  2. $\overline{7} \in \mathbb{Z} / 120$
  3. $\overline{7} \in (\mathbb{Z} / 120) ^ *$
  4. The rotation matrix $ D_\alpha = (\begin{smallmatrix} \cos\alpha & - \sin \alpha \\ \sin \alpha & \cos \alpha \end{smallmatrix}) \in GL_2(\mathbb{R})$ for $\alpha = 30^\circ$ & $\alpha = 17^\circ$

Working so far:

First I consider the general definition of the order of a group element. According to this related question:

"the order of an element $g$ of a group $G$ is the smallest positive integer $n: g^n=e$" (1)

  1. Rewriting the permutation in cyclic notation produces $(1 \ 4 \ 8 \ 7 \ 3 \ 6)(2)(5 \ 9)$, Which allows me to easily see the number of elements in each cycle. Noticing that the number of elements in the largest cycle is divisible by the number of elements in each smaller cycle and using the defintion (1), we can write the permutation as g in the form $g^6=e$, i.e the order of the group element in 1. is 6.
  2. As has been pointed out in a previous question $\overline{7}$ is the residue/congruence class for the field $\mathbb{F_{120}}$ Calculating the inverse using the Euclidean algorithm produces $-17 \equiv 103$. From this point however I am not sure how to proceed. Could someone please verify that I am going in the right direction/give me some tips for how to proceed?
  3. Here I am really at a loss. I understand that $(\mathbb{Z} / 120)^*$ is the set of all numbers between 1 and 120 that are coprime with 120. Could someone please provide a hint?
  4. $GL_2(\mathbb{R})$ (alternative notation $GL(2,\mathbb{R})$) is the general linear group of degree 2. Here I consider each alpha value (which I will designate as $\alpha_1$ & $\alpha_2$ respectively). In order to find the order of the rotation matrices, I raise each matrix to the power of of $n \in \mathbb{N}$ And increment the value of n until I reach the original rotation matrix by a process of trial and error.

$\begin{equation} {D_{\alpha_1}}^1 = (\begin{smallmatrix} \frac{\sqrt{3}}{2} & \frac{-1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{smallmatrix}) \\ {D_{\alpha_1}}^2 = (\begin{smallmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{smallmatrix}) \\ ~ \\ ...\\ ~ \\ {D_{\alpha_1}}^{13} = {D_{\alpha_1}}^1\\ \end{equation}$

Thus the order for ${D_{\alpha_1}}$ is 13.

After having performed the above calculations, one of my peers said "of course its 13, because you just need to think about how many times you would need to rotate something (in a $360^\circ$) by $30^\circ$ until you get the same angle. Using the same logic would imply that the order of ${D_{\alpha_2}}$ is 361, as $gcd(17,360) = 1$. Whilst my peer's advice seems logical, the reasoning seems to lack rigor. Is it valid to accept this reasoning at face value?


For (1), you are correct: we write the permutation in cyclic notation and then find the lowest common multiple of all the cycle lengths (here $\text{lcm}(6,1,2) = 6$).

For (2), we are looking for the minimum positive integer $n$ such that $7n\equiv 0 \pmod{120}$. Doing this by trial and error would take a while, so instead there is a simple formula for this that you may have been told or may be able to work out for yourself.

For (3), note that the identity of the group is $\overline{1}$ (because the group operation is multiplication). Can you compare this to the situation in (2), where the identity was $\overline{0}$ and the operation was addition?

For (4), the order is actually 12 and not 13. This is because ${D_{\alpha_1}}^{12} = I$, and your equation is this multiplied by $D_{\alpha_1}$. Your peer is correct, however, that the best way to go about this is to consider geometrically how many rotations of $30^\circ$ it will take to reach a $360^\circ$ rotation (which is the identity here). This is a perfectly valid method. Can you work out the smallest positive integer for which $k\times 17^\circ$ is a multiple of $360^\circ$? Do you see how this relates to the method we used for a previous part?

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    $\begingroup$ For (1), your feedback is helpful, thanks. For (2), the first formula that comes to mind is the Euclidean algorithm. I will need to think about this. For (3), Again, I will need to think about this. For (4), from what we know about the gcd, it must follow that the smallest possible value for k is 360. Yes, I think I follow. $\endgroup$ – Oscar Apr 17 '18 at 22:46
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    $\begingroup$ The Euclidean algorithm is not necessary here. For (4), your answer is correct - and the problem you've just solved is finding the minimum $n$ such that $17n \equiv 0 \pmod{360}$. This is actually the same type of problem we're looking to solve in (2). $\endgroup$ – Adam Apr 17 '18 at 22:53
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    $\begingroup$ The formula for finding the order of $\overline{x}$ in $\mathbb{Z} / n$ is $n/\text{hcf}(x,n)$. So the answer for (2) is $120$. To give another example, the order of $\overline{42}$ in $\mathbb{Z}/120$ is $20$. Does this make sense? $\endgroup$ – Adam Apr 17 '18 at 23:10
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    $\begingroup$ For $\overline{7}$: $n/hcf(x,n) = 120/ hcf(7,120) = 120 / 1$. For $\overline{42}$: $n/hcf(x,n) = 120/6$. Yes, it makes perfect sense, thanks! Now only (3) remains to be slain. $\endgroup$ – Oscar Apr 17 '18 at 23:19
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    $\begingroup$ if I am interpreting your hint correctly, the answer to part 3 is "how many times must 7 be multiplied by itself before it is congruent to 1 mod 120", in which case I would conclude the solution is order n = 4. $\endgroup$ – Oscar Apr 19 '18 at 16:25
  1. Your answer is correct.

  2. You seem to be completely off base here. $\mathbf Z_{120}$ is the additive group of integers modulo $120$. (By the way, there is no field with $120$ elements. The number of elements in a finite field is always a prime power.) So you should be looking for the smallest positive integer such that $7n\equiv 0\pmod{120}$

  3. The order of an element always divides the group order. How many elements are there? This cuts down the search.

  4. Your peer is correct, but I don't see how he gets $13$, nor how Adam gets 12 in his answer. The smallest positive $k$ such that $17k\equiv 0\pmod{360}$ is $360$, for the reason you give.

  • $\begingroup$ $12$ was my answer for $30^\circ$. And indeed I agree that $360$ is the answer for $17^\circ$. $\endgroup$ – Adam Apr 17 '18 at 22:36
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    $\begingroup$ For 4: There are two different rotation matrices i.e $\alpha = 30^\circ$ and $\alpha = 17^\circ$. For 3: I appreciate the tip $\endgroup$ – Oscar Apr 17 '18 at 22:40
  • $\begingroup$ @Adam Okay, I see now. I thought $D_{\alpha_1} was the given matrix, and I couldn't figure out what you were doing. $\endgroup$ – saulspatz Apr 17 '18 at 22:40
  • $\begingroup$ @Oscar Oh, I didn't read that question closely at all. I just skipped over the $30^\circ$ completely. No wonder I couldn't figure out what was happening. (Still I don't see how your friend gets $13$). $\endgroup$ – saulspatz Apr 17 '18 at 22:47
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    $\begingroup$ @saulspatz That was my error. I made the mistake of searching for an n value such that $g^n = g^1$ instead of $g^n = e$ $\endgroup$ – Oscar Apr 17 '18 at 22:49

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