# Find the order of the cyclic subgroup of the given group generated by the indicated element

$\mathbf{29.}$ The subgroup of $U_6$ generated by $\cos\frac{2\pi}3+i\sin\frac{2\pi}3.$

$\mathbf{30.}$ The subgroup of $U_5$ generated by $\cos\frac{4\pi}5+i\sin\frac{4\pi}5.$

$\mathbf{31.}$ The subgroup of $U_8$ generated by $\cos\frac{3\pi}2+i\sin\frac{3\pi}2.$

where $U_n = \{z \in \mathbb{C} : z^n = 1 \}$ (nth roots of unity)

For example in (29), I have to multiply $e^{\frac{2\pi i}{3}}$ three times to get back 1. So the order is 1. However I don't see how the number "6" under the U plays a role here.

Also for instance, for the subgroup of $U_8$ generated by $e^{\frac{5\pi i}{4}}$, the order is 8 because $(e^{\frac{5\pi i}{4}})^8 = 1$. The answer given was

The 1st number which makes $e^{\frac{5\pi i}{4}} = 1$ is 8 because the number must be either 2,4, or 8. 2 and 4 are not working so $|\langle e^{\frac{5\pi i}{4}}\rangle| = 8$

I don't understand how 2 and 4 come up here or even related here?

Also, sorry for the inconvenience I caused

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Hi, jak. It would really help if you could start transcribing some of your exercises directly as text in your questions, e.g. in this case, it wouldn't be too terribly time-consuming to do so. Perhaps practice a bit during some "down time". – amWhy Nov 5 '12 at 1:19
I would guess its the subgroup of the $6$th roots of unity. – EuYu Nov 5 '12 at 1:21
@jak A few reasons pictures are undesirable. Firstly it's perceived as lazy and some people find the lack of effort a bit rude. But the more important reason is that we wish for these questions to be an archive for future visitors and pictures are unreliable for that purpose. – EuYu Nov 5 '12 at 1:23
And believe it or not, there are still a few of us on dial-up, for whom images take a lot longer to load than text does. – Brian M. Scott Nov 5 '12 at 1:25
(1) It's $e^{5 \pi i / 4}$, not $e^{5 \pi /4}$. (2) The order is not 8 "because $(e^{\frac{5\pi i}{4}})^8 = 1$". There is another condition you need. (3) $(e^{\frac{5\pi i}{4}})^4$ is not equal to 1. (4) You did not clog up the comments section, others did. – wj32 Nov 5 '12 at 1:34

It seems you are confusing some things from group theory. Most of your questions are answered from Lagrange's theorem.

The order of a group is the number of elements in a group. The order of an element $a$ is the smallest positive integer $x$ for which $a^x = 1$.

Lagrange's theorem tells us that the size of a subgroup $H$ of $G$ must divide the size of $G$. That is if $H\le G$ then we must have $|H|$ divides $|G|$. This in particular implies that the order of an element must also divide the order of the group because the set $\{a,\ a^2,\ \cdots,\ a^x\}$ forms a subgroup called the cyclic subgroup of $a$ denoted $\langle a\rangle$. This is what we mean by the subgroup generated by $a$.

For your first question, your group is $U_6$ the $6$th roots of unity. The order of this group is $6$ and so every subgroup must be of order $1,\ 2,\ 3,\ 6$. You can see that the order of your element is $3$ and so the subgroup generated by it must be of order $3$ as well. In particular the subgroup is in fact $U_3$. The number $6$ has significance here in that it limits the orders we must try.

For the second question, the order of the group is $5$, so what must the order of the element be?

And for the last, again you have to try the divisors of $8$ which are $1,\ 2,\ 4,\ 8$. Clearly the element is not $1$ itself, so it remains to try $2,\ 4,\ 8$. This is what your solution suggests.

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that's an interesting Theorem. I'll look into it. But just going through this trick right now, I found this "example" in a group multiplicative group of invertible matrices generated by $\begin{bmatrix} 0 &1 &0 &0 \\ 0 &0 &0 &1 \\ 0&0 &1 &0 \\ 1 &0 &0 &0 \end{bmatrix}$ the order is 3, and the it doesn't divide the size of G, which is 4 – Hawk Nov 5 '12 at 2:12
@jak What do you mean the size of $G$. What is $G$ here? – EuYu Nov 5 '12 at 2:16
It's the multiplicative group of invertible $4 \times 4$ matrices (I misplaced G in my question) – Hawk Nov 5 '12 at 2:22
The general linear group is not of order $4$... Even if we consider the group over $\mathbb{F}_2$ the order is $20160$. – EuYu Nov 5 '12 at 2:25

You are given $U_n = \{z \in \mathbb{C} : z^n = 1 \}$, so for (29), you have

$$z^6=1=e^0 \implies z^6 = e^{i2k\pi} \implies z=e^{i2k\pi/6}= e^{ik\pi/3} \,,k=0,1,2\,,$$

where the fact $e^{i2k\pi}=1\,\,, k\in \mathbb{Z}$ has been used.

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I just noticed that the possible candidates are the factors of the subscripts, that's how the solution came up with these candidates so quickly. Is that a coincidence? It works for (29) too. – Hawk Nov 5 '12 at 1:57
@jak: This is the formal way to find the roots of a complex number. – Mhenni Benghorbal Nov 5 '12 at 2:11