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$\mathbf{29.}$ 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".2012-11-05
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    I would guess its the subgroup of the $6$th roots of unity.2012-11-05
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    @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.2012-11-05
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    I don't understand what the question is. The title says "finding a generator", but your picture seems to be saying "find the subgroup generated by something".2012-11-05
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    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.2012-11-05
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    @BrianM.Scott I don't believe you :) dial-up?2012-11-05
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    @Jennifer: ’Fraid so, though perhaps not for too much longer.2012-11-05
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    @jak I wasn't implying that images are *wrong*, per se, and there are circumstances in which there's no way around an image (e.g. some geometric problems). Another reason text is preferable to images is the greater accessibility rendered by text.2012-11-05
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    (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.2012-11-05

2 Answers 2

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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 42012-11-05
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    @jak What do you mean the size of $G$. What is $G$ here?2012-11-05
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    It's the multiplicative group of invertible $4 \times 4$ matrices (I misplaced G in my question)2012-11-05
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    The general linear group is not of order $4$... Even if we consider the group over $\mathbb{F}_2$ the order is $20160$.2012-11-05
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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.2012-11-05
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    @jak: This is the formal way to find the roots of a complex number.2012-11-05