About orders of elements(Algebraic structures-subgroups)

**Hyungmin** · January 10th 2010, 04:46 PM

Hi i just stuck with finding orders of the following elements.
It's Question 3(a,b,c,d)

By the way, i have an answer but dont know how to solve it so please
teach me about process. The answers are [ (a)8, 6, 12 (b)9, 12, 2, 12 (c) 12, 2, 20 (d) 6, infinite , 4]
Thank you very much.

I just attatched the files.(Adobe Reader) or
http://personalpages.manchester.ac.u...lgStnewEx4.pdf

The address is here

**Jhevon** · January 10th 2010, 05:38 PM

Originally Posted by Hyungmin

Hi i just stuck with finding orders of the following elements.
It's Question 3(a,b,c,d)

By the way, i have an answer but dont know how to solve it so please
teach me about process. The answers are [ (a)8, 6, 12 (b)9, 12, 2, 12 (c) 12, 2, 20 (d) 6, infinite , 4]
Thank you very much.

I just attatched the files.(Adobe Reader) or
http://personalpages.manchester.ac.u...lgStnewEx4.pdf

The address is here

recall that the order of an element $a$ of a group $G$ , is the smallest positive integer $n$ such that $a^n = e$ , where $e$ is the identity element of $G$ .

Hopefully you know how to interpret $a^n$ . it is just applying the operation of $G$ to $a$ $n$ times.

So, for example, for 3(i)(a). The element is $z = \frac {\sqrt 2}2 - \frac {\sqrt 2}2i$ in $\mathbb C ^*$ , which is the group of non-zero complex numbers with the operation multiplication.

Notice that $z^8 = 1$ , and 8 is the smallest positive integer power for which this happens (keep multiplying $z$ by itself until you get to 1, you'll notice you had to do this 8 times). hence, the order of $z$ is 8. the others are similar. (i'd probably use the cis notation for the complex number problems. or try to employ any other trick you know that allows you to multiply quickly).

**Hyungmin** · January 12th 2010, 06:12 AM

then what about part C and D
do i have to keep doing standard notation ?

**Swlabr** · January 12th 2010, 06:38 AM

Originally Posted by Hyungmin

then what about part C and D
do i have to keep doing standard notation ?

You can make part (c) easier by proving that if $\alpha, \beta$ are permutations then $o(\alpha \beta) = gcd(o(\alpha), (\beta))$ , where $o(\gamma)$ denotes the order of $\gamma$ .

For (d) I don't know of any "trick" to finding the orders of matrices. It really is just multiply them together and see where it goes.

**HallsofIvy** · January 12th 2010, 06:45 AM

Originally Posted by Hyungmin

then what about part C and D
do i have to keep doing standard notation ?

?? There is no "part C and D" (and no part A and B!). Do you mean parts (iii) and (iv)?
Use whatever notation you prefer as long you get the right answer.

$\tau= (19)(28)(37)(46)$ is easy- it consists of two-cycles, each of which will return to the original, identity, after two repetitions. The order is 2.

It is probably simplest to put $\sigma= \begin{pmatrix}1& 2& 3& 4& 5& 6& 7& 8& 9\\ 2& 3& 1& 5& 6& 7& 4& 9& 8\end{pmatrix}$ in cycle notation: 1 changes into 2 which changes into 3 which changes back to 1: (123). 4 changes into 5 which changes into 6 which changes into 7 which changes back to 4: (4567). 8 changes into 9 which changes back into 8: (89). $\sigma= (123)(4567)(89)$ . (123) changes back to itself after 3 cycles, (4567) after 4 and (89) in 2. They will all change back to the identity simultaneously in the least common multiple of 3, 4, and 2: 12. $\sigma$ has order 12.

$\theta= (13579)(2468)$ , cycles of order 5 and 4. The least common multiple of 5 and 4 is 20. $\theta$ has order 20.

**Swlabr** · January 12th 2010, 06:49 AM

Originally Posted by HallsofIvy

An important property of matrices is that "the determinant of a product is the product of the determinants". In particular, a product can only be 0 if one of the factors is 0 so only a power of 0 is 0. A matrix has a finite order if and only if its determinant is 0. None of the determinants in iv has 0 determinant.

The questions are on group theory not linear algebra (we are working in $GL_2(\mathbb{R})$ ). The identity element here is the identity matrix not the zero matrix.

**HallsofIvy** · January 12th 2010, 07:24 AM

Originally Posted by Swlabr

The questions are on group theory not linear algebra (we are working in $GL_2(\mathbb{R})$ ). The identity element here is the identity matrix not the zero matrix.

That suddenly dawned on me and I deleted that part of my answer! You people are just too fast for me to pretend I didn't make that mistake!

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