1. ## About orders of elements(Algebraic structures-subgroups)

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.

http://personalpages.manchester.ac.u...lgStnewEx4.pdf

2. 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.

http://personalpages.manchester.ac.u...lgStnewEx4.pdf

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).

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

4. 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.

5. 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.

6. 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.

7. 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!