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Math Help - About orders of elements(Algebraic structures-subgroups)

  1. #1
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    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.


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

    The address is here
    Attached Files Attached Files
    Last edited by Hyungmin; January 10th 2010 at 05:07 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Hyungmin View Post
    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).
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    then what about part C and D
    do i have to keep doing standard notation ?
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Hyungmin View Post
    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.
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  5. #5
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    Quote Originally Posted by Hyungmin View Post
    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.
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    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.
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    Quote Originally Posted by Swlabr View Post
    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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