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walters

  • one year ago

Let G be a group such that (a*b)^2=a^2*b^2 can pls show the associativity

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  1. walters
    • one year ago
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    |dw:1362323673416:dw|

  2. walters
    • one year ago
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    i am not sure

  3. terenzreignz
    • one year ago
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    Isn't the associativity of the binary operation one of the requirements fore being a group?

  4. walters
    • one year ago
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    yes it is

  5. terenzreignz
    • one year ago
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    Then why do you have to prove it when G is already a group? :/

  6. walters
    • one year ago
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    actually my question was about showing that the group is abelian so i am not sure whether should i show commutativity only or i have to show them all

  7. terenzreignz
    • one year ago
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    :) Given that G is a group, to show that it's abelian, all that's necessary is to show commutativity. Is it given in the question that G is a group?

  8. walters
    • one year ago
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    yes

  9. terenzreignz
    • one year ago
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    Then it's easy to show commutativity :) I'm certain you've already done it XD

  10. walters
    • one year ago
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    yes ok now i understand the question

  11. walters
    • one year ago
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    wait..

  12. terenzreignz
    • one year ago
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    NO rush :) I'm going to savour this question because I love group theory :D

  13. walters
    • one year ago
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    let the mapping \[\tau _{ab}\] for a,b element R ,maps the real into the reals by the rule \[\tau _{ab}:x rightarrowa x+b\] and g={\[\tau _{ab}\]:a\[\neq\]=0} be a group under the composition of mappings Prove that N={\[\tau _{1b}\] element G} is a subgroup

  14. walters
    • one year ago
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    |dw:1362324922660:dw|

  15. terenzreignz
    • one year ago
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    Hang on, I'm going to need a while to digest that :D

  16. terenzreignz
    • one year ago
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    Okay, so, you know the subgroup test?

  17. terenzreignz
    • one year ago
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    A subset H of a group G is a subgroup of G if for any elements a and b in H, ab⁻¹ is in H.

  18. walters
    • one year ago
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    A subset H Group is a subgroup iff 1 the identity element in G is also in H 2 if h1 ,h2 element in G is also in G 3 if h element H,then h^-1 element H

  19. terenzreignz
    • one year ago
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    Oh. No shortcuts, eh? :D Okay, first of all, is the identity element in N? The identity element is the one that would send x to x, IE the identity function :D

  20. walters
    • one year ago
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    i am not sure

  21. terenzreignz
    • one year ago
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    Okay. Maybe we should translate the definition of N from gibberish to English :D sloowly :) \[\large N= \left\{ \tau_{1b}\ \right\}\] It means all the mappings that send x to ax + b where a = 1 So, in English :D (not really English, but in simpler terms) \[\large N = \left\{\tau|\tau \in G, \tau(x)=x+b, \ \ b\in \mathbb{R} \right\}\]

  22. terenzreignz
    • one year ago
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    So, is the identity function... \[\large id(x)=x\] in N?

  23. walters
    • one year ago
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    yes,

  24. terenzreignz
    • one year ago
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    So that's the identity element down :D Now, let's try the second necessity, in that given any two elements of the set N, if we compose them, the composition is still in N. Any idea how to proceed?

  25. walters
    • one year ago
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    i think we have to show that it is the group under the binary operation |dw:1362326767652:dw|

  26. terenzreignz
    • one year ago
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    Fair enough :) Let's take the set G, first., shall we?

  27. terenzreignz
    • one year ago
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    Let there be two functions f and g such that f and g are in G. In other words f(x) = ax + b g(x) = cx + d Where a,b,c, and d are real numbers Catch me so far?

  28. walters
    • one year ago
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    yes

  29. terenzreignz
    • one year ago
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    Well, consider fog(x) fog(x) = a(cx + d) + b = acx + ad + b and is this still in G?

  30. walters
    • one year ago
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    yes

  31. terenzreignz
    • one year ago
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    So we've shown closure. The rest is easy, right? :D

  32. walters
    • one year ago
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    are u not suppose to use |dw:1362327123975:dw|

  33. terenzreignz
    • one year ago
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    yeah, but that's just notation :D I'll leave that to you :> \[\huge f(x) = \tau_{ab} \ \ \ \ g(x) = \tau_{cd}\]

  34. walters
    • one year ago
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    ok

  35. terenzreignz
    • one year ago
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    So, back to the question about N being a subgroup?

  36. terenzreignz
    • one year ago
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    We've shown the identity, and now we need to show closure.

  37. walters
    • one year ago
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    yes i see it

  38. walters
    • one year ago
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    @experimentX

  39. walters
    • one year ago
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    for that one of subgroup

  40. experimentX
    • one year ago
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    I am not so good with abstract algebra either ... but let's see it.

  41. walters
    • one year ago
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    this is what i did since G is a group ,let 1,b element of real number then |dw:1362353281934:dw| then i don't know wat to do next

  42. experimentX
    • one year ago
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    hmm ... if G is a group then why do you have to show associativity? isn't it associative by hypothesis?

  43. walters
    • one year ago
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    yes i see that since it is the group it is not neccessary one thing that i can't do it is that of showing N is a subgroup

  44. experimentX
    • one year ago
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    when is N, normalizer?

  45. walters
    • one year ago
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    is not N for natural numbers?

  46. experimentX
    • one year ago
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    sorry, is N normalizer?

  47. walters
    • one year ago
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    yes it is

  48. experimentX
    • one year ago
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    given that a, b in G, then (a*b)^2=a^2*b^2 looks like we could use this http://en.wikipedia.org/wiki/Normal_subgroup

  49. walters
    • one year ago
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    there are 2 different questions (given that a, b in G, then (a*b)^2=a^2*b^2) and that onee of subgroup

  50. walters
    • one year ago
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    let me retype my questions

  51. walters
    • one year ago
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    1 Let G be a group such that (a*b)^2=a^2*b^2 for all a,b element G show that <G,*> is abelian

  52. walters
    • one year ago
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    2 let the mapping \[\tau _{ab}\] for a,b element of R ,maps the reals into the reals by the rule \[\tau _{ab}: x \rightarrow ax +b\] and G ={\[\tau _{ab}:a \neq0\]} be a group under the composition of mappings Proe that N={\[\tau _{1b}\ G]} s a subgroup of G

  53. experimentX
    • one year ago
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    |dw:1362355116918:dw| use associativity.

  54. walters
    • one year ago
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    is already a group i only have to show that is abelian so to show comutativity only

  55. experimentX
    • one year ago
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    for any two element, the above picture shows commutativity.

  56. walters
    • one year ago
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    |dw:1362355288828:dw|

  57. walters
    • one year ago
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    but this is confusing because the * operation does not mean is multiplication wat if this operation is under (-) LHS will not be equal to RHS

  58. experimentX
    • one year ago
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    yes it's not multiplication ... but associativity and inverse are properties of group, we are only allowed to exploit that.

  59. walters
    • one year ago
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    SO IS IT GOING TO BE|dw:1362355812287:dw|

  60. walters
    • one year ago
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    SO IF I CAN PUT THE SQUARE ROOT both side Lhs is it going to be equal to RHS

  61. experimentX
    • one year ago
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    No, let assume \(a\) , \(b \in G\) \[ (a*b)^2 = (a*b)*(a*b)\\ (a*b)*(a*b) = a^2*b^2 = (a*a)*(b*b)\\ a^{-1}*(a*b)*(a*b) = a^2*b^2 = a^{-1}*(a*a)*(b*b)\\ (a^{-1}*a)*b*(a*b) = a^2*b^2 = (a^{-1}*a)*a*(b*b)\\ (e)*b*(a*b) = a^2*b^2 = (e)*a*(b*b) \\ b*(a*b) = a*(b*b)\\ b*(a*b)*b^{-1} = a*(b*b)*b^{-1}\\ b*a*(b*b^{-1}) = a*b*(b*b^{-1})\\ b*a = a*b\] Since two arbitrary elements of G are commutative, the group G is albelian.

  62. experimentX
    • one year ago
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    http://planetmath.org/Normalizer.html http://planetmath.org/NormalSubgroup.html Let N be normalizer of G i,e. \( H \subseteq G \) Since it's abelian \( aNa^{-1} = aa^{-1}N = eN = N\; \forall a \in G \) Hence N is normal subgroup. However I am not sure ,,, I am a physicist, abstract algebra is one of my weakest point.

  63. terenzreignz
    • one year ago
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    Oh... I missed it :( Why did I have to fall asleep?! Hey, @experimentX I like abstract algebra and all, but does this group stuff have any applications, say, in Physics?

  64. experimentX
    • one year ago
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    yeah ... in QM it is used to check symmetry of operators. And there is Lie algebra supposedly used in string theories ... but i haven't advanced to that point.

  65. terenzreignz
    • one year ago
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    LOL I thought abstract algebra was just invented when Mathematicians got bored from Calculus XD But yeah... Abstract Algebra = My Favourite branch of Maths :)

  66. experimentX
    • one year ago
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    yeah ... i admit, it's elegant (although when i first encountered, i couldn't comprehend) ... there is a free lecture series on Harvard Extension ... but i hardly get time.

  67. experimentX
    • one year ago
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    and of course ... just viewing lecture pretty useless without doing exercise (perhaps pratice is the best way of understanding)

  68. terenzreignz
    • one year ago
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    I hope this guy sticks around :) I'm itchin' for more abstract algebra :D

  69. experimentX
    • one year ago
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    I hope i'll get enough time to do this http://www.extension.harvard.edu/open-learning-initiative/abstract-algebra

  70. terenzreignz
    • one year ago
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    Linear Algebra comes before Abstract Algebra?

  71. experimentX
    • one year ago
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    also @terenzreignz have you check this site? http://math.stackexchange.com/ http://math.stackexchange.com/questions/tagged/abstract-algebra yeah, it's easy that way ... since you are oriented with matrices since early school. http://math.stackexchange.com/questions/tagged/abstract-algebra

  72. terenzreignz
    • one year ago
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    Don't vector spaces fall under Linear Algebra? I don't think I'd have understood it nearly as well had I not read up on Groups first...

  73. experimentX
    • one year ago
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    i think linear algebra is subset of Abstract algebra ... yeah vector spaces fall under linear algebra ... (but linear algebra is a bit broad) ... it's both applied and abstract.

  74. experimentX
    • one year ago
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    i was reading this book ... http://www.amazon.com/Linear-Algebra-Undergraduate-Texts-Mathematics/dp/0387964126 it had (little part) some group theory inside it. I think linear algebra falls under Abstract algebra ... since both they are linear algebra ... just add the condition of linearity, and vector spaces ... they are linear by their definition. And matrices are vector spaces of which has max application on applied sciences.

  75. experimentX
    • one year ago
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    *since both they are algebra

  76. terenzreignz
    • one year ago
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    Vector spaces seem to be between Groups and Rings, and yet, above rings, since they involve Fields... It's a complicated relationship :D The Vector Space is an abelian group, it doesn't have multiplication defined on it (unless it be an inner product space, and even then, the products aren't vectors), but it does have scalar multiplication, which distributes in the same manner as in rings, but vector spaces involve fields... Best not to think about this stuff :D

  77. experimentX
    • one year ago
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    I haven't seen that way since I am still novice on abstract algebra ... looks like youare right... and i think inner product is generalization of scalar product.

  78. terenzreignz
    • one year ago
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    Just appreciate the beauty of algebra... it's an awesome break from the rigors of calculus :D

  79. experimentX
    • one year ago
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    yeah ... but somehow i enjoy and work more on calculus. Still i would like to learn abstract algebra and Real Analysis (in dept).

  80. terenzreignz
    • one year ago
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    Ahh... Real Analysis... Just when you thought Calculus couldn't possibly get worse XD

  81. experimentX
    • one year ago
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    When calculus gets worse ... it's Real Analysis, When real analysis goes worse ... I am hopeless ... i guess set theory then.

  82. terenzreignz
    • one year ago
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    I thought Complex Analysis comes after Real Analysis?

  83. experimentX
    • one year ago
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    The complex analysis I am doing is mostly applied. I am well oriented with Real Analysis (nearly up to Lebesgue integral)... but the methods of real analysis are very hard.

  84. experimentX
    • one year ago
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    you can easily think of solution on calculus ... you begin at something and use these particular theorems ... and arrive at result. this is pretty straight forward. But Real Analysis is different thing ... nearly all theorem requires a long paragraph of explanation. and it has to be locally correct. Real Analysis lies somewhere between set theory and calculus.

  85. terenzreignz
    • one year ago
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    Now I'm scared :D

  86. experimentX
    • one year ago
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    From mathematical perspective elementary calculus is a child play.

  87. experimentX
    • one year ago
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    to sum up ... in real analysis, it might be easy to understand a theorem ... but hard to solve problem.

  88. experimentX
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    and also note that ,, when calculus get's worse ... we often use techniques of Real Analysis to solve problem.

  89. terenzreignz
    • one year ago
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    Hmmm... that makes sense... but does little to console me :D Now I'm rethinking my purpose in life XD

  90. experimentX
    • one year ago
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    haha ... that is the ultimate question ... if you trust WolframAlpha then it calculates as 42

  91. experimentX
    • one year ago
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    http://www.wolframalpha.com/input/?i=the+ultimate+meaning+of+life%2C+universe+and+everything

  92. terenzreignz
    • one year ago
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    Well, if they say so :D

  93. experimentX
    • one year ago
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    well ... cya around :)

  94. terenzreignz
    • one year ago
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    Yeap :) Thanks for... uhh.... stuff :D

  95. experimentX
    • one year ago
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    yw!!

  96. terenzreignz
    • one year ago
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    We kind of used this thread as a chat-medium.... sorry about that. What do you need help with?

  97. walters
    • one year ago
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    by subgroup i know that i must show that Let H <= G(group) H=G iff H is not empty ie e elment H 2 gh^-1 element of H for every g,h element of H

  98. terenzreignz
    • one year ago
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    That's the subgroup test :D

  99. walters
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    so this means i must first check the identity of mappings and the identity should be in N

  100. walters
    • one year ago
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    to conclude that N is not enpty

  101. terenzreignz
    • one year ago
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    You've thus concluded that N is not empty, right? To recap, N is the set of all functions that send x to x+b, for some real number b, right?

  102. walters
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    but should i not suppose to get the exact value as that one in mappings

  103. terenzreignz
    • one year ago
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    Exact value of what?

  104. walters
    • one year ago
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    identity element in G

  105. terenzreignz
    • one year ago
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    I guess so. Let's let e(x) be the identity element in G, shall we?

  106. terenzreignz
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    Now, by definition, for any tau in G \[\huge \tau\circ e(x)=\tau(x)=e\circ\tau(x) \]

  107. terenzreignz
    • one year ago
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    Are you following me so far? :)

  108. terenzreignz
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    Now, tau always takes the form \[\large \tau(x)=ax+b \ \ \ \ a,b\in\mathbb{R}\] Following the first equation, we have \[\huge \tau(e(x))=a\cdot e(x)+b=ax+b=\tau(x)\]

  109. terenzreignz
    • one year ago
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    And it just goes to show that \[\huge e(x)=x\] So the identity element, is, after all, the identity function :D

  110. walters
    • one year ago
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    i've let \[\tau _{pg}\] element of G be identity with respect to 0 opereration yes i also got x on my rough work

  111. terenzreignz
    • one year ago
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    So, with that done, clearly e(x) is in N, meaning N is not empty :) If you're still not convinced, let's recap N the 'proper Mathematical way' \[\huge N=\left\{\tau(x) \ | \ \tau(x)=x+b , \ \ b\in\mathbb{R}\right\}\]

  112. terenzreignz
    • one year ago
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    And clearly, the identity map is in N, just take b=0 Convinced now that the identity is in N? :)

  113. walters
    • one year ago
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    ok now ths become more interesting how am i gonna answer this question should i start by showing that it is a group "especially by showing the identity of G" or should i start by answering the subgroup 's question

  114. terenzreignz
    • one year ago
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    Use the subgroup test. Show it's not empty (we've already done that) and show that for any two elements in N \[\huge \alpha,\beta\in N\]\[\huge \alpha\circ\beta^{-1}\in N\]

  115. terenzreignz
    • one year ago
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    This was what you told me earlier, right?

  116. walters
    • one year ago
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    am i going to use \[\tau _{cd}\] and \[\tau _{ef}\] because of my condition is of type |dw:1362487393261:dw|

  117. terenzreignz
    • one year ago
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    You can, but we're dealing with elements of N, which means the coefficient of x is 1.

  118. terenzreignz
    • one year ago
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    Remember, again, the original definition of N \[\huge N=\left\{ \tau_{1b} \ \ | \ \ b\in\mathbb{R} \right\}\]

  119. walters
    • one year ago
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    so it will look like this let |dw:1362487871485:dw| but thre is something i don't understand on the subscripts how am i going to writeit because i have cd and ef (am i gonna have c and e equal to 1 or what)?

  120. terenzreignz
    • one year ago
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    Just write \[\huge \tau_{1c} \ \ \ and \ \ \ \tau_{1d}\]

  121. terenzreignz
    • one year ago
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    Are you still there? Let's get back to the subgroup test... Let \[\huge \alpha,\beta\in N\]

  122. walters
    • one year ago
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    yes i am still trying to work it aside

  123. terenzreignz
    • one year ago
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    Then \[\huge \alpha(x)=x+a\]\[\huge\beta(x)=x+b\]for some\[\huge a,b\in \mathbb{R} \]

  124. walters
    • one year ago
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    so wat will be this |dw:1362488547392:dw|

  125. terenzreignz
    • one year ago
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    No... Okay, let's look for the inverse of beta... \[\huge \beta^{-1}(x)\]

  126. terenzreignz
    • one year ago
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    By definition of the inverse... \[\huge \beta\circ\beta^{-1}(x)=e(x)\]

  127. terenzreignz
    • one year ago
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    So... \[\huge \beta(\beta^{-1}(x))=\beta^{-1}(x)+b=x\]

  128. terenzreignz
    • one year ago
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    Which can only mean \[\huge \beta^{-1}(x)=x-b\] So far so good?

  129. walters
    • one year ago
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    what ?

  130. terenzreignz
    • one year ago
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    We've found the value of the inverse of beta.

  131. terenzreignz
    • one year ago
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    So, let's calculate \[\huge \alpha\circ\beta^{-1}(x)=\alpha(\beta^{-1}(x))\]

  132. walters
    • one year ago
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    but emenyana(wait) is |dw:1362489338312:dw|

  133. terenzreignz
    • one year ago
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    \[\huge =\beta^{-1}(x)+a\]\[\huge =x-b+a\]

  134. terenzreignz
    • one year ago
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    No, see, we're not taking the multiplicative inverse, we're taking the inverse of the function. Remember that the binary operation is composition, not muliplication

  135. walters
    • one year ago
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    oh i see

  136. terenzreignz
    • one year ago
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    so, \[\huge \alpha\circ\beta^{-1}(x)=x+(a-b)\]And since a and b are both real, so is a-b And therefore \[\huge \alpha\circ\beta^{-1}\in N\] And further implying \[\huge N\le G\] ∎ That's it, and thanks, it was fun D

  137. walters
    • one year ago
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    thanks very much ,see u on the other question i am still busy with it i will consult wen i get stuck it is the proof

  138. terenzreignz
    • one year ago
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    Ok :)

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