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terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Isn't the associativity of the binary operation one of the requirements fore being a group?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Then why do you have to prove it when G is already a group? :/

walters
 one year ago
Best ResponseYou've already chosen the best response.0actually 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

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2:) 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?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Then it's easy to show commutativity :) I'm certain you've already done it XD

walters
 one year ago
Best ResponseYou've already chosen the best response.0yes ok now i understand the question

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2NO rush :) I'm going to savour this question because I love group theory :D

walters
 one year ago
Best ResponseYou've already chosen the best response.0let 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

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Hang on, I'm going to need a while to digest that :D

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Okay, so, you know the subgroup test?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2A 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.

walters
 one year ago
Best ResponseYou've already chosen the best response.0A 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

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Oh. 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

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Okay. 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) \[\large N = \left\{\tau\tau \in G, \tau(x)=x+b, \ \ b\in \mathbb{R} \right\}\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2So, is the identity function... \[\large id(x)=x\] in N?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2So 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?

walters
 one year ago
Best ResponseYou've already chosen the best response.0i think we have to show that it is the group under the binary operation dw:1362326767652:dw

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Fair enough :) Let's take the set G, first., shall we?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Let 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?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Well, consider fog(x) fog(x) = a(cx + d) + b = acx + ad + b and is this still in G?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2So we've shown closure. The rest is easy, right? :D

walters
 one year ago
Best ResponseYou've already chosen the best response.0are u not suppose to use dw:1362327123975:dw

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2yeah, but that's just notation :D I'll leave that to you :> \[\huge f(x) = \tau_{ab} \ \ \ \ g(x) = \tau_{cd}\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2So, back to the question about N being a subgroup?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2We've shown the identity, and now we need to show closure.

walters
 one year ago
Best ResponseYou've already chosen the best response.0for that one of subgroup

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1I am not so good with abstract algebra either ... but let's see it.

walters
 one year ago
Best ResponseYou've already chosen the best response.0this 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

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1hmm ... if G is a group then why do you have to show associativity? isn't it associative by hypothesis?

walters
 one year ago
Best ResponseYou've already chosen the best response.0yes 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

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1when is N, normalizer?

walters
 one year ago
Best ResponseYou've already chosen the best response.0is not N for natural numbers?

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1sorry, is N normalizer?

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1given 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

walters
 one year ago
Best ResponseYou've already chosen the best response.0there are 2 different questions (given that a, b in G, then (a*b)^2=a^2*b^2) and that onee of subgroup

walters
 one year ago
Best ResponseYou've already chosen the best response.0let me retype my questions

walters
 one year ago
Best ResponseYou've already chosen the best response.01 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

walters
 one year ago
Best ResponseYou've already chosen the best response.02 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

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1dw:1362355116918:dw use associativity.

walters
 one year ago
Best ResponseYou've already chosen the best response.0is already a group i only have to show that is abelian so to show comutativity only

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1for any two element, the above picture shows commutativity.

walters
 one year ago
Best ResponseYou've already chosen the best response.0but this is confusing because the * operation does not mean is multiplication wat if this operation is under () LHS will not be equal to RHS

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1yes it's not multiplication ... but associativity and inverse are properties of group, we are only allowed to exploit that.

walters
 one year ago
Best ResponseYou've already chosen the best response.0SO IS IT GOING TO BEdw:1362355812287:dw

walters
 one year ago
Best ResponseYou've already chosen the best response.0SO IF I CAN PUT THE SQUARE ROOT both side Lhs is it going to be equal to RHS

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1No, 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.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1http://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.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Oh... 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?

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1yeah ... 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.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2LOL I thought abstract algebra was just invented when Mathematicians got bored from Calculus XD But yeah... Abstract Algebra = My Favourite branch of Maths :)

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1yeah ... 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.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1and of course ... just viewing lecture pretty useless without doing exercise (perhaps pratice is the best way of understanding)

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2I hope this guy sticks around :) I'm itchin' for more abstract algebra :D

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1I hope i'll get enough time to do this http://www.extension.harvard.edu/openlearninginitiative/abstractalgebra

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Linear Algebra comes before Abstract Algebra?

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1also @terenzreignz have you check this site? http://math.stackexchange.com/ http://math.stackexchange.com/questions/tagged/abstractalgebra yeah, it's easy that way ... since you are oriented with matrices since early school. http://math.stackexchange.com/questions/tagged/abstractalgebra

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Don'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...

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1i 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.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1i was reading this book ... http://www.amazon.com/LinearAlgebraUndergraduateTextsMathematics/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.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1*since both they are algebra

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Vector 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

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Just appreciate the beauty of algebra... it's an awesome break from the rigors of calculus :D

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1yeah ... but somehow i enjoy and work more on calculus. Still i would like to learn abstract algebra and Real Analysis (in dept).

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Ahh... Real Analysis... Just when you thought Calculus couldn't possibly get worse XD

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1When calculus gets worse ... it's Real Analysis, When real analysis goes worse ... I am hopeless ... i guess set theory then.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2I thought Complex Analysis comes after Real Analysis?

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1The 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.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1you 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.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Now I'm scared :D

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1From mathematical perspective elementary calculus is a child play.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1to sum up ... in real analysis, it might be easy to understand a theorem ... but hard to solve problem.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1and also note that ,, when calculus get's worse ... we often use techniques of Real Analysis to solve problem.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Hmmm... that makes sense... but does little to console me :D Now I'm rethinking my purpose in life XD

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1haha ... that is the ultimate question ... if you trust WolframAlpha then it calculates as 42

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=the+ultimate+meaning+of+life%2C+universe+and+everything

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Well, if they say so :D

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1well ... cya around :)

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Yeap :) Thanks for... uhh.... stuff :D

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2We kind of used this thread as a chatmedium.... sorry about that. What do you need help with?

walters
 one year ago
Best ResponseYou've already chosen the best response.0by 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

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2That's the subgroup test :D

walters
 one year ago
Best ResponseYou've already chosen the best response.0so this means i must first check the identity of mappings and the identity should be in N

walters
 one year ago
Best ResponseYou've already chosen the best response.0to conclude that N is not enpty

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2You'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?

walters
 one year ago
Best ResponseYou've already chosen the best response.0but should i not suppose to get the exact value as that one in mappings

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Exact value of what?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2I guess so. Let's let e(x) be the identity element in G, shall we?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Now, by definition, for any tau in G \[\huge \tau\circ e(x)=\tau(x)=e\circ\tau(x) \]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Are you following me so far? :)

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Now, tau always takes the form \[\large \tau(x)=ax+b \ \ \ \ a,b\in\mathbb{R}\] Following the first equation, we have \[\huge \tau(e(x))=a\cdot e(x)+b=ax+b=\tau(x)\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2And it just goes to show that \[\huge e(x)=x\] So the identity element, is, after all, the identity function :D

walters
 one year ago
Best ResponseYou've already chosen the best response.0i've let \[\tau _{pg}\] element of G be identity with respect to 0 opereration yes i also got x on my rough work

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2So, 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' \[\huge N=\left\{\tau(x) \  \ \tau(x)=x+b , \ \ b\in\mathbb{R}\right\}\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2And clearly, the identity map is in N, just take b=0 Convinced now that the identity is in N? :)

walters
 one year ago
Best ResponseYou've already chosen the best response.0ok 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

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Use the subgroup test. Show it's not empty (we've already done that) and show that for any two elements in N \[\huge \alpha,\beta\in N\]\[\huge \alpha\circ\beta^{1}\in N\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2This was what you told me earlier, right?

walters
 one year ago
Best ResponseYou've already chosen the best response.0am i going to use \[\tau _{cd}\] and \[\tau _{ef}\] because of my condition is of type dw:1362487393261:dw

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2You can, but we're dealing with elements of N, which means the coefficient of x is 1.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Remember, again, the original definition of N \[\huge N=\left\{ \tau_{1b} \ \  \ \ b\in\mathbb{R} \right\}\]

walters
 one year ago
Best ResponseYou've already chosen the best response.0so 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)?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Just write \[\huge \tau_{1c} \ \ \ and \ \ \ \tau_{1d}\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Are you still there? Let's get back to the subgroup test... Let \[\huge \alpha,\beta\in N\]

walters
 one year ago
Best ResponseYou've already chosen the best response.0yes i am still trying to work it aside

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Then \[\huge \alpha(x)=x+a\]\[\huge\beta(x)=x+b\]for some\[\huge a,b\in \mathbb{R} \]

walters
 one year ago
Best ResponseYou've already chosen the best response.0so wat will be this dw:1362488547392:dw

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2No... Okay, let's look for the inverse of beta... \[\huge \beta^{1}(x)\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2By definition of the inverse... \[\huge \beta\circ\beta^{1}(x)=e(x)\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2So... \[\huge \beta(\beta^{1}(x))=\beta^{1}(x)+b=x\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2Which can only mean \[\huge \beta^{1}(x)=xb\] So far so good?

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2We've found the value of the inverse of beta.

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2So, let's calculate \[\huge \alpha\circ\beta^{1}(x)=\alpha(\beta^{1}(x))\]

walters
 one year ago
Best ResponseYou've already chosen the best response.0but emenyana(wait) is dw:1362489338312:dw

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2\[\huge =\beta^{1}(x)+a\]\[\huge =xb+a\]

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2No, 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

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.2so, \[\huge \alpha\circ\beta^{1}(x)=x+(ab)\]And since a and b are both real, so is ab 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

walters
 one year ago
Best ResponseYou've already chosen the best response.0thanks 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
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