walters
  • walters
Let G be a group such that (a*b)^2=a^2*b^2 can pls show the associativity
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
walters
  • walters
|dw:1362323673416:dw|
walters
  • walters
i am not sure
terenzreignz
  • terenzreignz
Isn't the associativity of the binary operation one of the requirements fore being a group?

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walters
  • walters
yes it is
terenzreignz
  • terenzreignz
Then why do you have to prove it when G is already a group? :/
walters
  • walters
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
terenzreignz
  • terenzreignz
:) 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?
walters
  • walters
yes
terenzreignz
  • terenzreignz
Then it's easy to show commutativity :) I'm certain you've already done it XD
walters
  • walters
yes ok now i understand the question
walters
  • walters
wait..
terenzreignz
  • terenzreignz
NO rush :) I'm going to savour this question because I love group theory :D
walters
  • walters
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
walters
  • walters
|dw:1362324922660:dw|
terenzreignz
  • terenzreignz
Hang on, I'm going to need a while to digest that :D
terenzreignz
  • terenzreignz
Okay, so, you know the subgroup test?
terenzreignz
  • terenzreignz
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.
walters
  • walters
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
terenzreignz
  • terenzreignz
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
walters
  • walters
i am not sure
terenzreignz
  • terenzreignz
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) \[\large N = \left\{\tau|\tau \in G, \tau(x)=x+b, \ \ b\in \mathbb{R} \right\}\]
terenzreignz
  • terenzreignz
So, is the identity function... \[\large id(x)=x\] in N?
walters
  • walters
yes,
terenzreignz
  • terenzreignz
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?
walters
  • walters
i think we have to show that it is the group under the binary operation |dw:1362326767652:dw|
terenzreignz
  • terenzreignz
Fair enough :) Let's take the set G, first., shall we?
terenzreignz
  • terenzreignz
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?
walters
  • walters
yes
terenzreignz
  • terenzreignz
Well, consider fog(x) fog(x) = a(cx + d) + b = acx + ad + b and is this still in G?
walters
  • walters
yes
terenzreignz
  • terenzreignz
So we've shown closure. The rest is easy, right? :D
walters
  • walters
are u not suppose to use |dw:1362327123975:dw|
terenzreignz
  • terenzreignz
yeah, but that's just notation :D I'll leave that to you :> \[\huge f(x) = \tau_{ab} \ \ \ \ g(x) = \tau_{cd}\]
walters
  • walters
ok
terenzreignz
  • terenzreignz
So, back to the question about N being a subgroup?
terenzreignz
  • terenzreignz
We've shown the identity, and now we need to show closure.
walters
  • walters
yes i see it
walters
  • walters
@experimentX
walters
  • walters
for that one of subgroup
experimentX
  • experimentX
I am not so good with abstract algebra either ... but let's see it.
walters
  • walters
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
experimentX
  • experimentX
hmm ... if G is a group then why do you have to show associativity? isn't it associative by hypothesis?
walters
  • walters
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
experimentX
  • experimentX
when is N, normalizer?
walters
  • walters
is not N for natural numbers?
experimentX
  • experimentX
sorry, is N normalizer?
walters
  • walters
yes it is
experimentX
  • experimentX
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
walters
  • walters
there are 2 different questions (given that a, b in G, then (a*b)^2=a^2*b^2) and that onee of subgroup
walters
  • walters
let me retype my questions
walters
  • walters
1 Let G be a group such that (a*b)^2=a^2*b^2 for all a,b element G show that is abelian
walters
  • walters
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
experimentX
  • experimentX
|dw:1362355116918:dw| use associativity.
walters
  • walters
is already a group i only have to show that is abelian so to show comutativity only
experimentX
  • experimentX
for any two element, the above picture shows commutativity.
walters
  • walters
|dw:1362355288828:dw|
walters
  • walters
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
experimentX
  • experimentX
yes it's not multiplication ... but associativity and inverse are properties of group, we are only allowed to exploit that.
walters
  • walters
SO IS IT GOING TO BE|dw:1362355812287:dw|
walters
  • walters
SO IF I CAN PUT THE SQUARE ROOT both side Lhs is it going to be equal to RHS
experimentX
  • experimentX
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.
experimentX
  • experimentX
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.
terenzreignz
  • terenzreignz
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?
experimentX
  • experimentX
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.
terenzreignz
  • terenzreignz
LOL I thought abstract algebra was just invented when Mathematicians got bored from Calculus XD But yeah... Abstract Algebra = My Favourite branch of Maths :)
experimentX
  • experimentX
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.
experimentX
  • experimentX
and of course ... just viewing lecture pretty useless without doing exercise (perhaps pratice is the best way of understanding)
terenzreignz
  • terenzreignz
I hope this guy sticks around :) I'm itchin' for more abstract algebra :D
experimentX
  • experimentX
I hope i'll get enough time to do this http://www.extension.harvard.edu/open-learning-initiative/abstract-algebra
terenzreignz
  • terenzreignz
Linear Algebra comes before Abstract Algebra?
experimentX
  • experimentX
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
terenzreignz
  • terenzreignz
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...
experimentX
  • experimentX
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.
experimentX
  • experimentX
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.
experimentX
  • experimentX
*since both they are algebra
terenzreignz
  • terenzreignz
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
experimentX
  • experimentX
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.
terenzreignz
  • terenzreignz
Just appreciate the beauty of algebra... it's an awesome break from the rigors of calculus :D
experimentX
  • experimentX
yeah ... but somehow i enjoy and work more on calculus. Still i would like to learn abstract algebra and Real Analysis (in dept).
terenzreignz
  • terenzreignz
Ahh... Real Analysis... Just when you thought Calculus couldn't possibly get worse XD
experimentX
  • experimentX
When calculus gets worse ... it's Real Analysis, When real analysis goes worse ... I am hopeless ... i guess set theory then.
terenzreignz
  • terenzreignz
I thought Complex Analysis comes after Real Analysis?
experimentX
  • experimentX
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.
experimentX
  • experimentX
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.
terenzreignz
  • terenzreignz
Now I'm scared :D
experimentX
  • experimentX
From mathematical perspective elementary calculus is a child play.
experimentX
  • experimentX
to sum up ... in real analysis, it might be easy to understand a theorem ... but hard to solve problem.
experimentX
  • experimentX
and also note that ,, when calculus get's worse ... we often use techniques of Real Analysis to solve problem.
terenzreignz
  • terenzreignz
Hmmm... that makes sense... but does little to console me :D Now I'm rethinking my purpose in life XD
experimentX
  • experimentX
haha ... that is the ultimate question ... if you trust WolframAlpha then it calculates as 42
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=the+ultimate+meaning+of+life%2C+universe+and+everything
terenzreignz
  • terenzreignz
Well, if they say so :D
experimentX
  • experimentX
well ... cya around :)
terenzreignz
  • terenzreignz
Yeap :) Thanks for... uhh.... stuff :D
experimentX
  • experimentX
yw!!
terenzreignz
  • terenzreignz
We kind of used this thread as a chat-medium.... sorry about that. What do you need help with?
walters
  • walters
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
terenzreignz
  • terenzreignz
That's the subgroup test :D
walters
  • walters
so this means i must first check the identity of mappings and the identity should be in N
walters
  • walters
to conclude that N is not enpty
terenzreignz
  • terenzreignz
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?
walters
  • walters
but should i not suppose to get the exact value as that one in mappings
terenzreignz
  • terenzreignz
Exact value of what?
walters
  • walters
identity element in G
terenzreignz
  • terenzreignz
I guess so. Let's let e(x) be the identity element in G, shall we?
terenzreignz
  • terenzreignz
Now, by definition, for any tau in G \[\huge \tau\circ e(x)=\tau(x)=e\circ\tau(x) \]
terenzreignz
  • terenzreignz
Are you following me so far? :)
terenzreignz
  • terenzreignz
Now, tau always takes the form \[\large \tau(x)=ax+b \ \ \ \ a,b\in\mathbb{R}\] Following the first equation, we have \[\huge \tau(e(x))=a\cdot e(x)+b=ax+b=\tau(x)\]
terenzreignz
  • terenzreignz
And it just goes to show that \[\huge e(x)=x\] So the identity element, is, after all, the identity function :D
walters
  • walters
i've let \[\tau _{pg}\] element of G be identity with respect to 0 opereration yes i also got x on my rough work
terenzreignz
  • terenzreignz
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' \[\huge N=\left\{\tau(x) \ | \ \tau(x)=x+b , \ \ b\in\mathbb{R}\right\}\]
terenzreignz
  • terenzreignz
And clearly, the identity map is in N, just take b=0 Convinced now that the identity is in N? :)
walters
  • walters
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
terenzreignz
  • terenzreignz
Use the subgroup test. Show it's not empty (we've already done that) and show that for any two elements in N \[\huge \alpha,\beta\in N\]\[\huge \alpha\circ\beta^{-1}\in N\]
terenzreignz
  • terenzreignz
This was what you told me earlier, right?
walters
  • walters
am i going to use \[\tau _{cd}\] and \[\tau _{ef}\] because of my condition is of type |dw:1362487393261:dw|
terenzreignz
  • terenzreignz
You can, but we're dealing with elements of N, which means the coefficient of x is 1.
terenzreignz
  • terenzreignz
Remember, again, the original definition of N \[\huge N=\left\{ \tau_{1b} \ \ | \ \ b\in\mathbb{R} \right\}\]
walters
  • walters
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)?
terenzreignz
  • terenzreignz
Just write \[\huge \tau_{1c} \ \ \ and \ \ \ \tau_{1d}\]
terenzreignz
  • terenzreignz
Are you still there? Let's get back to the subgroup test... Let \[\huge \alpha,\beta\in N\]
walters
  • walters
yes i am still trying to work it aside
terenzreignz
  • terenzreignz
Then \[\huge \alpha(x)=x+a\]\[\huge\beta(x)=x+b\]for some\[\huge a,b\in \mathbb{R} \]
walters
  • walters
so wat will be this |dw:1362488547392:dw|
terenzreignz
  • terenzreignz
No... Okay, let's look for the inverse of beta... \[\huge \beta^{-1}(x)\]
terenzreignz
  • terenzreignz
By definition of the inverse... \[\huge \beta\circ\beta^{-1}(x)=e(x)\]
terenzreignz
  • terenzreignz
So... \[\huge \beta(\beta^{-1}(x))=\beta^{-1}(x)+b=x\]
terenzreignz
  • terenzreignz
Which can only mean \[\huge \beta^{-1}(x)=x-b\] So far so good?
walters
  • walters
what ?
terenzreignz
  • terenzreignz
We've found the value of the inverse of beta.
terenzreignz
  • terenzreignz
So, let's calculate \[\huge \alpha\circ\beta^{-1}(x)=\alpha(\beta^{-1}(x))\]
walters
  • walters
but emenyana(wait) is |dw:1362489338312:dw|
terenzreignz
  • terenzreignz
\[\huge =\beta^{-1}(x)+a\]\[\huge =x-b+a\]
terenzreignz
  • terenzreignz
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
walters
  • walters
oh i see
terenzreignz
  • terenzreignz
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
walters
  • walters
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
terenzreignz
  • terenzreignz
Ok :)

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