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

Mathematics
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|dw:1362323673416:dw|
i am not sure
Isn't the associativity of the binary operation one of the requirements fore being a group?

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

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