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

- walters

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- walters

|dw:1362323673416:dw|

- walters

i am not sure

- terenzreignz

Isn't the associativity of the binary operation one of the requirements fore being a group?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- walters

yes it is

- terenzreignz

Then why do you have to prove it when G is already a group? :/

- 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

:)
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

yes

- terenzreignz

Then it's easy to show commutativity :)
I'm certain you've already done it XD

- walters

yes ok now i understand the question

- walters

wait..

- terenzreignz

NO rush :)
I'm going to savour this question because I love group theory :D

- 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

|dw:1362324922660:dw|

- terenzreignz

Hang on, I'm going to need a while to digest that :D

- terenzreignz

Okay, so, you know the subgroup test?

- 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

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

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

i am not sure

- 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

So, is the identity function...
\[\large id(x)=x\] in N?

- walters

yes,

- 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

i think we have to show that it is the group under the binary operation |dw:1362326767652:dw|

- terenzreignz

Fair enough :)
Let's take the set G, first., shall we?

- 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

yes

- terenzreignz

Well, consider fog(x)
fog(x) = a(cx + d) + b
= acx + ad + b
and is this still in G?

- walters

yes

- terenzreignz

So we've shown closure. The rest is easy, right? :D

- walters

are u not suppose to use |dw:1362327123975:dw|

- terenzreignz

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

- walters

ok

- terenzreignz

So, back to the question about N being a subgroup?

- terenzreignz

We've shown the identity, and now we need to show closure.

- walters

yes i see it

- walters

@experimentX

- walters

for that one of subgroup

- experimentX

I am not so good with abstract algebra either ... but let's see it.

- 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

hmm ... if G is a group then why do you have to show associativity? isn't it associative by hypothesis?

- 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

when is N, normalizer?

- walters

is not N for natural numbers?

- experimentX

sorry, is N normalizer?

- walters

yes it is

- 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

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

let me retype my questions

- 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

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

|dw:1362355116918:dw|
use associativity.

- walters

is already a group i only have to show that is abelian so to show comutativity only

- experimentX

for any two element, the above picture shows commutativity.

- walters

|dw:1362355288828:dw|

- 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

yes it's not multiplication ... but associativity and inverse are properties of group, we are only allowed to exploit that.

- walters

SO IS IT GOING TO BE|dw:1362355812287:dw|

- walters

SO IF I CAN PUT THE SQUARE ROOT both side Lhs is it going to be equal to RHS

- 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

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

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

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

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

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

and of course ... just viewing lecture pretty useless without doing exercise (perhaps pratice is the best way of understanding)

- terenzreignz

I hope this guy sticks around :)
I'm itchin' for more abstract algebra :D

- experimentX

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

- terenzreignz

Linear Algebra comes before Abstract Algebra?

- 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

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

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

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

*since both they are algebra

- 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

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

Just appreciate the beauty of algebra... it's an awesome break from the rigors of calculus :D

- 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

Ahh... Real Analysis...
Just when you thought Calculus couldn't possibly get worse XD

- experimentX

When calculus gets worse ... it's Real Analysis, When real analysis goes worse ... I am hopeless ... i guess set theory then.

- terenzreignz

I thought Complex Analysis comes after Real Analysis?

- 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

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

Now I'm scared :D

- experimentX

From mathematical perspective elementary calculus is a child play.

- experimentX

to sum up ... in real analysis, it might be easy to understand a theorem ... but hard to solve problem.

- experimentX

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

- terenzreignz

Hmmm... that makes sense... but does little to console me :D
Now I'm rethinking my purpose in life XD

- experimentX

haha ... that is the ultimate question ... if you trust WolframAlpha then it calculates as 42

- experimentX

http://www.wolframalpha.com/input/?i=the+ultimate+meaning+of+life%2C+universe+and+everything

- terenzreignz

Well, if they say so :D

- experimentX

well ... cya around :)

- terenzreignz

Yeap :)
Thanks for... uhh.... stuff :D

- experimentX

yw!!

- terenzreignz

We kind of used this thread as a chat-medium.... sorry about that.
What do you need help with?

- 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

That's the subgroup test :D

- walters

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

- walters

to conclude that N is not enpty

- 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

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

- terenzreignz

Exact value of what?

- walters

identity element in G

- terenzreignz

I guess so. Let's let e(x) be the identity element in G, shall we?

- terenzreignz

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

- terenzreignz

Are you following me so far? :)

- 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

And it just goes to show that
\[\huge e(x)=x\]
So the identity element, is, after all, the identity function :D

- 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

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

And clearly, the identity map is in N, just take b=0
Convinced now that the identity is in N? :)

- 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

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

This was what you told me earlier, right?

- walters

am i going to use \[\tau _{cd}\] and \[\tau _{ef}\] because of my condition is of type
|dw:1362487393261:dw|

- terenzreignz

You can, but we're dealing with elements of N, which means the coefficient of x is 1.

- terenzreignz

Remember, again, the original definition of N
\[\huge N=\left\{ \tau_{1b} \ \ | \ \ b\in\mathbb{R} \right\}\]

- 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

Just write \[\huge \tau_{1c} \ \ \ and \ \ \ \tau_{1d}\]

- terenzreignz

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

- walters

yes i am still trying to work it aside

- terenzreignz

Then
\[\huge \alpha(x)=x+a\]\[\huge\beta(x)=x+b\]for some\[\huge a,b\in \mathbb{R} \]

- walters

so wat will be this |dw:1362488547392:dw|

- terenzreignz

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

- terenzreignz

By definition of the inverse...
\[\huge \beta\circ\beta^{-1}(x)=e(x)\]

- terenzreignz

So...
\[\huge \beta(\beta^{-1}(x))=\beta^{-1}(x)+b=x\]

- terenzreignz

Which can only mean
\[\huge \beta^{-1}(x)=x-b\]
So far so good?

- walters

what ?

- terenzreignz

We've found the value of the inverse of beta.

- terenzreignz

So, let's calculate
\[\huge \alpha\circ\beta^{-1}(x)=\alpha(\beta^{-1}(x))\]

- walters

but emenyana(wait) is
|dw:1362489338312:dw|

- terenzreignz

\[\huge =\beta^{-1}(x)+a\]\[\huge =x-b+a\]

- 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

oh i see

- 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

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

Ok :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.