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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?

yes it is

Then why do you have to prove it when G is already 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

|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.

i am not sure

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

yes,

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?

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|

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.

when is N, normalizer?

is not N for natural numbers?

sorry, is N normalizer?

yes it is

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

|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|

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

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

Linear Algebra comes before Abstract Algebra?

*since both they are algebra

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

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

I thought Complex Analysis comes after Real Analysis?

Now I'm scared :D

From mathematical perspective elementary calculus is a child play.

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?

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

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? :)

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

This was what you told me earlier, right?

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

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|

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

oh i see

Ok :)