## walters Group Title Let G be a group such that (a*b)^2=a^2*b^2 can pls show the associativity one year ago one year ago

1. walters Group Title

|dw:1362323673416:dw|

2. walters Group Title

i am not sure

3. terenzreignz Group Title

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

4. walters Group Title

yes it is

5. terenzreignz Group Title

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

6. walters Group Title

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

7. terenzreignz Group Title

:) 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?

8. walters Group Title

yes

9. terenzreignz Group Title

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

10. walters Group Title

yes ok now i understand the question

11. walters Group Title

wait..

12. terenzreignz Group Title

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

13. walters Group Title

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

14. walters Group Title

|dw:1362324922660:dw|

15. terenzreignz Group Title

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

16. terenzreignz Group Title

Okay, so, you know the subgroup test?

17. terenzreignz Group Title

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.

18. walters Group Title

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

19. terenzreignz Group Title

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

20. walters Group Title

i am not sure

21. terenzreignz Group Title

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) $\large N = \left\{\tau|\tau \in G, \tau(x)=x+b, \ \ b\in \mathbb{R} \right\}$

22. terenzreignz Group Title

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

23. walters Group Title

yes,

24. terenzreignz Group Title

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?

25. walters Group Title

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

26. terenzreignz Group Title

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

27. terenzreignz Group Title

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?

28. walters Group Title

yes

29. terenzreignz Group Title

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

30. walters Group Title

yes

31. terenzreignz Group Title

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

32. walters Group Title

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

33. terenzreignz Group Title

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

34. walters Group Title

ok

35. terenzreignz Group Title

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

36. terenzreignz Group Title

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

37. walters Group Title

yes i see it

38. walters Group Title

@experimentX

39. walters Group Title

for that one of subgroup

40. experimentX Group Title

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

41. walters Group Title

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

42. experimentX Group Title

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

43. walters Group Title

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

44. experimentX Group Title

when is N, normalizer?

45. walters Group Title

is not N for natural numbers?

46. experimentX Group Title

sorry, is N normalizer?

47. walters Group Title

yes it is

48. experimentX Group Title

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

49. walters Group Title

there are 2 different questions (given that a, b in G, then (a*b)^2=a^2*b^2) and that onee of subgroup

50. walters Group Title

let me retype my questions

51. walters Group Title

1 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

52. walters Group Title

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 53. experimentX Group Title |dw:1362355116918:dw| use associativity. 54. walters Group Title is already a group i only have to show that is abelian so to show comutativity only 55. experimentX Group Title for any two element, the above picture shows commutativity. 56. walters Group Title |dw:1362355288828:dw| 57. walters Group Title 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 58. experimentX Group Title yes it's not multiplication ... but associativity and inverse are properties of group, we are only allowed to exploit that. 59. walters Group Title SO IS IT GOING TO BE|dw:1362355812287:dw| 60. walters Group Title SO IF I CAN PUT THE SQUARE ROOT both side Lhs is it going to be equal to RHS 61. experimentX Group Title 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.

62. experimentX Group Title

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.

63. terenzreignz Group Title

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?

64. experimentX Group Title

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.

65. terenzreignz Group Title

LOL I thought abstract algebra was just invented when Mathematicians got bored from Calculus XD But yeah... Abstract Algebra = My Favourite branch of Maths :)

66. experimentX Group Title

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.

67. experimentX Group Title

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

68. terenzreignz Group Title

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

69. experimentX Group Title

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

70. terenzreignz Group Title

Linear Algebra comes before Abstract Algebra?

71. experimentX Group Title

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

72. terenzreignz Group Title

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...

73. experimentX Group Title

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.

74. experimentX Group Title

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.

75. experimentX Group Title

*since both they are algebra

76. terenzreignz Group Title

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

77. experimentX Group Title

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.

78. terenzreignz Group Title

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

79. experimentX Group Title

yeah ... but somehow i enjoy and work more on calculus. Still i would like to learn abstract algebra and Real Analysis (in dept).

80. terenzreignz Group Title

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

81. experimentX Group Title

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

82. terenzreignz Group Title

I thought Complex Analysis comes after Real Analysis?

83. experimentX Group Title

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.

84. experimentX Group Title

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.

85. terenzreignz Group Title

Now I'm scared :D

86. experimentX Group Title

From mathematical perspective elementary calculus is a child play.

87. experimentX Group Title

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

88. experimentX Group Title

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

89. terenzreignz Group Title

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

90. experimentX Group Title

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

91. experimentX Group Title
92. terenzreignz Group Title

Well, if they say so :D

93. experimentX Group Title

well ... cya around :)

94. terenzreignz Group Title

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

95. experimentX Group Title

yw!!

96. terenzreignz Group Title

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

97. walters Group Title

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

98. terenzreignz Group Title

That's the subgroup test :D

99. walters Group Title

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

100. walters Group Title

to conclude that N is not enpty

101. terenzreignz Group Title

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?

102. walters Group Title

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

103. terenzreignz Group Title

Exact value of what?

104. walters Group Title

identity element in G

105. terenzreignz Group Title

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

106. terenzreignz Group Title

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

107. terenzreignz Group Title

Are you following me so far? :)

108. terenzreignz Group Title

Now, tau always takes the form $\large \tau(x)=ax+b \ \ \ \ a,b\in\mathbb{R}$ Following the first equation, we have $\huge \tau(e(x))=a\cdot e(x)+b=ax+b=\tau(x)$

109. terenzreignz Group Title

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

110. walters Group Title

i've let $\tau _{pg}$ element of G be identity with respect to 0 opereration yes i also got x on my rough work

111. terenzreignz Group Title

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' $\huge N=\left\{\tau(x) \ | \ \tau(x)=x+b , \ \ b\in\mathbb{R}\right\}$

112. terenzreignz Group Title

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

113. walters Group Title

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

114. terenzreignz Group Title

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$

115. terenzreignz Group Title

This was what you told me earlier, right?

116. walters Group Title

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

117. terenzreignz Group Title

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

118. terenzreignz Group Title

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

119. walters Group Title

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)?

120. terenzreignz Group Title

Just write $\huge \tau_{1c} \ \ \ and \ \ \ \tau_{1d}$

121. terenzreignz Group Title

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

122. walters Group Title

yes i am still trying to work it aside

123. terenzreignz Group Title

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

124. walters Group Title

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

125. terenzreignz Group Title

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

126. terenzreignz Group Title

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

127. terenzreignz Group Title

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

128. terenzreignz Group Title

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

129. walters Group Title

what ?

130. terenzreignz Group Title

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

131. terenzreignz Group Title

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

132. walters Group Title

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

133. terenzreignz Group Title

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

134. terenzreignz Group Title

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

135. walters Group Title

oh i see

136. terenzreignz Group Title

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

137. walters Group Title

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

138. terenzreignz Group Title

Ok :)