## Zaara 2 years ago Group Theory: any one pls guide me with a link to find the definition of these......... 1.(Zm,+) 2.(Zm,*)

1. experimentX

what is Zm?

2. Zarkon

they prob mean $$Z_m=\{0,1,2,\ldots,m-1\}$$

3. Zarkon

the set of integers mod $$m$$

4. Zaara

Yeah!!!! i need complete sentences to explain it... do u have any stuff???

5. experimentX

show that it satisfies the property to be called a group.

6. Zaara

Exactly!!! can i get that proof or its details... i mean any link???

7. experimentX

let 'a' and 'b' be two elements of Zm, show that it is closed in +, ie, $$(a+b) \mod b \in Z_m$$

8. experimentX

addition operation is associative, and find the identity element of Zm

9. experimentX

Edit:: $(a+b) \mod m \in Z_m$

10. experimentX

this is always true becuse $$(a+b) \mod m = c < m, c \in Z\m$$

11. Zaara

What i have written below is correct right?? or should there be any change????? Definitions: Let Zm be the set of non negative integers less than m: {0,1, ., m−1} The operation +m is defined as a +m b = (a + b) mod m. This is addition modulo m. The operation ∙m is defined as a ∙m b = (a + b) mod m. This is multiplication modulo m. Using these operations is said to be doing arithmetic modulo m. Example: Find 7 +11 9 and 7 ·11 9. Solution: Using the definitions above: – 7 +11 9 = (7 + 9) mod11 = 16 mod11 = 5 – 7 ·11 9 = (7 ∙ 9) mod11 = 63 mod 11 = 8

12. experimentX

huh!! what exactly is the question?

13. Zaara

I wanna define it with examples!

14. experimentX

aren't those two different distinct groups?

15. Zaara

u mean the addition modulo and the other???

16. experimentX

yep!! just this one .. (Zm,+), this is a group.

17. Zaara

Thats it...what is confusing u??

18. experimentX

you are mixing two things ... doing both of them at same time.

19. Zaara

where did i mix them?? i have defined Zm,then defined addition and multiplication modulo and have given example 4 each...

20. experimentX

let's work out one at each ... we do with (Zm,+) first.

21. Zaara

ohhh... u mean i have not seperately done it..ryt??? ok got it.... i have jst given it together for my ease....

22. experimentX

but why do you want to give example?

23. Zaara

definition with example is most expected... so that one can easily understand it....

24. experimentX

choose m=10, or 5 for sake of ease. then {0,1,2,3,4} are the elements of group!! show that it remains closed, and is associative and has identity '0', and has inverse.

25. Zaara

26. experimentX

no i don't think so ... you put stuff in one place, and it hurt my eyes.

27. Zaara

is it ok now??? Let Zm be the set of nonnegative integers less than m: {0,1, ., m−1} 1)The operation +m is defined as a +m b = (a + b) mod m. This is addition modulo m.(Zm,+) Example: Find 7 +11 9 Solution: Using the definitions above: 7 +11 9 = (7 + 9) mod11 = 16 mod11 = 5 2)The operation ∙m is defined as a ∙m b = (a + b) mod m. This is multiplication modulo m. Example: Find 7 ·11 9. Solution: Using the definitions above: 7 ·11 9 = (7 ∙ 9) mod11 = 63 mod 11 = 8

28. Zaara

@experimentX

29. experimentX

sorry ... i was busy!!

30. Zaara

if possible make some alterations and input the needed stuffs... and edit my answer...pls:)

31. experimentX

okay .. you showed that it is closed. you need to test for these http://en.wikipedia.org/wiki/Group_(mathematics)#Definition

32. Zaara

shud i show associativity,identity element and inverse element????

33. Zaara

@experimentX

34. experimentX

yes!! you need to show them!!

35. Zaara

bt in the link u gave... they have shown for multiplication modulo only.... what can i do for addition modulo???????

36. experimentX

same ... show that (a+b) mod m is in the set!!.

37. experimentX

38. terenzreignz

Never seen a Group Theory question before... this should be fun... :D

39. terenzreignz

Once you've shown closure, next step is to show that there is an element e of Zm (a non-negative integer less than m) such that for any k in Zm e + k = k + e = k

40. Zaara

similar to that??? r u sure???

41. terenzreignz

+ here is + modulo m, ok

42. terenzreignz

Quite sure.

43. terenzreignz

Okay, I'll do this (because it's easier :P) and you do the "existence of inverse" part. Let k be in Zm Consider 0. 0 is a non-negative integer. Since m is positive, 0 < m So 0 is in Zm 0 + k = 0 + k (mod m) = k(mod m) = k, since k is in Zm (and is therefore a nonnegative integer less than m). Similarly, k + 0 = k + 0 (mod m) = k(mod m) = k. Thus, we have shown that 0 + k = k + 0 = k Hence, 0 is the identity element in <Zm, +>.

44. Zaara

what abt inverse???

45. terenzreignz

-.- Here's the gist of it. Consider an arbitrary element of Zm, and show that there is something that you can add (modulo m) to it so that the sum is 0 (the additive identity).

46. experimentX

show that for a, m-a is the it's inverse, and m-a is also in the set.

47. terenzreignz

@experimentX A spoiler alert, next time? LOL JK

48. experimentX

well .. sorry!!

49. Zaara

why???

50. terenzreignz

Never mind, lol... Anyway, let a be in Zm. Your task is to show that an element m-a exists, and m-a is in Zm, and a + (m - a) = 0, where "+" here is + (mod m)

51. Zaara

why m-a??? i dnt get it...

52. terenzreignz

Well, remember that in the world of Zm, every multiple of m is basically zero. So, when I said "what do you need to add to a so that you get zero", I really mean what do you need to add to a to get m...

53. terenzreignz

Okay, first of all, the inverse of the additive identity is itself... 0 + 0 = 0, so, yeah... Now, let a be in Zm, where a is not equal to zero. Consider m-a. since a is in Zm, a < m a - a < m - a m - a > 0, meaning m-a is non-negative, (positive, even) m < m + a m - a < m + a - a m - a < m, meaning m-a is less than m. Thus, m-a is non-negative and is less than m, therefore m-a is in Zm. Consider a + (m-a) = a + m - a = m = 0(mod m) Thus, a + (m-a) = 0(mod m) a + (m-a) = 0, in Zm (m-a) is the inverse of a. Thus we have shown that for any element a in Zm, there is an element (m-a) such that a + (m-a) = 0, and therefore, all elements in <Zm, +> have inverses :D

54. terenzreignz

And there you go, we've shown closure, associativity, existence of an identity element and existence of inverses, we can conclude that <Zm, +(mod m)> forms a group. (An Abelian group, too, but more on some other time, I guess)

55. Zaara

great!!!!!!!!!!!!!!

56. terenzreignz

I have to go now, work on the other one: <Zm, *(mod m)> But since we're having spoilers anyway, it's not a group :P The reason for that is for you to find... *cough*oneisthemultiplicativeidentity*cough*theresnothingyoucanmultiplytozerotogetone *cough*zerohasnomultiplicativeinverse *cough* LOL anyway Terence out :D