Question

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

Answer 1

what is Zm?

Answer 2

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

Answer 3

the set of integers mod \(m\)

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

huh!! what exactly is the question?

Answer 13

I wanna define it with examples!

Answer 14

aren't those two different distinct groups?

Answer 15

u mean the addition modulo and the other???

Answer 16

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

Answer 17

Thats it...what is confusing u??

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

but why do you want to give example?

Answer 23

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

Answer 24

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.

Answer 25

then is my answer wrong????

Answer 26

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

Answer 27

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

Answer 28

@experimentX

Answer 29

sorry ... i was busy!!

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

@experimentX

Answer 34

yes!! you need to show them!!

Answer 35

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

Answer 36

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

Answer 37

instead of multiplying, we add ,,, since operation is +.

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

+ here is + modulo m, ok

Answer 42

Quite sure.

Answer 43

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, +>.

Answer 44

what abt inverse???

Answer 45

-.-
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).

Answer 46

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

Answer 47

@experimentX
A spoiler alert, next time? LOL JK

Answer 48

well .. sorry!!

Answer 49

why???

Answer 50

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)

Answer 51

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

Answer 52

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...

Answer 53

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

Answer 54

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)

Answer 55

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

Answer 56

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