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

- anonymous

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- schrodinger

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- experimentX

what is Zm?

- Zarkon

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

- Zarkon

the set of integers mod \(m\)

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## More answers

- anonymous

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

- experimentX

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

- anonymous

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

- experimentX

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

- experimentX

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

- experimentX

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

- experimentX

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

- anonymous

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

- experimentX

huh!! what exactly is the question?

- anonymous

I wanna define it with examples!

- experimentX

aren't those two different distinct groups?

- anonymous

u mean the addition modulo and the other???

- experimentX

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

- anonymous

Thats it...what is confusing u??

- experimentX

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

- anonymous

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

- experimentX

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

- anonymous

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

- experimentX

but why do you want to give example?

- anonymous

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

- 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.

- anonymous

then is my answer wrong????

- experimentX

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

- anonymous

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

- anonymous

@experimentX

- experimentX

sorry ... i was busy!!

- anonymous

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

- experimentX

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

- anonymous

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

- anonymous

@experimentX

- experimentX

yes!! you need to show them!!

- anonymous

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

- experimentX

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

- experimentX

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

- terenzreignz

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

- 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

- anonymous

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

- terenzreignz

+ here is + modulo m, ok

- terenzreignz

Quite sure.

- 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 .

- anonymous

what abt inverse???

- 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).

- experimentX

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

- terenzreignz

@experimentX
A spoiler alert, next time? LOL JK

- experimentX

well .. sorry!!

- anonymous

why???

- 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)

- anonymous

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

- 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...

- 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 have inverses
:D

- terenzreignz

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

- anonymous

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

- terenzreignz

I have to go now, work on the other one:
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

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