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what is Zm?

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

the set of integers mod \(m\)

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

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

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

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

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

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

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

huh!! what exactly is the question?

I wanna define it with examples!

aren't those two different distinct groups?

u mean the addition modulo and the other???

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

Thats it...what is confusing u??

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

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

but why do you want to give example?

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

then is my answer wrong????

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

sorry ... i was busy!!

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

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

yes!! you need to show them!!

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

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

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

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

+ here is + modulo m, ok

Quite sure.

what abt inverse???

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

@experimentX
A spoiler alert, next time? LOL JK

well .. sorry!!

why???

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

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