Question

Prove that O(G) mod O(H) =0 ???
Well i'm guessing that i have somehow to use the lagrange theorem to prove the above but am not too sure how to apply it any help will be greatly appreciated thanks :).

Answer 1

@.Sam. @DLS @Eleven17 @jdoe0001 @Jhannybean @jhonyy9 @nubeer @phi

Answer 2

um, what is G and H?

Answer 3

are these sets or groups?
is this number theory, set theory, or group theory?

Answer 4

@Stephane1200

Answer 5

Sorry for the delayed reply, well G would be the group and H a subgroup.

Answer 6

ok so G is a group C is a subgroup
then ord(C) | ord(G)

Answer 7

thus ord(G) modulo ord(C) = 0
i.e there is no remainder when we divide ord(G) with ord(C)

Answer 8

and yes this is lagrange theorem.

Answer 9

we are assuming G is finite....

Answer 10

@Stephane1200 understand?

Answer 11

Ok i think i get what you mean so basically i can use euclid principle to prove the above...

Answer 12

hmm, Lagrange theorem says if G has finite order and H belongs to G, then |h| | |g|
and since a | b b mod a = 0

Answer 13

I forget what Euclid principle is...sec

Answer 14

oh sorry my bad

Answer 15

is that not dealing with geometry?

Answer 16

you are right from definition

Answer 17

ok :)

Answer 18

if H is a subgroup of G then order of H should divide that of G

Answer 19

will divide*

Answer 20

lol yh will divide :P

Answer 21

hence if it divides there's no remainder the reason why it is zero ?

Answer 22

yeah 8mod2=0
8mod3=2 7mod6=1
it gives the remainder

Answer 23

9mod18=9

Answer 24

hmmm with examples it got easier now :)

Answer 25

have you done number theory?

Answer 26

well we covered it a bit

Answer 27

thx dude now i figured out i will have to use the lagrange theorem maybe not the full proof but at least the definition to demonstrate the claim that O(G) mod O(H) =0

Answer 28

yeah, I doupt you need to learn the proof, but its easy. Just know the theorem.

Answer 29

p.s. definitions are man made, theorems are just true(if proved).

Answer 30

lol i already learned the proof and it's quite long xD

Answer 31

its easy with cosets:)

Answer 32

yh cosets are easy :)

Answer 33

looks like you already did abstract algebra :p

Answer 34

im in it now:)

Answer 35

but done with number theory, and your question was more number theory.

Answer 36

oh ok i see well i was rounding up the bits and pieces i don't really know given i have exams in that on monday :P

Answer 37

have you already done ring and field