Question

group theory. Am I correct in saying (R,o) with x o y: lx . y| has no identity element?

Answer 1

i was thinking because i can only think of 0 being the neutral identity but \( |x * 0| = |0 * x| = |x| \) if as it is over the reals, if x is -ve then -x does not equal |x|.

Answer 2

can you explain me why you do that?

Answer 3

well the neural element is an element, say \(e\) such that \(a*e = e*a = a \)

Answer 4

by definition of neural element, you don't have that form for (R,0) because R*0 =0 it's not =R ---> it doesn't have identity.

Answer 5

oops i meant 1 not 0, ie x*1

Answer 6

whatever, the problem is 0 , how can you get R when multiple R with 0?

Answer 7

there is no multiplying by 0, thats what i mean it should be multiplying by 1.

Answer 8

I understand what you mean. the condition to be neural identity is the first term * the second term = the first term, ok?

Answer 9

if your couple is (3,1) then, it is "neural identity"

Answer 10

no you have misunderstood

Answer 11

ok, explain me

Answer 12

there should be one element, such that any other element 'multiplied' by that element equals itself

Answer 13

TedG, after explaining, you master your problem and you can answer it by yourself. hahahaa...

Answer 14

now i say 'multiplied' because it in group theory it doesn't mean your normal multiplication. it is some defined binary operation.

Answer 15

well i do sort of know i just wasnt sure haha

Answer 16

I guess!!! your group has more than 2 members than just R and 0

Answer 17

TedG, since your question is in group theory, I strongly recommend you make question at other site.

Answer 18

well the set in question is (R, o) with x o y defined as |x . y|, where . is normal multiplication, and o is the defined binary operation.

Answer 19

want to have the link?

Answer 20

no its fine i know the site.

Answer 21

ok, good luck