Question

Group theory Q: G is the set of the real numbers with operation x*y=x+y+1. find an isomorphism f: R-> G and show that it is an isomorphism.

Answer 1

I'd say, first show that G is a group under that operation. Identify the identity element, the inverse of a general element x, and whether or not the operation is associative (which it clearly seems to be).

Answer 2

do I need to define a function for the isomorphism?

Answer 3

Yup yup, you're looking for a specific isomorphism, i.e. a function

Answer 4

that is where i got stuck...i'm not sure how to find that function

Answer 5

Question for you, what's the identity in G under that operation?

Answer 6

0

Answer 7

Ok, so what's the inverse of an element in G?

Answer 8

-1

Answer 9

-x-y-1?

Answer 10

said otherwise,

-(x+y+1).

So, what would be an operation where 0 is the identity, and the negative of an element is its inverse (like nothing too complicated)?

Answer 11

addition

Answer 12

Are you sure that 0 is the identity element of the group? \[x*0=x+0+1=x+1\neq x\]

Answer 13

no, i'm not sure. i am so confused.

Answer 14

for addition, the identity should be zero, right?

Answer 15

so x+y+1 + something = 0, right?

Answer 16

if the operation is regular addition, then yes, but you are given a new operation:\[x*y=x+y+1\]You aren't using addition anymore. You are using *.

Answer 17

So when we say identity, you need to find a number e such that:\[x*e=e*x=x\]for all x.

Answer 18

ok. i see...so e=-x-1?

Answer 19

but e can't be dependent on x

Answer 20

Lets see. We are going to try to solve the equation:\[x*e=x\]

Answer 21

by definition:\[x*e=x+e+1\]

Answer 22

so e=-1

Answer 23

So we obtain:\[x*e=x\Longrightarrow x+e+1=x\Longrightarrow e+1=0\]yeah you got it.

Answer 24

So -1 is the identity element. You can test it just to make sure:\[3*-1=3+(-1)+1=3\]

Answer 25

so, how do i find a function that defines this isomorphism?

Answer 26

That i'm still thinking about lol.

Answer 27

Because of the operation *, your group G has a strange structure.

Answer 28

yes, i'm not sure how to proceed

Answer 29

i appreciate your help, tho!