Show (H,+) is isomorphic to (C,+)

[HaLAA]

Homework Statement

Let, M={ (a -b) (b a):a,b∈ℝ}, show (H,+) is isomorphic as a binary structure to (C,+)

Homework Equations

Isomorphism, Group Theory, Binary Operation

The Attempt at a Solution

Let a,b,c,d∈ℝ
Define f : M→ℂ by f( (a -b) (b a) ) = a+bi

1-1:
Suppose f( (a -b) (b a) )= f( (c -d) (d c)), then a+bi =c+di, thus a=c and b=d, thus f is one to one.

Onto:
Let a+bi∈ℂ , then (a -b) (b a)∈M, so f((a -b) (b a))=a+bi, thus f is onto.

Homomorphic:

f((a -b) (b a) + (c -d ) (d c)))
=f( ((a+c) -(b+d)) ((b+d) -(a+c)))
= (a+c)+(b+d)i
= f((a -b) (b a)) +f((c -d) (d c))

I don't think I show f: M→ℂ is 1-1 and onto correctly because my instructor didn't talk about them to much, can any show me how show f: M→ℂ is 1-1 and onto? thanks!

 

[fourier jr]

you define M but the title has (H,+) so I'm a bit confused. is the set M & then when you add the operation it becomes the group (H,+)? & when you say the elements of M are (a - b)*(b*a) do you mean that's ordinary multiplication & subtraction in ℝ? I wonder because I can't get that f is a homomorphism. I get
f( (a-b)(ba) ) + f( (c-d)(dc) ) = (a+bi) + (c+di) = (a+c) + (b+d)i = f( [(a+c) - (b+d)](b+d)(a+c) ) which I don't think equals f( (a-b)(ba) + (c-d)(dc) ) but maybe I'm missing something.

for 1-1 I think you missed a step. you got a+bi = c+di but before that I think you need to say (a-c) + (b-d)i = 0, then a-c=0 & b-d=0. an equivalent way is to start with a+bi ≠ c+di (etc) & conclude that f( (a-b)*(b*a) ) ≠ f( (c-d)*(d*c) )

for onto you got a+bi, & since a & b are real numbers, there is an element (a-b)*(b*a) in M such that f( (a-b)*(b*a) ) = a+bi. that is if I understood M correctly.

 

[HaLAA]

you define M but the title has (H,+) so I'm a bit confused. is the set M & then when you add the operation it becomes the group (H,+)? & when you say the elements of M are (a - b)*(b*a) do you mean that's ordinary multiplication & subtraction in ℝ? I wonder because I can't get that f is a homomorphism. I get
f( (a-b)(ba) ) + f( (c-d)(dc) ) = (a+bi) + (c+di) = (a+c) + (b+d)i = f( [(a+c) - (b+d)](b+d)(a+c) ) which I don't think equals f( (a-b)(ba) + (c-d)(dc) ) but maybe I'm missing something.

for 1-1 I think you missed a step. you got a+bi = c+di but before that I think you need to say (a-c) + (b-d)i = 0, then a-c=0 & b-d=0. an equivalent way is to start with a+bi ≠ c+di (etc) & conclude that f( (a-b)*(b*a) ) ≠ f( (c-d)*(d*c) )

for onto you got a+bi, & since a & b are real numbers, there is an element (a-b)*(b*a) in M such that f( (a-b)*(b*a) ) = a+bi. that is if I understood M correctly.

That H is a typo, and M is a 2 by 2 matrix that row one is a and -b, row 2 is b and a

 

[fourier jr]

wow I was really missing something... it makes a lot more sense now especially the homomorphism proof. i think what i said before still applies though if you replace everything with matrices.

 

[HaLAA]

wow I was really missing something... it makes a lot more sense now especially the homomorphism proof. i think what i said before still applies though if you replace everything with matrices.

Can you show me where I did indirect? I don't quite follow your word. Thanks

 

[fourier jr]

everything was ok I only misunderstood what the elements of M were. I would only add that if a+bi = c+di then (a-c) + (b-d)i = 0 then a-c=0 & b-d=0 to get that a=c & b=d, and also that if you're given a+bi in ℂ it is possible to find an element ## \Big( \begin{matrix} a & -b \\ b & a \end{matrix}\Big)## in M such that ## f\Big( \begin{matrix} a & -b \\ b & a \end{matrix}\Big) = a+bi##