# Anyone up to double check an isomorphism proof?

1. Sep 13, 2011

### Syrus

1. The problem statement, all variables and given/known data

*Attached is the problem statement, along with a definition which is to be used (I feel like some of the definitions in this text are somewhat untraditional, so I am including this one for clarity).

Edit* sorry, it's problem #31.

2. Relevant equations

3. The attempt at a solution

Let <T, !> be any binary structure isomorphic with <S,*>. This means there is some bijection Ψ: S → T such that (∀s1,s2 ∈ S)(Ψ(s1*s2) = Ψ(s1)!Ψ(s2)). Now let d ∈ T. Since Ψ is surjective, there is some c ∈ S such that Ψ(c) = d. But then there is some x ∈ S such that x*x = c. As a result, Ψ(x*x) = Ψ(c) = d = Ψ(x)!Ψ(x).

Q.E.D.

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2. Sep 13, 2011

### micromass

Staff Emeritus
This is ok, Syrus!!!!

3. Sep 13, 2011

### Syrus

Thank you, micromass =)

4. Sep 19, 2011

### Syrus

I was recently assigned more problems similar to that above. I figured I may as well post these as well since they correlate with the original post and I have time until they are due.

The proofs for exercises 32 and 33 are below. The original questions and other related materials are attached as thumbnails.

32: Let <T, !> be any binary structure isomorphic with <S,*>. This means there is some bijection Ψ: S → T such that (∀s1,s2 ∈ S)(Ψ(s1*s2) = Ψ(s1)!Ψ(s2)). Take b to be the identity element of S under *. Clearly, then, b*b = b. As a result, Ψ(b*b) = Ψ(b)!Ψ(b). By lemma 3.14 (attached above), however, Ψ(b) is the identity element of T under !. Hence, Ψ(b)!Ψ(b) = Ψ(b).

Q.E.D.

*from here on, C denotes the set of complex numbers, and R the set of reals.. the funky brackets are my best attempt at 2 X 2 matrices (i.e. in [a b] [a b] the a's should be in a line vertically).

33a: Let Ψ: C → H defined as follows: for any complex number, say x, of the form a + bi, where a,b ∈ R and i2 = -1, Ψ(x) = [a -b]
[b a].
Let j ∈ H. So j is a matrix of the form above ^, for some a,b ∈ R. But then we have k = a +bi ∈ C, and so Ψ(k) = [a -b]
[b a]. Thus, Ψ is surjective.
Now let m,n ∈ C. Then m = c +di and n = f + gi for some c,d,f,g ∈ R. Now suppose Ψ(m) = Ψ(n). Then equating matrices shows that c = f and d = g, and so m = c +di = f + gi = n. Hence, Ψ is injective.
Lastly, let j,k ∈ C. Then j = a + bi for some real nubmers a and b. Also, k = c + di for some real numbers c and d. Thus, Ψ(j + k) = Ψ((a + bi) + (c + di)) = Ψ((a + c) + (b+d)i) = Ψ(j) + Ψ(k). Thus, Ψ satisfies the homomorphism property.

33b: The proofs for injectivity and surjectivity are identical to those in part a. The proof of the homomorphism property is as follows: Let Ψ be defined as above. Let x,y ∈ C. Then x = j + ki and y = m + ni for some j,k,n,m in R. Thus, Ψ(x*y) = Ψ((j + ki) * (m +ni)) = Ψ(jm + jni + kmi - kn) = Ψ((jm - kn) + (jn + km)i) = Ψ(j) * Ψ(k).

Q.E.D.

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5. Sep 19, 2011

### micromass

Staff Emeritus
Shouldn't that be: as a result $\psi(b)=\psi(b*b)=\psi(b) ! \psi(b)$. And doesn't the proof end here??

I don't see what you're doing here. And I don't see why you use lemma 3.14. Nobody said that (S,*) has an identity. So lemma 3.14 isn't even applicable!!

This looks quite ok!!

6. Sep 19, 2011

### Syrus

Yes, youre correct micromass. It was my slip up, I can use the equivalence between (b*b) and b, along with the definition of isomorphism to show that psi(b) = psi(b)!psi(b).