Is <Z,+> with <Z,+> a valid isomorphism? Why or why not?

[SqrachMasda]

1-3) determine whether the given map (theta) is an isomorphism of the first binary structure with the second. If it is not an isomorphism, why not?

1) <Z,+> with <Z,+> where theta(n)= -n for n elements in Z

2) <Z,+> with <Z,+> where theta(n)= 2n for n elements in Z

3) <Z,+> with <Z,+> where theta(n)= n+1 for n elements in Z



4) Let alpha: <Z4,+> -> <Z#5,X> by alpha(0)=1, alpha(1)=2, alpha(2)=4, alpha(3)=3. Prove alphais an isomorphism of groups. (verify 16 equations).


5) Let gamma:<C*,X> -> <R+,X> by gamma(a+bi)= a^2 + b^2. Prove gamma is a homomorphism.

6) Let theta: <Z,+> -> <{+/-1,+/-1},X> by theta(k)=i^k. Prove theta is homomorpism.

7) Let G be a group. Prove that alpha:G->G by alpha(a)=a^2 is a homomorphism if and only if G is abelian.

 

[nnnnnnnn]

1-3) I don't know what "first binary structure with the second" means, but the first two are isomorphisms. The third isnt: try theta(a+b)=theta(a)+theta(b)...

 

[nnnnnnnn]

5) gamma(1) = 1

gamma( (a+bi)*(c+di) ) = ... = gamma(a+bi)*gamma(c+di)
I didn't work this out so I'm not sure if it works...

(a+bi)' = 1/(a+bi) = a/(a^2+b^2) - bi/(a^2+b^2)
gamma( a/(a^2+b^2) - bi/(a^2+b^2) ) = ... = 1/(a^2 + b^2)
Again, I didn't work this out so I'm not sure if it works...

Also I'm not sure what C* means so I'm not sure what affect this has on the problem...

 

[SqrachMasda]

i think by binary structure it's just the structure <Z,+> where + is the binary operation, mapping it to itself by the function f(n)=-n , thanks again. I can see the answers but i am not good with proofs, for these three

the 2nd one is not, the ans in the book reads
No, because theta does not map Z onto Z'. theta(n) is not equal to 1 for all n in Z.

 

[SqrachMasda]

thanks again, it's very helpful,...i'll have 4 more in an hour
thanks alot, appreciate it.

 

[nnnnnnnn]

the 2nd one is not, the ans in the book reads
No, because theta does not map Z onto Z'. theta(n) is not equal to 1 for all n in Z.

I'm not sure what that means, but:

f(x) = 2x

injective: f(x) = y & f(x') = y -> x = x' -> 2x = 2x' ->...-> x=x'
surjective: you know that if y is an even integer then it is equal to 2x for some x, where x is an integer...

the last part is showing that f(x+y) = f(x) + f(y), pretty straight forward...

 

[SqrachMasda]

ok thanks again

 

[Dogtanian]

The second on is not an iso. since it is not surjective.
Take, for example, 1 in Z. If (2) was surjective, then we could find an n in Z such that 2n = 1. In otherwords we need n=0.5 which is not in Z. So not surjective and so not an iso.

(7) is quite straight forward.
Let a, b be in G. Then alpha(a)alpha(b) = (a^2)(b^2) and alpha(ab) = (ab)^2 =abab. Clearly this only equals (a^2)(b^2) (and hence meaning alpha is a homomorphism) if you can "swap" the order of multiplication of the elements around, ie, if the group is abelain.

 

[nnnnnnnn]

The second on is not an iso. since it is not surjective.
Take, for example, 1 in Z. If (2) was surjective, then we could find an n in Z such that 2n = 1. In otherwords we need n=0.5 which is not in Z. So not surjective and so not an iso.

f:A->B
f(x)=2x

surjective: for every y of B, there exists an x of A such that f(x)=y

Your example doesn't fit this criteria as 1 is not in B. However, you know that if y is an even integer then it is equal to 2x for some x, where x is an integer...

I may be missing something or misreading the notation, but under normal circumstances, f(x)=2x is an isomorphism for integers under addition.

 

[nnnnnnnn]

of the first binary structure with the second.

I don't know what this means, but I get the impression that it is not asking for normal isomorphisms, so what I previously posted may not apply...

 

[Dogtanian]

f:A->B
f(x)=2x

surjective: for every y of B, there exists an x of A such that f(x)=y

Your example doesn't fit this criteria as 1 is not in B. However, you know that if y is an even integer then it is equal to 2x for some x, where x is an integer...

I may be missing something or misreading the notation, but under normal circumstances, f(x)=2x is an isomorphism for integers under addition.

Unless I'm the one to have missed something, then you are the one that is wrong. In (2) we a function f: Z ---> Z where f(n) = 2n. For this map to be an isomorphism from Z to Z we need it to be surjective. That is fro any integer y in Z we have an x in Z such that f(x) = y. Since f maps only to the even integers (i.e. the image of f is the even integers), there is no such x that maps to the odd integers. As such the map is not surjective.

If I'm right you are making the mistake of trying to show the map is surjective onto the image of the map, which clearly is the case, but this is not what the questions wants. Instead you need to show f maps ontot he whole of Z.

 

[nnnnnnnn]

If I'm right you are making the mistake of trying to show the map is surjective onto the image of the map, which clearly is the case, but this is not what the questions wants. Instead you need to show f maps ontot he whole of Z.

Yes, that's what I was doing... I see my mistake now, I was trying to prove it for Z->2Z.