What familiar group is isomorphic to the group of units in ℤ[i]?

[catherinenanc]

find a group isomorphic to ℤ

1. Knowing the below proof, The group of units of ℤ is isomorphic to a familiar group. Which one?



2. We have already shown: "Let R be a ring with unity, and let U denote the set of units in R. Show that U is a group under the multiplication in R."
Also, ℤ has been defined as the set of all complex numbers of the form a+bi, where a,b∈ℤ.



3. Does it have something to do do with the homomorphism Norm(a+bi)=a^2+b^2? I don't think so, because Norm is not an isomorphism, right?

 

[Dick]

Well, what are the group elements? How many are there? You can certainly use your Norm to say something about that. Can you find a generator?

 

[catherinenanc]

The things we have already proven are:
1) (Problem #16.23, as referred to later) Let R be a ring with unity, and let U denote the set of units in R. Show that U is a group under the multiplication in R.
2) Let ℤ denote the set of all complex numbers of the form a+bi, where a,b∈ℤ. Show that ℤ is a commutative ring with unity under ordinary addition and multiplication of complex numbers. ℤ is called the ring of Gaussian integers.
3) For r=a+bi∈ℤ, define the norm N(r) of r by N(r)=a²+b². Show that if r,s∈ℤ, then N(rs)=N(r)N(s).
4) Show that r=a+bi is a unit in ℤ iff N(r)=1. Using this information, find all the units in ℤ.
***5) (See Exercise 16.23.) The group of units of ℤ is isomorphic to a familiar group. Which one?

#5 is the only one I have left that I have not proven, and I am assuming that #1-4 are supposed to lead me to a conclusion in #5. So, the answer to your questions are: The group elements are all a+bi such that a,b∈ℤ and N(a+bi)=1 (which is what tells you a+bi is a unit). It seems to me like there are only two elements then, 0+1i and 1+0i. Is that right? I do not have any idea what a generator might be.

 

[Dick]

The things we have already proven are:
1) (Problem #16.23, as referred to later) Let R be a ring with unity, and let U denote the set of units in R. Show that U is a group under the multiplication in R.
2) Let ℤ denote the set of all complex numbers of the form a+bi, where a,b∈ℤ. Show that ℤ is a commutative ring with unity under ordinary addition and multiplication of complex numbers. ℤ is called the ring of Gaussian integers.
3) For r=a+bi∈ℤ, define the norm N(r) of r by N(r)=a²+b². Show that if r,s∈ℤ, then N(rs)=N(r)N(s).
4) Show that r=a+bi is a unit in ℤ iff N(r)=1. Using this information, find all the units in ℤ.
***5) (See Exercise 16.23.) The group of units of ℤ is isomorphic to a familiar group. Which one?

#5 is the only one I have left that I have not proven, and I am assuming that #1-4 are supposed to lead me to a conclusion in #5. So, the answer to your questions are: The group elements are all a+bi such that a,b∈ℤ and N(a+bi)=1 (which is what tells you a+bi is a unit). It seems to me like there are only two elements then, 0+1i and 1+0i. Is that right? I do not have any idea what a generator might be.

Well, no. There are four elements in Z that have Norm equal to 1. 1, -1, i, and -i.

 

[catherinenanc]

Good point. So can I create an isomorphism?? that is defined not with a function, but a table like the following:
1 --> (1,0)
-1 --> (0,0)
-i --> (0,0)
i --> (0,1)

Then I get f:U->ℤ2xℤ2.
They both have order 4, and it's one-to-one and onto...

 

[Dick]

Good point. So can I create an isomorphism?? that is defined not with a function, but a table like the following:
1 --> (1,0)
-1 --> (0,0)
-i --> (0,0)
i --> (0,1)

Then I get f:U->ℤ2xℤ2.
They both have order 4, and it's one-to-one and onto...

It's one-to-on and onto but it's NOT a homomorphism. f(i)f(-i) should be f(1) and it's not. Start writing down powers of i.

 

[catherinenanc]

duh..powers! That makes sense. What about ϕ:(ℤ₄,+)→(U,*) such that ϕ(z)=(0+1i)^{z} which is one-to-one and onto, and is a homomorphism, right?Thanks... I feel like it should have been obvious. )))

 

[Dick]

duh..powers! That makes sense. What about ϕ:(ℤ₄,+)→(U,*) such that ϕ(z)=(0+1i)^{z} which is one-to-one and onto, and is a homomorphism, right?Thanks... I feel like it should have been obvious. )))

Sure. When you are trying to figure out what the structure of a group is it's always a good idea to look for generators.

 

[catherinenanc]

good point. I will remember that. Thanks!