Prove Isomorphic Groups: (\mathbb Z_4,_{+4}) and (\langle i\rangle, \cdot)

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gruba
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Homework Statement


Show that the group [itex](\mathbb Z_4,_{+4})[/itex] is isomorphic to [itex](\langle i\rangle,\cdot)[/itex]?

Homework Equations


-Group isomorphism

The Attempt at a Solution



Let [itex]\mathbb Z_4=\{0,1,2,3\}[/itex].
[itex](\mathbb Z_4,_{+4})[/itex] can be represented using Cayley's table:
[tex] \begin{array}{c|lcr}<br /> {_{+4}} & 0 & 1 & 2 & 3 \\<br /> \hline<br /> 0 & 0 & 1 & 2 & 3 \\<br /> 1 & 1 & 2 & 3 & 0 \\<br /> 2 & 2 & 3 & 0 & 1 \\<br /> 3 & 3 & 0 & 1 & 2 \\<br /> \end{array}[/tex]

What is the set [itex]\langle i\rangle[/itex]?
How to define [itex](\langle i\rangle,\cdot)[/itex]?
 
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micromass said:
The subgroup of ##(\mathbb{C},\cdot)## generated by ##i##.
What should be the order of that subgroup, and how to represent it using Cayley's table?
 
gruba said:

Homework Statement


Show that the group [itex](\mathbb Z_4,_{+4})[/itex] is isomorphic to [itex](\langle i\rangle,\cdot)[/itex]?
Should this be ##(\mathbb{Z_4}, +)##?
A group is defined by a set of elements of the group, together with an operation.

gruba said:

Homework Equations


-Group isomorphism

The Attempt at a Solution



Let [itex]\mathbb Z_4=\{0,1,2,3\}[/itex].
[itex](\mathbb Z_4,_{+4})[/itex] can be represented using Cayley's table:
[tex] \begin{array}{c|lcr}<br /> {_{+4}} & 0 & 1 & 2 & 3 \\<br /> \hline<br /> 0 & 0 & 1 & 2 & 3 \\<br /> 1 & 1 & 2 & 3 & 0 \\<br /> 2 & 2 & 3 & 0 & 1 \\<br /> 3 & 3 & 0 & 1 & 2 \\<br /> \end{array}[/tex]

What is the set [itex]\langle i\rangle[/itex]?
How to define [itex](\langle i\rangle,\cdot)[/itex]?
 
Could someone explain this problem (using Cayley's tables - easier)? How to form Cayley's table for the group [itex](\langle i\rangle,\cdot)[/itex]?

One method to show the groups are isomorphic is to create Cayley's tables and compare them (that is only useful for small groups).
I don't understand the method which requires finding the function (isomorphism) between these groups
 
gruba said:
Could someone explain this problem (using Cayley's tables - easier)? How to form Cayley's table for the group [itex](\langle i\rangle,\cdot)[/itex]?
Why don't you try what Orodruin suggested -- find i2, i3, and so on. This is not a hard problem.
gruba said:
One method to show the groups are isomorphic is to create Cayley's tables and compare them (that is only useful for small groups).
I don't understand the method which requires finding the function (isomorphism) between these groups
 
Mark44 said:
Why don't you try what Orodruin suggested -- find i2, i3, and so on. This is not a hard problem.

Let [itex]f:\mathbb Z_4\rightarrow \langle i\rangle=\{i^0,i^1,i^2,i^3\}=\{1,i,-1,-i\}[/itex] where [itex]f[/itex] is an isomorphism.
From here, how to explicitly define a function [itex]f[/itex]?
 
Orodruin said:
What do you think? There are only four possibilities of defining a homomorphism (it is fully defined by specifying how f acts on the group generator). Two of them give isomorphisms!
[itex]f(x)=e^x[/itex] is one isomorphism.
 
Orodruin said:
Not between the given groups.
[itex]f(x)=e^{2\pi x i}[/itex]?
 
micromass said:
If I define ##f(0) = 1## and if I say ##f## is a homomorphism, can you figure out ##f(1)##, ##f(2)## and ##f(3)##? That is, can you describe ##f## completely??
[itex]f(0)=1,f(1)=i,f(2)=-1,f(3)=-i[/itex].

Using Lagrange interpolation polynomial on points [itex](0,1),(1,i),(2,-1),(3,-i)[/itex] gives
[itex]f(x)=-\frac{(x-1)(x-2)(x-3)}{6}+i\frac{x(x-2)(x-3)}{2}+\frac{x(x-1)(x-3)}{2}-i\frac{x(x-1)(x-2)}{6}[/itex].

But [itex]f(x)[/itex] is not one to one.

What is the actual method for describing an isomorphism, without taking a guess?
 
Last edited:
@gruba I'm guessing that the lack of further replies is caused by your last reply. It appears to me that you need more help than can be provided under the rules of this forum. My suggestion to is that you need to schedule a personal meeting with your teacher to clear up your misunderstandings on this topic.