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

In summary, We are asked to show that the group (\mathbb Z_4,_{+4}) is isomorphic to (\langle i\rangle,\cdot). To do this, we can use Cayley's tables to represent both groups. We can then define the group \langle i\rangle as the subgroup of (\mathbb{C},\cdot) generated by i and find its order. To show the groups are isomorphic, we can try to find a function (isomorphism) between them. One method is to define a homomorphism, which is fully defined by specifying how it acts on the group generator. There are only four possibilities, and two of them give isomorphisms. We can use Lagrange
  • #1
gruba
206
1

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}
{_{+4}} & 0 & 1 & 2 & 3 \\
\hline
0 & 0 & 1 & 2 & 3 \\
1 & 1 & 2 & 3 & 0 \\
2 & 2 & 3 & 0 & 1 \\
3 & 3 & 0 & 1 & 2 \\
\end{array}
[/tex]

What is the set [itex]\langle i\rangle[/itex]?
How to define [itex](\langle i\rangle,\cdot)[/itex]?
 
Physics news on Phys.org
  • #2
The subgroup of ##(\mathbb{C},\cdot)## generated by ##i##.
 
  • #3
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?
 
  • #4
gruba said:
What should be the order of that subgroup, and how to represent it using Cayley's table?
Why don't you try figuring it out? What is ##i^2##?
 
  • #5
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}
{_{+4}} & 0 & 1 & 2 & 3 \\
\hline
0 & 0 & 1 & 2 & 3 \\
1 & 1 & 2 & 3 & 0 \\
2 & 2 & 3 & 0 & 1 \\
3 & 3 & 0 & 1 & 2 \\
\end{array}
[/tex]

What is the set [itex]\langle i\rangle[/itex]?
How to define [itex](\langle i\rangle,\cdot)[/itex]?
 
  • #6
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
 
  • #7
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
 
  • #8
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]?
 
  • #9
gruba said:
From here, how to explicitly define a function fff?
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!
 
  • #10
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.
 
  • #11
gruba said:
[itex]f(x)=e^x[/itex] is one isomorphism.
Not between the given groups.
 
  • #12
Orodruin said:
Not between the given groups.
[itex]f(x)=e^{2\pi x i}[/itex]?
 
  • #13
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??
 
  • #14
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:
  • #15
@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.
 
  • #16
gruba said:
But f(x) is not one to one.

It is one-to-one on the relevant sets. You have specified f(x) for all elements of ##\mathbb Z_4##, there is absolutely no need to express it in terms of a polynomial (why would you think there was?).
 

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

1. What is an isomorphic group?

An isomorphic group is a mathematical concept that describes a relationship between two groups. Two groups are isomorphic if they have the same structure, meaning that their elements and the way they combine are the same, even if their notations are different.

2. How do you prove that two groups are isomorphic?

To prove that two groups are isomorphic, you need to show that there exists a bijective function (one-to-one and onto) between the two groups that preserves the group operation. This means that the function must map the elements of one group to the elements of the other group in a way that the group operation is still valid.

3. What are the elements of the groups (\mathbb Z_4,_{+4}) and (\langle i\rangle, \cdot) ?

The elements of the group (\mathbb Z_4,_{+4}) are {0, 1, 2, 3}, where the group operation is addition modulo 4. The elements of the group (\langle i\rangle, \cdot) are {1, i, -1, -i}, where the group operation is multiplication.

4. How can you show that (\mathbb Z_4,_{+4}) and (\langle i\rangle, \cdot) are isomorphic?

To show that (\mathbb Z_4,_{+4}) and (\langle i\rangle, \cdot) are isomorphic, you can define a function f: \mathbb Z_4 \rightarrow \langle i\rangle such that f(x) = i^x. This function is bijective and preserves the group operation, as f(x + y) = i^{x+y} = i^x \cdot i^y = f(x) \cdot f(y). Therefore, the two groups are isomorphic.

5. Can you give an example of how to prove isomorphic groups?

Yes, an example would be to show that the groups (\mathbb Z_3,_{+3}) and (\langle -i\rangle, \cdot) are isomorphic. We can define a function f: \mathbb Z_3 \rightarrow \langle -i\rangle such that f(x) = (-i)^x. This function is bijective and preserves the group operation, as f(x + y) = (-i)^{x+y} = (-i)^x \cdot (-i)^y = f(x) \cdot f(y). Therefore, the two groups are isomorphic.

Back
Top