# Isomorphic groups

gruba

## Homework Statement

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

## Homework Equations

-Group isomorphism

## The Attempt at a Solution

Let $\mathbb Z_4=\{0,1,2,3\}$.
$(\mathbb Z_4,_{+4})$ can be represented using Cayley's table:
$$\begin{array}{c|lcr} {_{+4}} & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3 \\ 1 & 1 & 2 & 3 & 0 \\ 2 & 2 & 3 & 0 & 1 \\ 3 & 3 & 0 & 1 & 2 \\ \end{array}$$

What is the set $\langle i\rangle$?
How to define $(\langle i\rangle,\cdot)$?

## Answers and Replies

Staff Emeritus
Homework Helper
The subgroup of ##(\mathbb{C},\cdot)## generated by ##i##.

gruba
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?

Staff Emeritus
Homework Helper
Gold Member
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##?

Mentor

## Homework Statement

Show that the group $(\mathbb Z_4,_{+4})$ is isomorphic to $(\langle i\rangle,\cdot)$?
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 $\mathbb Z_4=\{0,1,2,3\}$.
$(\mathbb Z_4,_{+4})$ can be represented using Cayley's table:
$$\begin{array}{c|lcr} {_{+4}} & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3 \\ 1 & 1 & 2 & 3 & 0 \\ 2 & 2 & 3 & 0 & 1 \\ 3 & 3 & 0 & 1 & 2 \\ \end{array}$$

What is the set $\langle i\rangle$?
How to define $(\langle i\rangle,\cdot)$?

gruba
Could someone explain this problem (using Cayley's tables - easier)? How to form Cayley's table for the group $(\langle i\rangle,\cdot)$?

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

Mentor
Could someone explain this problem (using Cayley's tables - easier)? How to form Cayley's table for the group $(\langle i\rangle,\cdot)$?
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

gruba
Why don't you try what Orodruin suggested -- find i2, i3, and so on. This is not a hard problem.

Let $f:\mathbb Z_4\rightarrow \langle i\rangle=\{i^0,i^1,i^2,i^3\}=\{1,i,-1,-i\}$ where $f$ is an isomorphism.
From here, how to explicitly define a function $f$?

Staff Emeritus
Homework Helper
Gold Member
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!

gruba
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!
$f(x)=e^x$ is one isomorphism.

Staff Emeritus
Homework Helper
Gold Member
$f(x)=e^x$ is one isomorphism.
Not between the given groups.

gruba
Not between the given groups.
$f(x)=e^{2\pi x i}$?

Staff Emeritus
Homework Helper
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??

gruba
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??
$f(0)=1,f(1)=i,f(2)=-1,f(3)=-i$.

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

But $f(x)$ is not one to one.

What is the actual method for describing an isomorphism, without taking a guess?

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