Show that this is a homomorphism

  • Thread starter ArcanaNoir
  • Start date
In summary, the conversation discusses proving that two groups, (\mathbb{Z} _4 , \oplus ) and (\mathbb{Z} ^{0}_{5} , \odot ), are isomorphic. The main challenge is defining the operation \theta in a way that preserves the operation and maps each element correctly. It is suggested to set up the entire multiplication table and verify that it matches the other table. However, there may be shortcuts that can be used.
  • #1
ArcanaNoir
779
4

Homework Statement


Show that [itex] (\mathbb{Z} _4 , \oplus ) \approx ( \mathbb{Z} ^{0}_{5} [/itex] [itex] , \odot ) [/itex]
Meaning, they are isomorphic. The 0 means the zero is deleted from the set. We are using circle plus and circle dot because we are not allowed to think of the operations as addition and multiplication yet. My trouble is with defining the operation [itex] \theta [/itex].

Homework Equations



I know [itex] \theta [/itex] must be operation preserving, that is, [itex] \theta (x \oplus y) = \theta (x) \odot \theta (y) [/itex]

The Attempt at a Solution



I tried defining [itex] \theta [/itex] as [itex] \theta ([x])= [x+1] [/itex]
so, [itex] \theta ([0]) = [1] [/itex] up to [itex] \theta ([3]) = [4] [/itex]
And now, [itex] \theta ([x] \oplus [y]) = \theta ([x \oplus y]) = [x \oplus y +1] [/itex]
Did I make the wrong steps there? I'm not sure how to arrive at [itex] \theta ([x]) \odot \theta ([y]) [/itex]
 
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  • #2
Hi Arcana!

θ needs to map each element.
So you can define it by defining θ(0), θ(1), θ(2), and θ(3).

Start with θ(0) and then make an arbitrary choice for θ(1).
Use the preservation of the operation to deduce what the other values must be.
 
  • #3
[itex] \theta ([0])=[1] [/itex]
[itex] \theta ([1])=[2] [/itex]
[itex] \theta ([2])=[3] [/itex]
[itex] \theta ([3])=[4] [/itex]

But how do I show it's operation preserving? Do I have to show it using each number, or is there some way to do it using arbitrary elements?
 
  • #4
Sorry, but this won't work.
Consider θ(1+1).
 
  • #5
ArcanaNoir said:
But how do I show it's operation preserving? Do I have to show it using each number, or is there some way to do it using arbitrary elements?

Basically what you would need to do is the show that each combination of a and b has its operation preserved.
That is, set up the entire multiplication table, map it, and verify that it matches the other entire multiplication table.

With the size of your current groups this is very doable.
 
  • #6
Okay, if that's the only way I can certainly do that. thanks for the tip-off about my mapping being wrong. I think with this information I'll be okay here. :)
 
  • #7
It's not the only way.
That is, you can make a couple of shortcuts.
But I guess that is better left for a later exercise.

Note that the definition of your isomorphism says: "for any elements a and b".
This means you have to show it "for any elements a and b".
 

1. What is a homomorphism?

A homomorphism is a function that preserves the algebraic structure of a mathematical object. In simpler terms, it is a function that maps elements from one set to another in a way that maintains the operations and relationships between those elements.

2. How can you show that a function is a homomorphism?

To show that a function is a homomorphism, you must demonstrate that it preserves the operations and relationships between the elements of the sets it is mapping. This can be done by performing the function on two elements and comparing the result to the function performed on the individual elements.

3. Why is it important to prove that a function is a homomorphism?

Proving that a function is a homomorphism is important because it ensures that the function is consistent and reliable in its mapping of elements between sets. This is especially important in mathematical and scientific contexts where precision and accuracy are crucial.

4. What types of mathematical objects can be homomorphisms?

Homomorphisms can be defined for a variety of mathematical objects, including groups, rings, fields, and vector spaces. They are also used in abstract algebra, topology, and other areas of mathematics.

5. Can a function be a homomorphism for some sets, but not others?

Yes, a function can be a homomorphism for some sets, but not others. This is because the properties and relationships between elements in different sets may vary, and a function may not be able to preserve those relationships in all cases.

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