[–] 2 points3 points ago

sorry, this has been archived and can no longer be voted on

Here is my take on the isomorphism. Two groups are isomorphic if you can change one to the other by relabeling its elements. I'll explain more completely what I mean.

Say you have groups G and H. You put sticky notes on the elements of G; on these sticky notes you have written the elements of H. If you can make G look like H this way, just in terms of sets, then G and H are in bijective correspondence -- the sticky notes give you a bijection from G to H. However, this is only the first part of being an isomorphism -- the second part is if the products behave properly.

Pick two elements of G and multiply them together. At this point you might ask, "Do you mean to multiply them together as elements of G, or to use their sticky note labels to figure out what the product is?" These two might be different things. However, when they're the SAME, that's when you know that your labels make an isomorphism; your bijection also makes them look the same as groups.

As an example, take the group Z_4 (which is the integers 0 through 3 with modular addition) and the complex numbers 1, i, -1, -i. These groups are isomorphic; I will label 1 as 0@ (so illustrated to remind us that our complex numbers have sticky notes on them), i as 1@, -1 as 2@, and -i as 3@. Then, for example, we might ask about the product (1@)(3@). If we're using the operation in Z_4, this would be 0@. If we look under the labels and use the operation in H, we would get (i)(-i) = 1, which is labeled 0@. Hence, it doesn't matter whether we looked under the label or not.

The (almost painfully) short definition, that an isomorphism is a bijection from G to H satisfying f(xy)=f(x)f(y) is equivalent. The above is an intuitive recasting. An automorphism is simply an isomorphism where the domain and range are equal, so you're relabeling the elements of G. For example, you might figure that to find an automorphism of Z_6, a cyclic group, the generator had better map to a generator; otherwise, you could tell that 1 isn't behaving right without looking under the sticky labels. But if 5 is labeled with the 1, then 5+5=4 had better be labeled with 1+1=2, etc, etc.

Edit: Formatting, and deleting a stray sentence that survived editing.

[–][S] 0 points1 point ago

sorry, this has been archived and can no longer be voted on

Wow, this helped a lot. Thank you so much.

[–] 1 point2 points ago

sorry, this has been archived and can no longer be voted on

An automorphism is simply an isomorphism from a group to itself, so if you understand isomorphisms you should be able to understand automorphisms.

So let's focus on isomorphisms.

Let [; G,J ;] be groups. Then if a function [; \phi:G \to J ;] is both a bijection and a homomorphism, it is an isomorphism. To be a bijection means it is both surjective and injective (since you understand permutations, I'm assuming you understand bijections), and to be a homomorphism simply means that for any [; x,y \in G;], [; \phi(xy) =\phi(x)\phi(y) ;]

So, basically, an isomorphism is a bijection between groups that preserves the structure of the group product. In other words, if a function is a bijection between two groups and it has the property that the product ab in the domain maps to the product of the images of a and b in the codomain, it is an isomorphism.

Now you ask "how would I figure out what the automorphism of Z6?" I'm not sure exactly what you mean, but taking a guess, I think what you need to notice is that there are many different automorphisms on a set. The identity function on a set is a simple example of an automorphism, but there can be many others. Again, just go back to the definitions: if it's a bijection and a homomorphism and it is from a set to itself, it is an automorphism.

[–] 0 points1 point ago

sorry, this has been archived and can no longer be voted on

A morphism is a map that respects structure, for example, the operation. An isomorphism is a map that identifies two structures as being the same. For example, we have the group (R, +) and (R+, *), and a map exp : (R, +) -> (R+, *) that identify the structures. In general, we think of two groups being the same if there is an isomorphism between them. In the context of groups, any group homomorphism which is also a bijective map of sets is an isomorphism.

An automorphism to is an isomorphism from an object (e.g. a group) to itself. So, to find the automorphisms of Z/6, find the maps Z/6 -> Z/6 that are bijections.

Hope that helps.

[–] 0 points1 point ago

sorry, this has been archived and can no longer be voted on

So, to find the automorphisms of Z/6, find the maps Z/6 -> Z/6 that are bijections.

Will every bijection from Z/6 -> Z/6 necessarily be an isomorphism?

[–] 0 points1 point ago

sorry, this has been archived and can no longer be voted on

No. It also has to be a group homomorphism: it has to respect the group laws.

[–] 0 points1 point ago

sorry, this has been archived and can no longer be voted on

In particular, a morphism Z/6 -> Z/6 is determined by where 1 goes. Therefore, there can be at most 6 homomorphisms. Which are automorphisms?

[–] 0 points1 point ago

sorry, this has been archived and can no longer be voted on

For Z6 as an additive group, every element is a sum of 1's (it's cyclically generated), so if you have a homomorphism from h:Z6->Z6, then h(1) tells you all the rest.

h(3)=h(1+1+1)=h(1)+h(1)+h(1)=3*h(1) (abusing notation a bit)

So every homomorphism has to be of the form that you have some fixed k=h(1) and it works by h(x)=k*x. And vice-versa,

h(x+y)=k*(x+y)=k*x+k*y=h(x)+h(y)

So every k gives a homomorphism. Now, for which k is h(x)=k*x also a bijection?
The text passage primarily discusses the concept of isomorphisms and automorphisms in group theory, providing definitions and examples to explain these concepts.
no_question

the passage does not contain any explicit mathematical questions or calculations that require an answer. it is primarily expository, explaining the concepts of isomorphisms and automorphisms in group theory.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
[–] 2 points3 points ago

sorry, this has been archived and can no longer be voted on

Here is my take on the isomorphism. Two groups are isomorphic if you can change one to the other by relabeling its elements. I'll explain more completely what I mean.

Say you have groups G and H. You put sticky notes on the elements of G; on these sticky notes you have written the elements of H. If you can make G look like H this way, just in terms of sets, then G and H are in bijective correspondence -- the sticky notes give you a bijection from G to H. However, this is only the first part of being an isomorphism -- the second part is if the products behave properly.

Pick two elements of G and multiply them together. At this point you might ask, "Do you mean to multiply them together as elements of G, or to use their sticky note labels to figure out what the product is?" These two might be different things. However, when they're the SAME, that's when you know that your labels make an isomorphism; your bijection also makes them look the same as groups.

As an example, take the group Z_4 (which is the integers 0 through 3 with modular addition) and the complex numbers 1, i, -1, -i. These groups are isomorphic; I will label 1 as 0@ (so illustrated to remind us that our complex numbers have sticky notes on them), i as 1@, -1 as 2@, and -i as 3@. Then, for example, we might ask about the product (1@)(3@). If we're using the operation in Z_4, this would be 0@. If we look under the labels and use the operation in H, we would get (i)(-i) = 1, which is labeled 0@. Hence, it doesn't matter whether we looked under the label or not.

The (almost painfully) short definition, that an isomorphism is a bijection from G to H satisfying f(xy)=f(x)f(y) is equivalent. The above is an intuitive recasting. An automorphism is simply an isomorphism where the domain and range are equal, so you're relabeling the elements of G. For example, you might figure that to find an automorphism of Z_6, a cyclic group, the generator had better map to a generator; otherwise, you could tell that 1 isn't behaving right without looking under the sticky labels. But if 5 is labeled with the 1, then 5+5=4 had better be labeled with 1+1=2, etc, etc.

Edit: Formatting, and deleting a stray sentence that survived editing.

[–][S] 0 points1 point ago

sorry, this has been archived and can no longer be voted on

Wow, this helped a lot. Thank you so much.

[–] 1 point2 points ago

sorry, this has been archived and can no longer be voted on

An automorphism is simply an isomorphism from a group to itself, so if you understand isomorphisms you should be able to understand automorphisms.

So let's focus on isomorphisms.

Let [; G,J ;] be groups. Then if a function [; \phi:G \to J ;] is both a bijection and a homomorphism, it is an isomorphism. To be a bijection means it is both surjective and injective (since you understand permutations, I'm assuming you understand bijections), and to be a homomorphism simply means that for any [; x,y \in G;], [; \phi(xy) =\phi(x)\phi(y) ;]

So, basically, an isomorphism is a bijection between groups that preserves the structure of the group product. In other words, if a function is a bijection between two groups and it has the property that the product ab in the domain maps to the product of the images of a and b in the codomain, it is an isomorphism.

Now you ask "how would I figure out what the automorphism of Z6?" I'm not sure exactly what you mean, but taking a guess, I think what you need to notice is that there are many different automorphisms on a set. The identity function on a set is a simple example of an automorphism, but there can be many others. Again, just go back to the definitions: if it's a bijection and a homomorphism and it is from a set to itself, it is an automorphism.

[–] 0 points1 point ago

sorry, this has been archived and can no longer be voted on

A morphism is a map that respects structure, for example, the operation. An isomorphism is a map that identifies two structures as being the same. For example, we have the group (R, +) and (R+, *), and a map exp : (R, +) -> (R+, *) that identify the structures. In general, we think of two groups being the same if there is an isomorphism between them. In the context of groups, any group homomorphism which is also a bijective map of sets is an isomorphism.

An automorphism to is an isomorphism from an object (e.g. a group) to itself. So, to find the automorphisms of Z/6, find the maps Z/6 -> Z/6 that are bijections.

Hope that helps.

[–] 0 points1 point ago

sorry, this has been archived and can no longer be voted on

So, to find the automorphisms of Z/6, find the maps Z/6 -> Z/6 that are bijections.

Will every bijection from Z/6 -> Z/6 necessarily be an isomorphism?

[–] 0 points1 point ago

sorry, this has been archived and can no longer be voted on

No. It also has to be a group homomorphism: it has to respect the group laws.

[–] 0 points1 point ago

sorry, this has been archived and can no longer be voted on

In particular, a morphism Z/6 -> Z/6 is determined by where 1 goes. Therefore, there can be at most 6 homomorphisms. Which are automorphisms?

[–] 0 points1 point ago

sorry, this has been archived and can no longer be voted on

For Z6 as an additive group, every element is a sum of 1's (it's cyclically generated), so if you have a homomorphism from h:Z6->Z6, then h(1) tells you all the rest.

h(3)=h(1+1+1)=h(1)+h(1)+h(1)=3*h(1) (abusing notation a bit)

So every homomorphism has to be of the form that you have some fixed k=h(1) and it works by h(x)=k*x. And vice-versa,

h(x+y)=k*(x+y)=k*x+k*y=h(x)+h(y)

So every k gives a homomorphism. Now, for which k is h(x)=k*x also a bijection?
user
The text passage primarily discusses the concept of isomorphisms and automorphisms in group theory, providing definitions and examples to explain these concepts.

TYPE: NO_QUESTION

The passage does not contain any explicit mathematical questions or calculations that require an answer. It is primarily expository, explaining the concepts of isomorphisms and automorphisms in group theory.