# Z_4 Homework Statement: Confirm My Book is Wrong

• ehrenfest
In summary: Since it's hard to write out a general coset, we'd usually rather use some other group we already know well to think about this one.In summary, the conversation discusses whether a specific presentation of Z_4 is correct or not. The expert provides a detailed explanation of why the presentation is correct and how it can be proven using group presentations. They also discuss the definition of a representation and how it differs from a presentation. Finally, they explain the normal subgroup and cosets of the group being presented.
ehrenfest

## Homework Statement

My book says that $(a,b:a^4=1,b=a^2)$ is a presentation of Z_4. I strongly disagree. If they want to get a presentation of Z_4, they need to get b as a consequence of their relations, but I only see that b^2 is a consequence of their relations. Please confirm that my book is wrong.

## The Attempt at a Solution

Sorry, but I agree with the book. The relation b=a^2 means that b is redundant as a generator, ie, given anything generated by a and b, use this relation to write it in terms of a alone, so that it is generated by a alone. Thus we have:

$$<a,b|a^4=1, b=a^2> \cong <a|a^4=1> \cong \mathbb{Z}_4$$

You can prove this (or, without much more effort, the obvious generalization to more generators) by resorting to the definition of:

$$<a_1,...,a_n | r_1,...,r_m>$$

as the quotient of the free group F generated by $a_1,...,a_n$ by the normal subgroup generated by $\{r_1,...,r_m\}$ (specifically, we write the relations in the form $r_i=1$, where $r_i$ is a word formed out of the $a_i$, ie, an element of F).

StatusX said:
Sorry, but I agree with the book. The relation b=a^2 means that b is redundant as a generator, ie, given anything generated by a and b, use this relation to write it in terms of a alone, so that it is generated by a alone. Thus we have:

$$<a,b|a^4=1, b=a^2> \cong <a|a^4=1>$$

I am not sure why that equation is true. I thought that the only way this group would be isomorphic to Z_4 would be if b were in the normal closure of a since it is clear that

$$<a,b|a^4=1, b=1> \cong <a|a^4=1> \cong \mathbb{Z}_4$$

Can explain that equation in terms of the definition of a group presentation? I am just learning what that is, so I haven't gotten far from the definition. So this group is the F[{a,b}] modded out by the normal closure N of the words {a^4,ba^{-2}}. What you wrote probably makes a lot of sense but group presentations are just really confusing me!

A presentation is just a list of generators and relations they satisfy. In this case the generators are a and b, and the relations are a^4=1 and a^2=b. An arbitrary element of the group with this presentation looks like a^n b^m, where n and m are integers. b=a^2 implies that we can write this element as a^n a^(2m) = a^(n+2m). What does a^4=1 imply?

I am not sure this is the same, but I am currently studying group representations. they are homomorphisms from a group into a linear space. Not isomorphisms like you suggested. Does this make the difference?

A representation is not the same thing as a presentation.

Let A be a set and let $\{r_i\} \subset F[A]$. Let R be the least normal subgroup of F[A] containing the $r_i$. An isomorphism $\phi$ of F[A]/R onto a group G is a presentation of G.

So, in our case $\{r_i\} = \{b^4, ba^{-2}\}$. If R is the least normal subgroup of F[A] that contains those elements, then can you please explain how you know that $F[A]/R$ is isomorphic to $Z_4$?

I guess those relations tells us that $bR=a^2R$ which implies that you can write any element of the quotient group F[A]/R only in terms of aR. So the elements of F[A]/R are all included in the set $\{a^mR : m \in \mathbb{Z}\}$. And the algebra of that set is just addition of exponents because that is how multiplication is defined in the free group. And we also know that a^4R=R, so we are modding out Z by 4. I think I see now. What bothers me is that I cannot figure out what R is exactly. I want to write it down so that I know what the cosets of F[A]/R are. Is that possible?

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ehrenfest said:
What bothers me is that I cannot figure out what R is exactly.
That depends, of course, on what "figure out exactly" means to you.

You already have a description of R as the kernel of a homomorphism $F[\{a,b\}] \to \mathbb{Z}_4$, and you can do a lot with that information.

Hurkyl said:
That depends, of course, on what "figure out exactly" means to you.

By "figure out exactly," I guess that if I give you an element of F[A], I want to know whether it is in R.

ehrenfest said:
So, in our case $\{r_i\} = \{b^4, ba^{-2}\}$. If R is the least normal subgroup of F[A] that contains those elements, then can you please explain how you know that $F[A]/R$ is isomorphic to $Z_4$?

I guess those relations tells us that $bR=a^2R$ which implies that you can write any element of the quotient group F[A]/R only in terms of aR. So the elements of F[A]/R are all included in the set $\{a^mR : m \in \mathbb{Z}\}$. And the algebra of that set is just addition of exponents because that is how multiplication is defined in the free group. And we also know that a^4R=R, so we are modding out Z by 4. I think I see now.

You're pretty much right. More rigorously, you want to form an isomorphism between $F_2/<\{a^4,ab^{-2}\}>$ and $F_1/<\{a^4\}>$. This is acheived by sending [a] to [a] (the first is an equivalence class in F2 and the second is an equivalence class in F1) and to [a2], which defines the map completely since these are generators. You've basically shown this is surjective, but you should really also show it's well-defined and injective, which shouldn't be too hard.

What bothers me is that I cannot figure out what R is exactly. I want to write it down so that I know what the cosets of F[A]/R are. Is that possible?

R is an ugly group, just like F[A]. It basically consists of all fancy ways of writing out the identity in the group being presented (eg, $(ab)^{50} a^4 a^2 b^{-1} a^4 (ab)^{-50} a^{-2}b$, etc).

## 1. What is Z4?

Z4 is a mathematical concept known as the cyclic group of order 4. It is a set of four elements (0, 1, 2, and 3) with a specific operation (addition modulo 4) that follows certain rules.

## 2. How can I confirm if my book is wrong about Z4?

You can confirm it by checking if the book's definition and properties of Z4 match with other reliable sources, such as textbooks or math websites. You can also consult with a math expert or teacher for clarification.

## 3. What are the common misconceptions about Z4?

One common misconception is that Z4 is the same as the set of integers (Z) from 0 to 4. Another misconception is that the elements in Z4 represent the remainder of division by 4, when in fact, they represent equivalence classes.

## 4. How is Z4 used in science?

Z4 has various applications in science, particularly in fields such as cryptography, coding theory, and signal processing. It is also used in the study of symmetry and in the classification of crystal structures.

## 5. What is the significance of Z4 in mathematics?

Z4 is a fundamental concept in abstract algebra and has important properties that make it useful in solving mathematical problems. It is also used in other branches of mathematics, such as number theory and graph theory.

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