Presentation of Z_4

1. Mar 7, 2008

ehrenfest

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution

2. Mar 7, 2008

StatusX

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).

3. Mar 7, 2008

ehrenfest

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!

4. Mar 8, 2008

morphism

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?

5. Mar 8, 2008

jacobrhcp

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?

6. Mar 8, 2008

morphism

A representation is not the same thing as a presentation.

7. Mar 8, 2008

ehrenfest

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?

Last edited: Mar 8, 2008
8. Mar 8, 2008

Hurkyl

Staff Emeritus
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.

9. Mar 8, 2008

ehrenfest

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.

10. Mar 9, 2008

StatusX

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.

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).