
#1
Mar708, 06:59 PM

P: 1,998

1. The problem statement, all variables and given/known data
My book says that [itex](a,b:a^4=1,b=a^2)[/itex] 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
Mar708, 08:09 PM

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P: 2,566

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:
[tex] <a,ba^4=1, b=a^2> \cong <aa^4=1> \cong \mathbb{Z}_4 [/tex] You can prove this (or, without much more effort, the obvious generalization to more generators) by resorting to the definition of: [tex] <a_1,...,a_n  r_1,...,r_m> [/tex] as the quotient of the free group F generated by [itex]a_1,...,a_n[/itex] by the normal subgroup generated by [itex]\{r_1,...,r_m\}[/itex] (specifically, we write the relations in the form [itex]r_i=1[/itex], where [itex]r_i[/itex] is a word formed out of the [itex]a_i[/itex], ie, an element of F). 



#3
Mar708, 08:39 PM

P: 1,998

[tex] <a,ba^4=1, b=1> \cong <aa^4=1> \cong \mathbb{Z}_4 [/tex] 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
Mar808, 01:15 AM

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P: 2,020

presentation of Z_4
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
Mar808, 02:02 AM

P: 169

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
Mar808, 02:41 AM

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A representation is not the same thing as a presentation.




#7
Mar808, 11:47 AM

P: 1,998

I guess those relations tells us that [itex]bR=a^2R[/itex] 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 [itex]\{a^mR : m \in \mathbb{Z}\}[/itex]. 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? 



#8
Mar808, 12:45 PM

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PF Gold
P: 16,101

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



#9
Mar808, 01:21 PM

P: 1,998





#10
Mar908, 04:54 AM

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P: 2,566




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