Automorphisms and Order of a Map

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Homework Help Overview

The problem involves the group G defined by pairs of integers under addition and explores the properties of a specific map φ that transforms elements of G. The tasks include demonstrating that φ is an automorphism and determining its order, as well as identifying elements that map to each other under φ.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the verification of φ as an automorphism through various properties such as one-to-one, onto, and operation preservation. Questions arise regarding the calculation of the order of the map and whether the same rules apply to maps as to groups.

Discussion Status

Some participants have confirmed the correctness of the first two parts of the problem, while others are exploring the concept of the order of a map and its implications. There is an ongoing dialogue about the calculations related to the order of φ, with some uncertainty expressed regarding the definitions and rules applicable to maps.

Contextual Notes

There is mention of a lack of resources in the participants' textbooks regarding the order of maps, which may be influencing their understanding and approach to the problem.

nobody56
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Homework Statement


Let G={(a,b)/ a,b\inZ} be a group with addition defined by (a,b)+(c,d)=(a+c,b+d).
a) Show that the map\phi:G\rightarrowG defined by \phi((a,b))=(-b,a) is an automorphism of G.

b) Determine the order of \phi.

c) determine all (a,b)\inG with \phi((a,b))=(b,a)


Homework Equations





The Attempt at a Solution


a) multiple parts
1) Let (a,b)\inG, then \phi(a,b)=(-b.a), which is an element in G since -b,a\inZ therefore \phi(a,b)=(-b,a)\inG
2) One to One: suppose \phi(a,b)=\phi(c,d). That implies (-b,a)=(-d,c), which implies a=c and b=d therefore one to one.
3) Onto: Let (c,d)\inG to find (X,X)\inG such that \phi(X,X)=(c,d)=\phi((d,-c)) therefore onto
4) Operation Preserving: Let (a,b),(c,d)\inG. \phi((a+c,b+d))=(-(b+d),a+c) and \phi((a,b))+\phi((c,d))=(-b,a)+(-d,c)=(-b-d,a+c)=(-(b+d),a+c)
therefore \phi is an isomophism, and since \phi is from G to G, by definition \phi is an automorphism of G

b) I am not sure where to start for the order of a map?

c) the set {(a,0)/a\inG} thus \phi((a,0))=(-0,a)=(0,a)
 
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for part (ii) the order of \phi is 5. (at least from my world over here, it looks like that).
I assume that the order of a mapping would be, whenever:

\phi^k=\phi so looking at your problem

\phi^5(a,b)=\phi(a,b)=(-b,a)

Check the calculations, i did it really fast...
 
are the other two parts ok?

and you said 5 because...
\phi((a,b))=(-b,a)
\phi((-b,a))=(-a,-b)
\phi((-a,-b))=(b,-a)
\phi((b,-a))=(a,b)
\phi((a,b))=(-b,a)
right?
(I can not find anything to do with order of maps in my book)
 
nobody56 said:
are the other two parts ok?
Yes, the first two parts are ok.

and you said 5 because...
\phi((a,b))=(-b,a)
\phi((-b,a))=(-a,-b)
\phi((-a,-b))=(b,-a)
\phi((b,-a))=(a,b)
\phi((a,b))=(-b,a)
right?
(I can not find anything to do with order of maps in my book)
The "order of a map" is, as sutupidmath said, the smallest positive integer, k, such that \phi^k= \phi. Did you look up "order" in the index of your book?
 
Yea, but I wasnt sure if I could apply the same rules to a map. In my book it just has for order is of a group or the order on an element.

Thanks for your help and for clarifing that.
 

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