Automorphisms and Order of a Map

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Homework Statement


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

b) Determine the order of [tex]\phi[/tex].

c) determine all (a,b)[tex]\in[/tex]G with [tex]\phi[/tex]((a,b))=(b,a)


Homework Equations





The Attempt at a Solution


a) multiple parts
1) Let (a,b)[tex]\in[/tex]G, then [tex]\phi[/tex](a,b)=(-b.a), which is an element in G since -b,a[tex]\in[/tex]Z therefore [tex]\phi[/tex](a,b)=(-b,a)[tex]\in[/tex]G
2) One to One: suppose [tex]\phi[/tex](a,b)=[tex]\phi[/tex](c,d). That implies (-b,a)=(-d,c), which implies a=c and b=d therefore one to one.
3) Onto: Let (c,d)[tex]\in[/tex]G to find (X,X)[tex]\in[/tex]G such that [tex]\phi[/tex](X,X)=(c,d)=[tex]\phi[/tex]((d,-c)) therefore onto
4) Operation Preserving: Let (a,b),(c,d)[tex]\in[/tex]G. [tex]\phi[/tex]((a+c,b+d))=(-(b+d),a+c) and [tex]\phi[/tex]((a,b))+[tex]\phi[/tex]((c,d))=(-b,a)+(-d,c)=(-b-d,a+c)=(-(b+d),a+c)
therefore [tex]\phi[/tex] is an isomophism, and since [tex]\phi[/tex] is from G to G, by definition [tex]\phi[/tex] 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[tex]\in[/tex]G} thus [tex]\phi[/tex]((a,0))=(-0,a)=(0,a)
 
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for part (ii) the order of [tex]\phi[/tex] is 5. (at least from my world over here, it looks like that).
I assume that the order of a mapping would be, whenever:

[tex]\phi^k=\phi[/tex] so looking at your problem

[tex]\phi^5(a,b)=\phi(a,b)=(-b,a)[/tex]

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

and you said 5 because...
[tex]\phi[/tex]((a,b))=(-b,a)
[tex]\phi[/tex]((-b,a))=(-a,-b)
[tex]\phi[/tex]((-a,-b))=(b,-a)
[tex]\phi[/tex]((b,-a))=(a,b)
[tex]\phi[/tex]((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...
[tex]\phi[/tex]((a,b))=(-b,a)
[tex]\phi[/tex]((-b,a))=(-a,-b)
[tex]\phi[/tex]((-a,-b))=(b,-a)
[tex]\phi[/tex]((b,-a))=(a,b)
[tex]\phi[/tex]((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 [itex]\phi^k= \phi[/itex]. 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.