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nobody56
Nov24-08, 01:07 AM
1. The problem statement, all variables and given/known data
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)


2. Relevant equations



3. 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)

sutupidmath
Nov24-08, 03:27 AM
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....

nobody56
Nov24-08, 03:59 AM
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)

HallsofIvy
Nov24-08, 05:25 AM
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?

nobody56
Nov24-08, 05:32 AM
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.