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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)

2. Relevant equations

3. 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]Ztherefore [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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# Homework Help: Automorphisms and Order of a Map

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