Proving Automorphism of Z_n with r in U(n)

[johnnyboy2005]

hey, was just hoping i could get a little help getting started on this one...

Let r be an element in U(n). Prove that the mapping (alpha):Zn-->Zn defined by (alpha)(s) = sr mod n for all s in Zn is an automorphism of Zn.

not expecting the answer, maybe to just push me in the right direction...thanks so much

 

[matt grime]

This is simply treating Zn as an additive group, right? (You should right Z/n or Z/nZ or Z/(n))

Show it is a homomorphism and show it is either invertible (easy given the other use of the word invertible in the question) or just show it is a bijection.

 

[math-chick_41]

Show that its 1-1 onto and operation preserving. since r is in U(n) then r^-1 exists and you will need that little tid bit to show 1-1 then since Zn are finite you get onto.

 

[johnnyboy2005]

so i actually left this question for a bit. This is my soln' so far...

to show it is an automorphism the groups must be one to one and onto (easy to show) and to show that the function is map preserving I'm saying that for any a and b in Z(n) you will have
(alpha)(a+b) = (alpha)(a) + (alpha)(b) = (a)r mod n + (b)r mod n = (a + b)rmodn which is in the automorphism

 

[matt grime]

so i actually left this question for a bit. This is my soln' so far...
to show it is an automorphism the groups must be one to one and onto

groups are not one to one and onto, even if we assume that you mean are in bijective correspondence this does not mean any map between them is an isomorphism. if you are proving a map is an isomorphism it might behove you to mention the map in the alleged proof.

(easy to show) and to show that the function is map preserving I'm saying that for any a and b in Z(n) you will have
(alpha)(a+b) = (alpha)(a) + (alpha)(b) = (a)r mod n + (b)r mod n = (a + b)rmodn which is in the automorphism

what do you mean by "in the automorphism"? Surely you aren't treating maps as sets.

It is a one line proof: (a+b)r=ar+br, and -ar=-(ar) are just general facts of mulitiplying numbers, so trivially it is a homomorphism, now why is it an isomorphism?

 

[johnnyboy2005]

it is an isomorphism because it is one to one and onto...

 

[matt grime]

and have you proved that? i don't think you can get away with saying it is easy to show without showing it; i would be sceptical that you had done so given the presentation of your argument, and the fact that if it were so easy why didn't you write it out? it'd be good if you didn't show one to one and onto but actually pointed out that the map is invertible.