Automorphism groups and determing a mapping

[irebat]

1. Suppose that Ø:Z(50)→Z(50) is an automorphism with Ø(11)=13. Determine a formula for Ø(x).
this is the problem I am getting, its chapter 6 problem 20 in Gallian's Abstract Algebra latest edition (you can find it on googlebooks) Am i wrong in thinking there's something wrong with the problem?
I think this because arent both 11 and 13 generators of U(50) because of coprimeness, am i missing something big? or is the book just giving me a bum problem?

edit: i have a final exam tommorow and the teacher stressed this chapters homework, please help!

 

[irebat]

hey guys I asked my TA and he walked me through it.

I seemed to understand when he explained it to me but when i got home for the life of me i can't seem to remember what he did.

this is the work i wrote down, can someone explain to me the thought process here?

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If this is respect to addition, note that
11n = 1 (mod 50)
==> 11n = 1 - 100 = -99 (mod 50)
==> n = -9 = 41 (mod 50).

Therefore, 41 * 11 = 1 (mod 50)
==> Ø(1)
= Ø(11 + 11 + ... + 11) [41 times]
= 41 * Ø(11), since Ø is a homomorphism
= 41 * 13 (mod 50)
= -9 * 13 (mod 50)
= -117 (mod 50)
= 33.

Therefore for any x in Z_50, we have
Ø(1) = 33 (mod 50) ==> Ø(x) = 33x (mod 50).

 

[Dick]

The thought process is that phi(x) is probably of the form a*x mod 50. You want to solve for a. You have that phi(11)=13, so you want a*11=13 mod 50. You would then solve for a just like in the real numbers. If you could find an inverse to 11 in Z_50 you could just multiply both sides by that inverse. So a*11*(11)^(-1)=a=13*(11)^(-1)=1. To actually find 11^(-1) mod 50 your TA used a clever trick to conclude 11^(-1) mod 50 is 41. I.e. 11*41=1 mod 50.