1. ## abstract algebra function

Let m, n be positive integers with m | n, and let k be any integer. Show that f :
Zn-> Zm defined by f([x]n) = [x]m^k, for all [x]n belong to Zn
is a well-defined function.
Hint: We know that g : Z×
m ! Z×
m defined by g([x]m) = [x]k
m is a function if k is a
positive integer.

2. Originally Posted by sd215
Let m, n be positive integers with m | n, and let k be any integer. Show that f :
Zn-> Zm defined by f([x]n) = [x]m^k, for all [x]n belong to Zn
is a well-defined function.
Hint: We know that g : Z×
m ! Z×
m defined by g([x]m) = [x]k
m is a function if k is a
positive integer.
I'm sorry, I have no idea what this says.

3. Originally Posted by sd215
Let m, n be positive integers with m | n, and let k be any integer. Show that f :
Zn-> Zm defined by f([x]n) = [x]m^k, for all [x]n belong to Zn
is a well-defined function.
Hint: We know that g : Z×
m ! Z×
m defined by g([x]m) = [x]k
m is a function if k is a
positive integer.
Do you mean that [x]n is the equivalence class of numbers that are equivalent mod n and [x]m^k is the equivalence class of numbers that are equivalent mod m^k? That is, [x]3 is an equivalence class mod 3 so that with n= 6, m= 3 (so that m|n) and k= 2, you are asked to show that the function f that maps any number, x, in $\displaystyle Z_6$ to $\displaystyle x(2^2)= 4x$ in $\displaystyle Z_4[$ is "well defined", that is that two numbers in the same equivalence class in $\displaystyle Z_n$ are mapped into numbers in the same equivalence class in $\displaystyle Z_m$.