Please Login to access more options.
Problem 36 (Permutations of $U(n)$)
Suppose $n$ is an integer greater than 1 and let $X=U(n)$, so the set of integers mod $n$ that have a multiplicative inverse mod $n$. For each $a\in U(n)$, we can define a permutation of $U(n)$ by $f_a(x)=xa\pmod n$ for each $x\in U(n)$. This problem asks you to show that this is indeed a permutation of $U(n)$. This requires that you show the following:
- For each integer $n\geq 2$, show that if $a\in U(n)$ and $x\in U(n)$, then the product $f_a(x)=xa\pmod n\in U(n)$. This shows that $f_a:U(n)\to U(n)$ is a map from $U(n)$ to $U(n)$.
- Given $n\geq 2$ and $a\in U(n)$, show that $f_a$ is one-to-one. (As a reminder, one-to-one means that if $f_a(x)=f_a(y)$, then $x=y$.)
- Given $n\geq 2$ and $a\in U(n)$, show that $f_a$ is onto. (As a reminder, onto means that if $y\in U(n)$, then there exists $x\in U(n)$ such that $f_a(x)=y$.)
The following pages link to this page.