This question in the reference to Number Theory (Modular Arithmetic). I need some help in understanding the proof of Proposition 6.
In the third line of the proof of Proposition 6,
"$\dots 0\leq k,l\leq (n-1)$. Therefore $|k-l|<n;\dots$"
Can someone please explain this. How did we get "$|k-l|<n$"?
In the 5th line,
"This means that $\{a,a+d,a+2d,\dots,a+(n-1)d\}=\{0,1,2,\dots,(n-1)\}$ modulo $n$. Further if $a+jd\equiv k \pmod{n}$ with $0\leq k<(n-1)$, then $(a+jd,n)=1$ iff $(k,n)=1.$ Hence the number of integers in $S$ which are prime to $n$ is precisely $\phi(n)$."
I do not understand the entire thing.
Why is "$\{a,a+d,a+2d,\dots,a+(n-1)d\}=\{0,1,2,\dots,(n-1)\}$ modulo $n$"?
Why is the inequality $0\leq k<(n-1)$ and not $0\leq k\leq (n-1)$?
Why is $(a+jd,n)=1$ iff $(k,n)=1$?
Please help.