Let $A=\{1,2,3..n\}$, $(n>10)$ following the rule:
for $x,y \in A$, we would say that $x\sim y,$ $\space$ iff $gcd(x,n) = gcd(y,n)$.
find the equivalence classes in case $n=6p$ while $p$ is prime number and $4 \lt p$ .
I'm kind of new to this, but if I select $[1]$ as the first class, doesn't it suppose to have all the numbers til $n$ because $gcd(x,1)=1$ for any intiger $x$?
If $n = 6p$ for $p$ a prime, two numbers $x,y$ are in the same equivalence class if $\gcd(x,6p) = \gcd(y,6p)$. So the equivalence classes correspond to the factors of $6p$. Since $p>4$, you will find that the equivalence classes are $\bar 1$, $\bar 2$, $\bar 3$, $\bar 6$, $\bar p$, $\overline{2p}$, $\overline{3p}$, and $\overline{6p}$.
To address your second question, if you select $\bar 1$ as the first class, you are looking at the set of all numbers $y$ such that $\gcd(y,n) = \gcd(1,n) = 1$. Take, for example, $n = 4$. Then the equivalence class $\bar 1 = \{1,3\}$.