If $d$ is a positive divisor of $n$, the number of elements of order $d$ in a cyclic subgroup of order $n$ is $\phi (d)$=the number of positive natural numbers less than $d$ which are coprime to $d$.
The question I have concerns a part of the proof:
If $d | n$ then there exists exactly one subgroup of order $d$ -- call it $\langle a \rangle$. Then every element of order $d$ also generates the subgroup $\langle a \rangle$ and an element $a^k$ generates $\langle a \rangle$ iff $gcd(k,d) = 1$ implies that the number of such elements is precisely $\phi (d)$.
How does every element of order $d$ also generate the subgroup $\langle a \rangle$, wouldn't it be only one $a$ since $|\langle a \rangle| = |a|$? And how does this fact imply $\phi(d)$ is the correct number?