Theorem:
If d is a positive divisor of n, the number of elements of order d in a cyclic group of order n is $\phi \left ( d \right )$ where $\phi$ is the euler-phi function defined as the number of elements less than d and relatively prime to d. This by definition is $gcd\left ( d,n\right ) \equiv \phi\left ( d \right )$
The author provides the proof as follows. Proof: By the Fundamental theorem of cyclic group, there exists exactly one subgroup of order d-say, $\left \langle a \right \rangle$.
Then, every element of order d also generates the subgroup $\left \langle a \right \rangle$
The proof continues.
There is an underlying subtlety with the bold. I am unsure if my understanding of the bold is correct.
Note: $\left | a \right |=\left | \left \langle a \right \rangle \right |=d$.
This implies $a=\left \langle a \right \rangle$. Thus, a has order d and any element of order d generates $\left \langle a \right \rangle$
If $C$ is the unique subgroup of order $d$ and $g$ is any element of order $d$, then $|\langle g\rangle|=|C|$ implies $\langle g\rangle=C$ implies $g\in C$. If this is what you mean, then yes.