Prove that $\phi \left(\frac{a^b-1}{c}\right)$ is divisible by $b$ What is the difference between the two questions below?

Prove that if $a,b,$ and $c$ are positive integers and $c \leq a$ where $\dfrac{a^b-1}{c}$ is an integer, then  $\phi \left(\frac{a^b-1}{c}\right)$ is divisible by $b$.
Consider the integer $d = \frac{a^b-1}{c}$, where $a, b$, and $c$ are positive integers and $c \le a.$ Prove that the set $G$ of integers that are between $1$ and $d$ and relatively prime to $d$ (the number of such integers is denoted by $\phi(d)$) can be partitioned into $n$ subsets, each of which consists of $b$ elements.

I am confused how to go about proving this since $\dfrac{a^b-1}{c}$ is a weird looking fraction and we are taking the totient of it.

Let $d=\frac{a^b-1}{c}$. Notice $\varphi(d)$ is the order of the multiplicative group of $\mathbb Z_d$.
What is the order of $a$ in $\mathbb Z_d^*$?
We have $a^b-1|a^b-1\implies \frac{a^b-1}{c}|a^b-1 \implies a^b \equiv 1 \bmod \frac{a^b-1}{c}$
Now notice $a^{b-1}-1\leq \frac{a^b}{c}-1< \frac{a^b-1}{c}$, so the order cannot be less than $b$.
So the order of $a$ is $b$ and so, by Lagrange's theorem $b|\varphi(d)$
