This is exercise 3B.6 of Isaacs' "Finite Group Theory"; more specifically, item $(a)$. It goes:
Let $N \lhd G$ and let $g \in G$, where $G$ is a finite group. Suppose $Ng$ has order $m$ in the group $G/N$. Prove there is an element $h \in Ng$ such that every prime divisor of $o(h)$ is also a prime divisor of $m$.
Isaacs actually gives a hint to this problem - he suggests finding a cyclic $\pi$-group $C$ such that $NC = N\langle g \rangle$, where $\pi$ is the set of prime divisors of $m$.
I managed to prove that this hint indeed solves the problem, but I don't know where to look for the group $C$. My idea was to somehow "cut away" the redundant part of $\langle g \rangle$, which is the intersection of $N$ and $\langle g \rangle$. For this, I tried to quotient $\langle g \rangle$ by $N \cap \langle g \rangle$ and use the correspondence theorem, but to no avail. I also tried a subgroup of $\langle g \rangle$ directly, since $m$ divides the order of $g$, but I also don't see how to proceed...
This is in the chapter about complements, and the Schur-Zassenhaus Theorem, but $N$ is not a Hall subgroup of $G$, so I don't even know if such a complement exists - my guess is that those results are going to be important for $(b)$ and $(c)$, but not for $(a)$.
Any help is appreciated! Thanks in advance!