This is a question from a problem set on representation theory.
Let $G$ be finite and $K \lhd G$. It is simple to show that irreducible characters of $G/K$ correspond to irreducible characters of $G$.
Now, given information about $G/K$, what can I say about the number of characters of degree 1 of $G$? By the proof in the link above, the number of characters of degree 1 of $G$ must be larger than or equal to the number of characters of degree one of $G/K$, but is there any more interesting relation between these two numbers? Considering we are dealing with quotient groups, I feel like there should be some sort of divisibility relation going on (something like: the number of char. of degree 1 of $G/K$ divides that of $G$), but I'm not sure how I would even approach showing this.
Your hypothesis is correct. The number of linear characters – ie characters of degree $1$ – of $G$ is equal to the index $[G:G']$ of the derived subgroup $G'\leqslant G$. (If you would like a proof of this let me know.) Now, taking derived subgroups commutes with quotients (why?), so the image of $G'$ in $G\big/K$ – ie $G'K\big/K$ – is precisely $\left(G\big/K\right)'$. Thus the number of linear characters of $G\big/K$ is $$m:=\frac{\left|G\big/K\right|}{\left|\left(G\big/K\right)'\right|}=\frac{\left|G\big/K\right|}{\left|G'K\big/K\right|}=\frac{|G|\big/|K|}{|G'K|\big/|K|}=\frac{|G|}{|G'K|},$$ while the number of linear characters of $G$ is $n:=|G|\big/|G'|$. Since $G'\leqslant G'K$, we have $|G'|\ \big|\ |G'K|$ by Lagrange's theorem, and so indeed $m\mid n$, as desired.