In the book Character Theory Of Finite Groups by I.Martin Issacs as exercise 2.14
Let $G$ be a finite group with commutator subgroup $G'$. Let $H \subset G' \cap Z(G)$ be cyclic of order n and let m be the maximum of the orders of elements of G/H. Assume that n is a prime power and show that $|G| \geq n²m$.
I have tried, following the hints given, to show, taking $\chi$ to be an irreducible character whose kernel intersects trivially with $H$, that $\chi(1) \geq n$. But, thus far as I am concerned, I see no idea of how to show the inequality, not to mention the latter part of the proof, as suggested, to deploy the Problem 2.9(b) which states that $\chi(1) \leq |G:A|$ for any abelian subgroup A of G.
I have asked my teacher, who replied that one is to show that $|G/H| \geq nm$, of which the proof I have no idea either. Of course, if one can demonstrate the existence of one subgroup of $G/H$ order $\geq nm$, then the problem should be resolved, as the teacher suggested. But this still bewilders me at present.
It was not clear to me from what you wrote whether you had succeeded in proving that $\chi(1) \geq n$ for a character of the type you considered, or whether this is a homework problem. This is the case. Consider an element of $H$ of order $n.$ Let $\sigma$ be an irreducible complex representation affording a character $\chi,$ and assume that $H \cap {\rm ker} \sigma = 1.$ THen $h\sigma$ is a matrix of order $n,$ but must also be a scalar matrix since $\sigma$ is irreducible and $H \leq Z(G).$ Furthermore, ${\rm det} \sigma$ is a $1$-dimensional representation of $G,$ so contains $G^{\prime}$ in its kernel. Since $H \leq G^{\prime},$ we can now conclude that $h\sigma$ is a scalar matrix of order $n$ and determinant $1.$ Let $h \sigma = \omega I$ for some primitive $n$-th root of unity $\omega.$ Then ${\rm det}(h\sigma) = \omega^{\chi(1)}.$ Hence $\omega^{\chi(1)} = 1$ and $\chi(1)$ must be divisible by $n.$ In particular, $\chi(1) \geq n.$ Now let $g \in G$ be an element such that $gH$ has order $m$ in G/H. Let $L = \langle g,H \rangle.$ Then $H \leq Z(L)$ and $L/H$ is cyclic, so $L$ is Abelian, and has order $mn.$ Hence with our character $\chi$ as before, we now have $n \leq \chi(1) \leq [G:L].$ Thus $|G| \geq n|L| \geq n^{2}m.$ I don't think the assumption that $n$ is a prime power is necessary, in fact.