How to bound the order of a finite group under the following hypotheses? 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.  
 A: 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.
A: Just to follow up on the assumption that $n$ is a prime-power, I think it is used here, as awllower said, to ensure that such an irreducible character $\chi$ of $G$ exists.
Let $h \in H$ be a generator of the cyclic group $H$ of order $n=p^k$. Assume for contradiction that $H \cap $ker$\chi > 1$ for every irreducible character $\chi \in$ Irr(G). For each $\chi$, this intersection is a cyclic subgroup of $H$ generated by $h^{p^l}$ for some $0 \leq l < k $. In fact, the subgroup generated with the maximum such $l$ is contained in every ker$\chi$. But this contradicts the fact that the intersection of all ker$\chi$ is trivial (Lemma 2.21.).
