Question

If we let $k(G)$ denote the number of conjugacy classes of the finite group $G$, then for any subgroup $H$ of $G$ (normal or not), a Theorem of P.X. Gallagher states that $[G:H]^{-1}k(H) \leq k(G) \leq [G:H]k(H)$ (this has probably been discovered and rediscovered many times). I find that the easiest way to see it is using irreducible complex characters. Of course, $k(G)$ is also the number of complex irreducible characters of $G$, and likewise for $H$. For each irreducible character $\chi$ of $G$, there is an irreducible character $\mu$ of $H$ such that $$\langle {\rm Res}^{G}{H}(\chi),\mu \rangle \neq 0.$$ Then by \mu$ occurs with non-zero multiplicity in the restriction of $\chi$ to $H$. By Frobenius reciprocity, $\chi$ is an irreducible constituent of the character of $${\rm Ind}^{G}{H}(\mu).$$ G$ induced from the character $\mu$ of $H$. On the other hand, using Frobenius reciprocity again, each irreducible constituent of ${\rm Ind}^{G}{H}(\mu)$ \mu$ induced to $G$ must have degree at least $\mu(1).$ Thus there are (even including multiplicities) at most $[G:H]$ irreducible constituents of $${\rm Ind}{H}^{G}(\mu).$$ \mu$ induced to $G$ Since this is true for each irreducible character of $H$, and since each irreducible character of $G$ must appear with non-zero multiplicity in at least one such character, we have $k(G) \leq [G:H]k(H).$ Going in the other direction, if $\chi$ is an irreducible character of $G$ and $\mu$ is an irreducible constituent of te restriction of $${\rm Res}^{G}{H}(\chi),$$ \chi$ to $H$, then we have $\chi(1) \leq [G:H]\mu(1)$, since $\chi$ occurs as a constituent of $${\rm Ind}{H}(G)(\mu)$$. \mu$ induced to $G$. Thus $\mu(1) \geq \frac{\chi(1)}{[G:H]}$. Hence there are at most $[G:H]$ distinct irreducible constituents of the restriction of $${\rm Res}^{G}_{H}(\chi).$$ \chi$ to $H$. Since each irreducible character of $H$ occurs as a constituent of some such irreducible character of $G$ (consider, for example, the restriction of the regular character), we have $k(H) \leq [G:H] k(G).$