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).$2 added 10 characters in body If we let$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 Frobenius reciprocity,$\chi$is an irreducible constituent of${\rm ${\rm Ind}^{G}{H}(\mu).$ H}(\mu).$$On the other hand, using Frobenius reciprocity again, each irreducible constituent of {\rm Ind}^{G}{H}(\mu) must have degree at least \mu(1). Thus there are (even including multiplicities) at most [G:H] irreducible constituents of {\rm {\rm Ind}{H}^{G}(\mu). H}^{G}(\mu).$$ 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 ${\rm${\rm Res}^{G}{H}(\chi)$, H}(\chi),$$then we have \chi(1) \leq [G:H]\mu(1), since \chi occurs as a constituent of {\rm {\rm Ind}{H}(G)(\mu). H}(G)(\mu)$$. Thus$\mu(1) \geq \frac{\chi(1)}{[G:H]}$. Hence there are at most$[G:H]$distinct irreducible constituents of${\rm Res}^{G}_{H}(\chi)$.${\rm Res}^{G}_{H}(\chi).$$Since each irreducible character of H occurs as a constituent of some such irreducible character of G (consider, for eaxmpleexample, the restriction of the regular character), we have k(H) \leq [G:H] k(G). 1 If we let k(G) denote the number of conjugacy classes of the finite group G, then for any subgroup H of G (normal or not), a Theorem of P.X. Gallagher states that [G:H]^{-1}k(H) \leq k(G) \leq [G:H]k(H) (this has probably been discovered and rediscovered many times). I find that the easiest way to see it is using irreducible complex characters. Of course, k(G) is also the number of complex irreducible characters of G, and likewise for H. For each irreducible character \chi of G, there is an irreducible character \mu of H such that$$\langle {\rm Res}^{G}{H}(\chi),\mu \rangle \neq 0. Then by Frobenius reciprocity, $\chi$ is an irreducible constituent of ${\rm Ind}^{G}{H}(\mu).$ On the other hand, using Frobenius reciprocity again, each irreducible constituent of ${\rm Ind}^{G}{H}(\mu)$ must have degree at least $\mu(1).$ Thus there are (even including multiplicities) at most $[G:H]$ irreducible constituents of ${\rm Ind}{H}^{G}(\mu)$. 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 ${\rm Res}^{G}{H}(\chi)$, then we have $\chi(1) \leq [G:H]\mu(1)$, since $\chi$ occurs as a constituent of ${\rm Ind}{H}(G)(\mu)$. Thus $\mu(1) \geq \frac{\chi(1)}{[G:H]}$. Hence there are at most $[G:H]$ distinct irreducible constituents of ${\rm Res}^{G}_{H}(\chi)$. Since each irreducible character of $H$ occurs as a constituent of some such irreducible character of $G$ (consider, for eaxmple, the restriction of the regular character), we have $k(H) \leq [G:H] k(G).$