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Notice that when $N$ is a normal subgroup of $G$ with $G/N$ non-Abelian, it may not be straightforward to obtain the precise value of $m$, the maximum number of conjugacy classes of any subgroup of $G/N$ (though of course we always have $m < [G:N]$ given that $G/N$ is non-Abelian).

Notice that when $N$ is a normal subgroup of $G$ with $G/N$ non-Abelian, it may not be straightforward to obtain the precise value of $m$, the maximum number of conjugacy classes of any subgroup of $G/N$ (though of course we always have $m < [G:N]$ given that $G/N$ is non-Abelian. Later edit: in fact, it is an easy exercise to check that a non-Abelian group $X$ has at most $|X|-3$ conjugacy classes, and if it has $|X|-3$ conjugacy classes, then $|X| \in \{6,8 \}).$

Applying Gallagher's inequalities with $H = I_{G}(\mu)/N$, we obtain that $k(I_{G}(\mu)/N) \leq [G:I_{G}(\mu)]k(G/N).$ Notice that $[G:I_{G}(\mu)]$ is the length of the $G$-orbit of $\mu$. Letting $\mu$ run through orbit representatives for the action of $G$ on irreducible characters of $N$, we obtain another inequality of P.X. Gallagher (also proved independently by H. Nagao around 1962): $k(G) \leq k(N)k(G/N).$

Applying Gallagher's inequalities with $I_{G}(\mu)/N$ in the role of $H$, and $G/N$ in the role of $G$, , we obtain that $k(I_{G}(\mu)/N) \leq [G:I_{G}(\mu)]k(G/N).$ Notice that $[G:I_{G}(\mu)]$ is the length of the $G$-orbit of $\mu$. Letting $\mu$ run through orbit representatives for the action of $G$ on irreducible characters of $N$, we obtain another inequality of P.X. Gallagher (also proved independently by H. Nagao around 1962): $k(G) \leq k(N)k(G/N).$

Recalling that $k_{G}(N)$ is the number of $G$-conjugacy classes contained in $N$, we see that whenever $N$ is a normal subgroup of $G$ with $G/N$ Abelian, then we have $k(G) \leq (k(G/N)$ $\times$ (the number of conjugacy classes of $G$ contained in $N$). This answers the question in the case $[G:N] =2$

Recalling that $k_{G}(N)$ is the number of $G$-conjugacy classes contained in $N$, we see that whenever $N$ is a normal subgroup of $G$ with $G/N$ Abelian, then we have $k(G) \leq k(G/N)$ $\times$ (the number of conjugacy classes of $G$ contained in $N$). This answers the question in the case $[G:N] =2$