Let $G$ be a finite group with normal subgroup $N$. Show that there are at least as many conjugacy classes in $G$ as in $G/N$.
If $a\in G$, then the conjugacy class of $a$ is the set $\{g^{-1}ag:g\in G\}$. I know that if $g_1^{-1}ag_1=g_2^{-1}ag_2$, then $g_2 g_1^{-1}\in C(a)$ (where $C(a)$ is the centralizer of $a$). In other words, $ g_1^{-1}ag_1=g_2^{-1}ag_2$ if $C(a)g_1=C(a)g_2$. Thus the number of elements in the conjugacy class of $a$ is $[G:C(a)]=\frac{|G|}{|C(a)|}$. I think that this fact would help me to answer this exercise, since it relates cosets (the possible elements of $G/N$) with the elements in the conjugacy class of $a$. I have tried to do this in various ways, but so far I have not succeeded. Can you help me please?
Editing (proof writing verification)
Let $C_G$ be the set of conjugacy classes of $G$ and let $C_{G/N}$ be the set of conjugacy classes of $G/N$. Define $\alpha: C_G \rightarrow C_{G/N}$ via $\alpha (C_a)=C_{aN}$.
$C_a=\{g^{-1}ag:g\in G\}, C_{aN}=\{g^{-1}NaNgN:gN\in G/N\}=\{g^{-1}agN:gN\in G/N\}$.
If $C_a\neq C_b$, then
$$ \alpha(C_a C_b)=\alpha(C_{ab})=C_{abN}=C_{aNbN}=C_{aN}C_{bN}= \alpha(C_a )\alpha( C_b)$$
thus $\alpha$ is a homomorphism. It is immediately that $\alpha$ is onto. Also, $\text{Ker}(\alpha)=N$, so if $N=\{e\}$, then $\alpha$ is one-to-one and $G$ and $G/N$ has the same number of conjugacy clases. Otherwise, $G$ has at least as many conjugacy classes as in $G/N$.
Is the proof correct? Could you change some of the notation or is it okay?