Group algebra of a quotient group 
[Definition] Let $K$ be a field and $G$ be a group. Group algebra $K[G]$ is the set of all linear combinations of finitely many elements of $G$ with coefficients in $K$.

If we are given a normal subgroup $N$ of $G$, since the quotient $G/N$ is also a group we can consider group algebra $K(G/N)$. I want to know the relationship between $KG$ and $K(G/N)$. I suppose $K(G/N)$ is a quotient group of $KG$, for example $K(G/N)\simeq KG/KN$. I'm not sure if my supposition is true. Can anyone explain this in detail?
 A: You can consider the projection map 
$\pi: G\to G/N$
that induces a natural morphism of algebra 
$\pi’: KG\to K(G/N)$ 
that maps each polynomial $p=\sum_{g\in G}k_gg$, $|\{k_g: k_g\neq 0\}|<\infty$, to the polynomial 
$\pi’(p):=\sum_{g\in G}k_ggN$ 
This map is clearly surjective. We consider $p\in K_N$, $\sum_{h\in N}k_hh$. Then 
$\pi’(p)= \sum_{h\in N}k_hhN=\left(\sum_{h\in N}k_h\right)N$
that is not zero in general. This means that we must consider the quotient 
$K\left(G/N\right)/(N)$
In this case we get that 
$KN\subseteq \ker(\pi’)$
where $\pi’$ is defined as 
$\pi’:KG\to K\left(G/N\right)/(N)$
Now we consider a polynomial $p\in KG$, $p=\sum_{g\in G}k_g g$,  such that $\pi’(p)=0$. Then there exists $\lambda \in K$ for which
$\sum_{g\in G}k_g gN=\lambda N$
So
$ \sum_{g\not \in N}k_ggN+\left(\sum_{g\in N}k_g-\lambda\right)N=$
$=\sum_{g\not \in N}k_ggN+\sum_{g\in N}k_gN-\lambda N$
$=\sum_{g\in G}k_g gN-\lambda N=0$
This means that 
$k_g=0$ for each $g\not \in N$ and 
$\sum_{g\in N}k_g=\lambda$
So 
$p=\sum_{g\in G}k_gg=\sum_{g\in N}k_gg\in KN$ 
In other way we have $\ker(\pi’)=KN$. This implies 
$KG/KN\cong K\left(G/N\right)/(N)$
