Show that any normal subgroup is of form $Ker \chi$ for some $\chi \in Char(G)$. $\textbf{The question is as follows:}$

Let $G$ be a finite group and $N$ a normal subgroup of $G$.
  $\rm a)$ Show that $N$ is of the form $ker \xi$ for some $\xi \in Char(G)$.
  $\rm b)$ In $a)$, can one always take $\xi$ to be irreducible?

$\textbf{Some attempt:}$
$\rm a)$ Let $G$ be a finite group and $N \unlhd G.$ Consider then the quotient group $\frac{G}{N}$ and its associated group algebra $A(\frac{G}{N})$. We define then $\rho : G \to A(\frac{G}{N})$ by $(\rho_g(f))(hN) : = f(hNgN)= f(hgN) $. This is a representation since 
$$(\rho_{g_1 g_2 }(f))(hN) = f(hNg_1 g_2 N) = f(hg_1 g_2 N)$$
$\hspace{10.44cm} =f(h g_1 N g_2 N)$
$\hspace{10.44cm} = \rho_{g_2} (f) (hg_1 N)$
$\hspace{10.44cm} = \rho_{g_2} (f) (h N g_1 N)$
$\hspace{10.44cm} = (\rho_{g_1} (\rho_{g_2} (f)))(hN) $
Moreover, it's evident that $N \subseteq Ker \rho$ and moreover if $g \in Ker \rho$, one has then $\rho_g(\delta_N) = \delta_N$ and so in particular $1 = \delta_N (N) = \delta_N(gN)$ and so $N = gN$ and this implies that $g \in N$. It follows that if $\chi_{\rho}$ is the associated character of $\rho$ then $Ker \chi = N.$ So there exists some character $\chi$ of $G$ for which $N = Ker \chi$.
$\rm b)$ About this I think not necessarily. But I am not sure!
Can someone please let me know if I am wrong and to give me a precise answer?(And even I am not sure about $a)$ also!)
Thanks! 
 A: Your answer for part a seems correct. If a group $G$ acts on a set $S$ then there is a representation (respectively a character) induced by this action. Consider the vector space $X$ with the basis $S$, i.e. $$X=\Big\{\sum_{s\in S} c_s\cdot s\mid c_s\in\mathbb{C},\mbox{all but finitely many $c_s$ are $0$}\Big\}.$$ Basically, $X$ is the vector space of all finite formal sums on the set $S$. Since $G$ acts on the basis of this vector space $X$, it also acts on the $X$ itself. You define the action of $G$ on $X$ by $$x\cdot g = \sum_{s\in S} c_s\cdot (s\cdot g)$$ for $x=\sum_{s\in S}c_s\cdot s$. This action is linear, so it induces a representation of $G$. 
Now consider $S=\{Ng\mid g\in G\}$, the set of all cosets of $N$. There is an action of $G$ on $S$ by right multiplication with kernel $N$. It is easy to observe that the induced representation also have the kernel $N$.
For part b it is not necessarily true that every normal subgroup of a group $G$ can be written as the kernel of some irreducible character. Consider the Klein $4$ subgroup $V=\mathbb{Z}_2\times\mathbb{Z}_2$. It has $5$ normal subgroups but only $4$ irreducible characters. In the case of $V$, the trivial subgroup is not the kernel of any irreducible character (that is because there is only one element of order $2$ in the multiplicative group $\mathbb{C}^\times$, which is $-1$, so you cannot embed $V$ into $\mathbb{C}^\times$).
