I would like to prove the following:
We have a group $G$ and: $$\chi:G\rightarrow \mathbb{C}$$$$ \tilde{\chi}: G/N \rightarrow\mathbb{C}$$ characters of groups $G$ and $G/N$ respectively, where $\tilde{\chi}$ is given by $\tilde{\chi}(gN)=\chi(g)$ and additionally: $N\leq \ker(\chi)$. Prove that $\chi$ is a irreducible character if and only if $\tilde{\chi}$ is a irreducible character of $G/N$.
My try:
$$\text{(for irreducible characters)}\quad 1=(\chi,\chi)=\frac{1}{|G|}\sum_\limits{g \in G} \chi(g)\overline{\chi(g)}= $$
$$\text{Here we use the definition } \tilde{\chi}(gN)=\chi(g) $$ $$=\frac{1}{|G|}\sum_\limits{g \in G} \tilde{\chi}(gN)\tilde{\chi}(gN)=\frac{1}{|G/N||N|}\sum_\limits{g \in G} \tilde{\chi}(gN)\tilde{\chi}(gN)=$$ $$\text{Here I'd like to make the substitution: } \sum_\limits{g \in G} \rightarrow \sum_\limits{gN \space \in \space G/N} $$
$$\text{My asumption:} \sum_\limits{g \in G}=|N|\sum_\limits{gN \space \in \space G/N}$$
$$=\frac{1}{|G/N||N|} |N|\sum_\limits{gN \space \in \space G} \tilde{\chi}(gN)\tilde{\chi}(gN)=\frac{1}{|G/N|} \sum_\limits{gN \space \in \space G} \tilde{\chi}(gN)\tilde{\chi}(gN)=(\tilde{\chi},\overline{\tilde{\chi}})=1$$
My Question:
Is my assumption regarding the substitution above correct? May I substitute a sum $\sum_\limits{g \in G}$ with $|N|\sum_\limits{gN \space \in \space G/N}$? Because intuitively I know that to each element $gN \in G/N$ in the sum $\sum_\limits{gN \space \in \space G/N}$ correspond $|N|$ elements $g \in G$ in > the sum $\sum_\limits{g \in G}$, see for example the table below:
Example borrowed from: https://kconrad.math.uconn.edu/blurbs/grouptheory/coset.pdf .
You need to use that $N \subseteq ker(\chi)$, which implies that $\chi$ is constant on cosets of $N$. For let $\frak{X}$ be a representation that affords $\chi$, then for all $g \in G$, $n \in N$, one has $\frak{X}$$(gn)$=$\frak{X}$$(g)$$\frak{X}$$(n)$=$\frak{X}$$(g)$$\frak{I}$=$\frak{X}$$(g)$, whence $\chi(gn)=\chi(g)=\chi(ng)$. Now choose a transversal set $T$ for $N$ in $G$ (meaning $T$ contains exactly one element of each coset of $N$). Observe that $G=\bigcup_{t \in T,n \in N}tn$ and write $\overline{g}$ for $gN$. Then $$[\tilde\chi,\tilde\chi]_{G/N}=\frac{1}{|G:N|}\sum_{\overline{g} \in G/N}|\tilde\chi(\overline{g})|^2=\frac{|N|}{|G|}\sum_{\overline{g} \in G/N}|\chi(g)|^2\\=\frac{|N|}{|G|}\sum_{t \in T}|\chi(tn)|^2=\frac{|N|}{|G|}\sum_{t \in T}|\chi(t)|^2=\frac{|N|}{|G|}\frac{1}{|N|}\sum_{n \in N}\sum_{t \in T}|\chi(t)|^2=\\\frac{1}{|G|}\sum_{n \in N}\sum_{t \in T} \chi(tn)=\frac{1}{|G|}\sum_{g \in G}|\chi(g)|^2=[\chi,\chi]_G=1.$$