"all-or-nothing" lemma

Proposition

Let $G$ be a finite group and $N$ be a normal subgroup of G. Let $\chi$ be an irreducible character of $G$ afforded by a representation $\mathfrak{X}$. Let $\hat{N}$ denote the element ${\sum }_{n\in N}n$ in the group algebra $\mathbb{C}[G]$.

If the normal subgroup $N$ is not contained in the kernel of χ , then the matrix $\mathfrak{X}(\hat{N})$ is the zero matrix.

\begin{proof} Since $N$ is a normal subgroup, the element $\hat{N}$ lies in the center of $\mathbb{C}[G]$. Thus $\mathfrak{X}(\hat{N})$ commutes with every element of $\mathbb{C}[G]$ and so $\mathfrak{X}(\hat{N})=\lambda I$ by Schur lemma.

Note that $\mathrm{tr}(\mathfrak{X}(\hat{N}))=\lambda \cdot \chi (1)$ and

$$\mathrm{tr}(\mathfrak{X}(\hat{N}))=\sum _{n\in N}\chi (n).$$

On the other hand, notice that

$$⟨{\chi }_N,{\chi }_1⟩=\frac{1}{|N|}\sum _{n\in N}\chi (n)$$

where ${\chi }_N=\chi {\downarrow }_N^G$ and ${\chi }_1$ is the trivial character of $N$. Then we have $\lambda \chi (1)=|N|⟨{\chi }_N,{\chi }_1⟩$.

Remark that $⟨{\chi }_N,{\chi }_1⟩>0$ iff there exists $v\in V$ such that $\mathfrak{X}(n)v=v$ for all $n\in N$ iff there exists non-trivial $W\subset V$ such that $\mathfrak{X}(n)w=w$ for all $w\in W$. Furthermore,

$$\mathfrak{X}(g^{-1}ng)w=w⟹\mathfrak{X}(n)(\mathfrak{X}(g)x)=\mathfrak{X}(g)x$$

for all $g\in G$ and so $N\subseteq \ker \mathfrak{X}$, leading to a contradiction. (Gemini insists to call it “all-or-nothing” lemma. )

Therefore, $⟨{\chi }_N,{\chi }_1⟩=0$ and so $\lambda =0$. Now we finish the proof. \end{proof}