# Show that every normal subgroup can be written as an intersection of $\ker \chi$'s for $\chi \in \operatorname{Irr}(G).$

The question is as follows:

This can be the third part of this question Linked to the results:

Let $G$ be a finite group and $N$ a normal subgroup of $G$.

c) Show that $N$ is the intersection of the sets of the form $\ker \xi$ that contain $N$ with $\xi \in \operatorname{Irr}(G)$.

Some attempts:

For the finite group $G$ and $\rho$ a representation with induced character $\chi_{\rho}$, we have $\ker \rho = \{ g \in G: \chi_{\rho}(g) = \chi_{\rho}(e) \}$ by Lemma 15.17; Isaacs "Algebra: A Graduate Course". Then it makes sense to define the kernel of the character $\chi$, denoted $\ker \chi$ by $\ker \chi = \{ g \in G: \chi(g) = \chi(e) \}$.

In particular, we know that for every character $\chi$ one has that $\ker \chi \unlhd G.$ For the irreducible characters $\chi^{(\alpha)}$, $\alpha \in \hat{G}$, we give the special symbols $N^{(\alpha)}$ for $\ker \chi^{(\alpha)}$. Now what we are going to show is that knowing $N^{(\alpha)}$ for every $\alpha \in \hat{G}$ enables one to know $\ker \chi$ for every character $\chi$. Indeed, if we let $\chi$ be a character with representation as a linear combination of the irreducible characters $\chi = \sum_{\alpha \in \hat{G}} m^{(\alpha)}\chi^{(\alpha)}$, then we have $$\ker \chi =\bigcap \{ N^{(\alpha)} : m^{(\alpha)} > 0 \}.$$ Because if $\chi^{(\alpha)} (g) = d_{\alpha}$ for every $\alpha$ such that $m^{(\alpha)} > 0$ one sees that $$\chi (g) = \sum_{\substack{\alpha \in \hat{G}\\ m^{(\alpha)} > 0 }} m^{(\alpha)} \chi^{(\alpha)} (g) = \sum_{\substack{\alpha \in \hat{G}\\ m^{(\alpha)} > 0 }} m^{(\alpha)} \chi^{(\alpha)} (e) = \chi (e)$$ and so $g \in \ker \chi$.

Conversely, since one evidently has that $|\chi^{(\alpha)} (g)| \le d_{\alpha}$ for every $\alpha \in \hat{G}$ we see that for $g \in \ker \chi$ one has that \begin{align} |\chi (g)| &= \left| \sum_{\substack{\alpha \in \hat{G}\\ m^{(\alpha)} > 0 }} m^{(\alpha)} \chi^{(\alpha)} (g) \right|\\ &\le \sum_{\substack{\alpha \in \hat{G}\\ m^{(\alpha)} > 0 }} m^{(\alpha)} \left| \chi^{(\alpha)} (g) \right| \\& \le \sum_{\substack{\alpha \in \hat{G}\\ m^{(\alpha)} > 0 }} m^{(\alpha)} d_{\alpha} \\ &= \chi(1) = \chi(g) \end{align} from where it follows from this that $\chi^{(\alpha)} (g)$ must be real and so if $\chi^{(\alpha)} (g) \le d_{\alpha}$ for any $\alpha \in \hat{G}$ then this would induce a strict inequality for $\chi(g)$ and $\chi(1)$. It follows that $\chi^{(\alpha)} (g) = d_{\alpha}$ for any $\alpha \in \hat{G}$ such that $m^{(\alpha)} >0$. Then the conclusion follows.

Can someone please let me know if I am wrong and we cannot get the result through what I wrote?

Thanks!

• 1. What is $\hat{G}$? If you use it for the group of all linear characters as Isaacs (1994, xii) does, then a character in general cannot be written as a linear combination of those in general. 2. What is $d_\alpha$? If $d_\alpha = \chi^{(\alpha)}(e)$ then why you have $\chi^{(\alpha)}(g) = d_\alpha$? See also Isaacs (1994, p.23). – Orat Jan 7 '18 at 4:48
• @Orat Many thanks! Yes I use Isaacs notation. Can you please correct and edit the proof which I wrote above? Thanks! – Nikita Jan 7 '18 at 5:31
• @Orat (1) Yes every character of a complex representation of a finite group can be written as an integer-linear combination of characters of irreducible representations. This is because every representation is a direct sum of irreducible representations. Why do you claim otherwise? (2) In which part of Nikita's argument are you referencing the $d_\alpha=\chi^{(\alpha)}(e)$ equation? In the first part, Nikita is assuming $g$ is in $\bigcap$ of the $N^{(\alpha)}$s (which automatically implies the equation you are asking about) and then concluding $g\in\ker\chi$. – anon Jan 7 '18 at 6:25
• @anon Well, for (1), I wrote linear, not irreducible. For (2), it seems that it's my misunderstanding of his argument (anyway he/she used $g$ without any explanation). – Orat Jan 7 '18 at 7:31

Your answer may be corrected; however it is easier for me to rewrite a proof because it lacks definitions (e.g. what are $d_\alpha$ and $\chi^{(\alpha)}$?).
Proof. Set $\operatorname{Irr}(G | N) := \{\, \xi \in \operatorname{Irr}(G) \mid \ker \xi \ge N \,\}$. It is well-known that $\operatorname{Irr}(G | N)$ can be identified with $\operatorname{Irr}(G/N)$ by $$\operatorname{Irr}(G/N) \to \operatorname{Irr}(G | N), \quad \psi \mapsto \tilde\psi$$ which is defined by $\tilde\psi(g) = \psi(gN)$ for $\psi \in \operatorname{Irr}(G/N)$ and $g \in G$. Thus we have $$\bigcap_{\xi \in \operatorname{Irr}(G | N)} \ker \xi = \bigcap_{\psi \in \operatorname{Irr}(G/N)} \ker \tilde\psi.$$
Applying second part of Lemma 2.21 of Isaacs (1994) to $G/N$ (i.e. $\bigcap_{\psi \in \operatorname{Irr}(G/N)} \ker \psi = \{ N \}$) yields $$\bigcap_{\psi \in \operatorname{Irr}(G/N)} \ker \tilde\psi = N.$$ Thus we get $N = \bigcap_{\xi \in \operatorname{Irr}(G | N)} \ker \xi$ as expected.