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?


  • $\begingroup$ 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). $\endgroup$ – Orat Jan 7 '18 at 4:48
  • $\begingroup$ @Orat Many thanks! Yes I use Isaacs notation. Can you please correct and edit the proof which I wrote above? Thanks! $\endgroup$ – Nikita Jan 7 '18 at 5:31
  • $\begingroup$ @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 $g\in\ker\chi$. $\endgroup$ – anon Jan 7 '18 at 6:25
  • $\begingroup$ @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). $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.