# Character related to a maximal subgroup

I am trying to prove the following statement:

If $H$ be a maximal subgroup of $G$ and $\xi=(1_H)^G$, where $1_H$ is principal character of $H$, and $\chi$ be a non-principal irreducible constituent of $\xi$, then $ker(\xi)=ker(\chi)$.

I can see that $ker(\xi)\le ker(\chi)$, but for the $ker(\xi)\ge ker(\chi)$ I could not get enough useful results.

I will be so thankful for your help in solving this problem.

• Why did you delete your earlier question today "About faithful irreducible representation"? – Derek Holt Jan 2 '16 at 17:52
• @DerekHolt: By your hint, I found that the question was so easy and I should not ask such questions. – A-213 Jan 2 '16 at 17:59
• @DerekHolt: If you think that the question should not be deleted, I can undelete it. – A-213 Jan 2 '16 at 18:24

Here is an outline solution. $\ker \xi$ is the core of $H$ in $G$ (i.e. the largest normal subgroup of $G$ that is contained in $H$), so if $\ker \chi \le H$ then $\ker \chi \le \ker \xi$.
Otherwise, since $H$ is maximal in $G$, we have $G=KH$ and so$\chi_H$ is irreducible. But by Frobenius Reciprocity, $\langle \chi_H,1_H \rangle >0$, so $\chi = 1_H$.
• Thank you so much for your helpful answer. Would you please explain this part a little more: we have $G=KH$ and so $\chi_H$ is irreducible ? – A-213 Jan 2 '16 at 20:52
• $KH$ is a subgroup of $G$ with $H < KH \le G$,so $G=KH$ follows from the fact that $H$ is a maximal subgroup of $G$. If $M$ is the ${\mathbb C}G$-module corresponding to the character $\chi$, then any ${\mathbb C}H$-submodule would be invaraint under the kernel $K$ and so it would also be a ${\mathbb C}G$-submodule. So $M$ must be irreducible (i.e. simple) as a ${\mathbb C}H$-subdule, which is equivalent to $\chi_H$ being irreducible. – Derek Holt Jan 2 '16 at 22:58
• May be I am missing something, but, isn't $\xi$ is same as $\mathbb{C}[G/H]$ as $G$-module ($G$ acts by permuting the cosets) and so $\ker \xi\subseteq G-H$ (since every element of $G$ other than $H$ moves every coset). – pritam Jan 4 '16 at 0:44
• On the contrary, as I said before $\ker \xi \le H$. The kernel consists of those elements of $G$ that fix every coset of $H$. – Derek Holt Jan 4 '16 at 20:00
• $\ker(\xi) = \{g \in G \mid \xi(g) = {\rm Id} \}$. – Derek Holt Jan 5 '16 at 9:26