Irreducible Characters

Question

I tried to solve the following questions but I got stuck. I need to some hints to continue.

1. If the induced character of an irreducible character of $H$ is again irreducible, then center of the induced character must lie in $H$. But how can I prove that?

2. $H$ and $K$ are two subgroups of $G$. $\varphi$ is a character of $H$. If restriction to $K$ of induced character of $\varphi$ is irreducible, $HK=G$.

Answer 1

Proposition Let $\vartheta$ be a character of $H \subseteq G$. Then $$Z(\vartheta^G)=core_G(Z(\vartheta))=\bigcap_{x \in G}Z(\vartheta)^x$$

Proof (all references are to the book of I.M. Isaacs, Character Theory of Finite Groups) Let $\chi=\vartheta^G$. Then $g \in Z(\chi)$ iff $|\sum_{x \in G}\vartheta^{\circ}(xgx^{-1})|=\sum_{x \in G}\vartheta(1)$. Since $|\vartheta^{\circ}(xgx^{-1})| \leq \vartheta(1)$ (Lemma (2.15)) we get by the triangle inequality, $$\sum_{x \in G}\vartheta(1)=|\sum_{x \in G}\vartheta^{\circ}(xgx^{-1})| \leq \sum_{x \in G}|\vartheta^{\circ}(xgx^{-1}| \leq\sum_{x \in G}\vartheta(1).$$ So we must have equality everywhere, yielding $|\vartheta^{\circ}(xgx^{-1})|= \vartheta(1)$ for all $x \in G$. This happens iff $g \in Z(\vartheta)^x$ for all $x \in G$.

Corollary Let $H$ be a subgroup of $G$, and $\chi \in Irr(G)$, such that $\chi=\vartheta^G$ for a character $\vartheta$ of $H$ (note that this character necessarily must be irreducible). Then $Z(G) \subseteq Z(\chi) \subseteq H$.

Proof Since $\chi$ is irreducible it follows from Corollary (2.28) that $Z(G) \subseteq Z(\chi)$. But $Z(\chi)=core_G(Z(\vartheta)) \subseteq Z(\vartheta) \subseteq H$. See also Problem (5.12) in Isaacs' book.

Different proof Write $Z=Z(\chi)$, recall that $Z \unlhd G$ and that by Lemma(2.27)(c) $\chi_Z=\chi(1)\lambda$ for some linear character $\lambda$ of $Z$. Note that $\vartheta^{HZ}$ is irreducible and lies under $\chi$. Then $\vartheta^{HZ}|_Z$ is a multiple of $\lambda$ and hence $\vartheta^{HZ}|_Z=\vartheta^{HZ}(1) \lambda$. It follows that $|HZ:H|\vartheta(1)=\vartheta^{HZ}(1)=[(\vartheta^{HZ})_Z,\lambda]=(Mackey)[(\vartheta_{H \cap Z})^Z, \lambda]=[\vartheta_{H \cap Z},\lambda_{H \cap Z}] \leq \vartheta(1)$, where the last inequality follows from $\lambda$ being an irreducible constituent of $\vartheta_{H \cap Z}$. We conclude that $HZ=H$, that is $Z(\chi) \subseteq H$.

Answer 2

Here is the proof of 2. I urge you to draw a diagram to depict the different groups and characters.

Let $H$ and $K$ be subgroups of $G$ and $\varphi$ a character of $H$. Assume that $(\varphi^G)_K \in Irr(K)$. Then $G=HK$.

Proof Put $\varphi^G=\chi$. Observe that since $(\varphi^G)_K$ is irreducible, both $\varphi$ and $\chi$ must be irreducible. Let $\psi \in Irr(H \cap K)$ be an irreducible constituent of $\varphi_{H \cap K}$.

By Frobenius Reciprocity, $\varphi$ must be an irreducible constituent of $\psi^H$, say $\psi^H=a\varphi+\Delta$, with $a$ a positive integer and $\Delta$ a character of $H$ with $[\Delta,\varphi]=0$ or $\Delta=0$. It follows that $\psi^G=(\psi^H)^G=a\varphi^G+\Delta^G=a\chi+\Delta^G$. So again by Frobenius Reciprocity, $[\psi^G,\chi]=[(\psi^K)^G,\chi]=[\psi^K,\chi_K] \geq a.$

On the other hand, $a=[\varphi,\psi^H]=[\varphi_{H \cap K},\psi]$, so $\varphi_{H \cap K}=a\psi+\Gamma$ with $\Gamma$ a character of $H \cap K$ with $[\Gamma,\psi]=0$ or $\Gamma=0$. Hence $(\varphi_{H \cap K})^K=a\psi^K+\Gamma^K$ and it follows that $[(\varphi_{H \cap K})^K, \chi_K]=[a\psi^K+\Gamma^K,\chi_K]=a[\psi^K,\chi_K]+[\Gamma^K,\chi_K] \geq a^2$, by the last formula of the previous paragraph. This means that the irreducible character $\chi_K$ has at least multiplicity $a^2$ as an irreducible constituent of $(\varphi_{H \cap K})^K$. This implies that $$(\varphi_{H \cap K})^K(1)=\varphi(1)|K:H \cap K| \geq a^2\chi(1) \geq \chi(1)=\varphi(1)|G:H|$$ It follows that $|G| \leq \frac{|H| \cdot |K|}{|H \cap K|}=|HK|$. Hence, $G=HK$, since $HK \subseteq G$ as a set.$\square$

Notes (1) This is Problem(5.7) in Isaacs' book. (2) From the last part of the proof it follows that in fact $a=1$.