Let $H$ be a proper subgroup of $G$ such that for all $\chi\in Irr(G)$, $\chi_H\in Irr(H)$.
That is, the restriction of every irreducible character of $G$ to $H$ is an irreducible character of $H$.
What can be said in that case ? Is there any theorem related to this case ?
Theorem Let $G$ be a finite group with a proper subgroup $H$, such that all irreducible characters of $G$ restrict irreducibly to $H$. Then the following hold.
$(a)$ $H$ is a normal subgroup and $G/H$ is abelian
$(b)$ Every irreducible character of $H$ is $G$-invariant
$(c)$ $k(G)=|G:H|k(H)$ (here $k(X)$ denotes the number of conjugacy classes of a group $X$)
$(d)$ $Z(G)=C_G(H)$ and $[G,H]=H'$
$(e)$ If $H$ is a Hall-subgroup, then $H$ is a direct factor of $G$.
Proof (a) Since $H$ is proper, $(1_H)^G=1_G+\sum a_{\chi}\chi$, with $a_{\chi}$ positive integers, with at least one $\chi \in Irr(G)$. Assume $\chi \in Irr(G|1_H)$, then by Frobenius Reciprocity, $0 \lt a_{\chi}=[(1_H)^G,\chi]=[1_H,\chi_H]$. Since $\chi_H$ is irreducible we must have $\chi_H=1_H$, so $a_{\chi}=1$, $\chi$ is linear and $H \subseteq ker(\chi)$. Hence, $(1_H)^G$ is the sum of all linear characters $\lambda$ of $G$ with $H \subseteq ker(\lambda)$. This implies that $ker((1_H)^G)=core_G(H)=\bigcap \{ker(\lambda): \lambda_H=1_H\} \supseteq H$. Hence $H=core_G(H)$, that is $H$ is normal and of course $G' \subseteq \bigcap \{ker(\lambda): \lambda_H=1_H\}$, since each $\lambda$ here is linear.
(b) This follows from (a) and Clifford Theory (see for example (6.2)Theorem in Isaacs' CTFG).
(c) We can count the irreducible characters of $H$ and $G$, applying Gallagher's Correspondence Theorem (see (6.17)Corollary in CTFG): let $\chi \in Irr(G)$ and $\chi_H=\vartheta \in Irr(H)$ then $Irr(G|\vartheta)=\{\lambda\chi: \lambda \in Irr(G/H)\}$. Since $G/H$ is abelian, this yields $\#Irr(G)=|G/H|\cdot \#Irr(H)$.
(d) It is easy to prove (using Schur's Lemma) that if $\chi \in Irr(G)$ with $\chi_H \in Irr(H)$, then $C_G(H) \subseteq Z(\chi)$. Since $\bigcap_{\chi \in Irr(G)} Z(\chi)=Z(G)$ and obviously $Z(G) \subseteq C_G(H)$, the assertion follows. For the second statement remember that every linear $\lambda \in Irr(H)$ is $G$-invariant by (b). So for all $g \in G, h \in H$ we have $\lambda(ghg^{-1}h^{-1})=\lambda^g(h)\lambda(h^{-1})=\lambda(h)\lambda(h^{-1})=1$. Hence $[G,H] \subseteq ker(\lambda)$. Since $H'=\bigcap \{ker(\lambda):\lambda \in Irr(H), \lambda(1)=1\}$ and of course $H' \subseteq [G,H]$, we are done.
(e) Assume that $H$ is a Hall-subgroup. Then owing to the Schur-Zassenhaus Theorem, $H$ has a complement $A$, with $G=HA$ and $H \cap A=1$. Note that $A$ is abelian. Let $P \in Syl_p(A)$, then by (a) $P$ acts (coprimely) on $H$ by conjugation. By (b) $P$ fixes every element of $Irr(H)$, and hence we can apply Brauer's Permutation Lemma (see (6.32)Theorem in CTFG) and it follows that $P$ fixes every $H$-conjugacy class. Since for every $h \in H$ gcd$(|Cl_H(h)|,p)=1$, $Cl_H(h)$ must contain a fixed point, so an element centralized by $P$. Or otheriwse put $Cl_H(h) \cap C_H(P) \neq \emptyset$. For every conjugacy class $C$ of $H$ pick an $h_C$ fixed by $P$ and put $K=\langle h_C: C \text{ is an } H-\text{conjugacy class} \rangle \leq H$. Then obvioulsy $H=\bigcup_{h \in H}K^h$ and a classical result implies that in fact $H=K$. Hence $H$ is centralized by $P$ and since $A$ is generated by its Sylow-subgroups, it follows that $A$ centralizes $H$, so $G=H \times A$.
Note Probably more can be said on the type of group pairs $(G,H)$ above. One thing that is worth mentioning is the "dual" (restriction of characters vs. induction) of the above property. The following was proved by Eric Kuisch (see Homogeneous Character Induction, J. Alg. 149, (1992), 454-471, Theorem(1.2))
Theorem Let $G$ be a group and $H$ a proper non-trivial subgroup. Then the following are equivalent.
$(a)$ Each non-trivial irreducible character of $H$ induces irreducibly to $G$
$(b)$ $G$ is a Frobenius group with kernel $H$.
This is quite surprising: $(b) \Rightarrow (a)$ was well-known (see for example (6.24)Theorem in CTFG), but the reverse implication was unknown at that time!