I've been studying the book of Isaacs of Character Theory of Finite Groups.
Clifford's Theorem (Theorem 6.2 of the book) states the following.
Theorem: Let $H\lhd G$ and let $\chi\in\text{Irr}(G)$. Let $\vartheta$ be an irreducible constituent of $\chi_H$ (the restriction of $\chi$ to the subgroup $H$) and suppose $\vartheta_1,\ldots,\vartheta_t$ are the distinct conjugates of $\vartheta$ in $G$. Then $\chi_H=$[$\chi_H$,$\vartheta$]$\sum_{i=1}^t\vartheta_i$.
The proof which Isaacs gives is the following
Proof: We compute $(\vartheta^G)_H$. For $h\in H$, we have $\vartheta^G(h)=\frac{1}{|H|}\sum_{x\in G}\vartheta^\circ(xhx^{-1})=\frac{1}{|H|}\sum_{x\in G}\vartheta^x(h)$ since $xhx^{-1}\in H$ for all $x\in G$. Thus $|H|(\vartheta^G)_H=\sum_{x\in G}\vartheta^x$ and hence if $\varphi\in\text{Irr}(H)$ and $\varphi\not\in\{\vartheta_i\}$, we have $0=[\sum\vartheta^x,\varphi]$ and therefore $[(\vartheta^G)_H,\varphi]=0$. Since $\chi$ is a constituent of $\vartheta^G$ by Frobenius reciprocity, it follows that $[\chi_H,\varphi]=0$. Thus all irreducible constituents of $\chi_H$ are among the $\vartheta_i$ and $\chi_H=\sum_{i=1}^t[\chi_H,\vartheta_i]\vartheta_i$. However, $[\chi_H,\vartheta_i]=[\chi_H, \vartheta]$ and the proof is complete.
But I don't see why the hypothesis of irreducibility of $\chi$ is necessary, since I see no difference with the proof taking $\chi$ an arbitrary character of $G$. I wonder what I'm missing.
Best Answer
The irreducibility of $\chi$ is necessary to assert that $\chi$ is a constituent of $\vartheta^G$. By Frobenius reciprocity we have that $0\not=[\chi_H,\vartheta]=[\chi,\vartheta^G]$. Then $\chi$ is a constituent of $\vartheta^G$ or $\vartheta^G$ is a constituent of $\chi$. If $\chi\in\text{Irr}(G)$ then necessarily $\chi$ is a constituent of $\vartheta^G$. If $\chi\not\in\text{Irr}(G)$ then $\vartheta^G$ may be a constituent of $\chi$, and in this case we can't guarantee $[\chi_H,\varphi]=0$. For example, consider $H=G$, a character $\chi\not\in\text{Irr}(G)$ which is not a multiple of an irreducible character, and $\vartheta,\varphi$ distinct irreducible constituents of $\chi_H=\chi$. Then $\vartheta^G=\vartheta$ and $\vartheta^G$ is a constituent of $\chi$. Since $\vartheta$ is $G$-invariant we have that $\varphi\not\in\{\vartheta_i\}$ and $[\chi_H,\varphi]=[\chi,\varphi]\not=0$.