I am studying the proof of Clifford's Theorem 2.2.3 presented in A. Zimmerman's "Representation Theory: an homological algebra point of view", pag. $\sim$ 171.
Let $N\le G$ a normal subgroup of a finite group and $I_G(M)$ be the inertia subgroup of the $KN$-indecomposable module $M$ in $G$. We assume further that $$M\uparrow^{I_G(M)}_N \simeq M_1\oplus\dots\oplus M_r$$
for $M_i$ indecomposable $KI_G(M)$-module.
It seems everything clear except one thing:
at a certain point he says that
$$M_i\uparrow_{I_G(M)}^G\downarrow_{N}^G\simeq \bigoplus_{gI_G(M)\in G/I_G(M)}(\mbox{}^gM)^{n_i}$$
for suitable $n_i$ integers. This is clear. Next he says that, since $\mbox{}^{g_1}M\simeq \mbox{}^{g_2}M$ if and only if $g_1I_G(M)=g_2I_G(M)$, we must have $M_i\downarrow_{N}^{I_G(M)}\simeq M^{n_i}$.
The problem is that I can prove that $M_i\downarrow_{N}^{I_G(M)}$ is indeed a certain power of $M$, but I can not understand why this power must be exactly the $n_i$ that we found above. It does not seems trivial to me: in particular, I can not understand how to use the fact he mentioned about cosets.
The easiest thing here is just to argue that $n_i$ and the multiplicity of $M$ in $M_i\downarrow _N ^{I_G(M)}$ equal $\dim M_i/ \dim M$.
First, $\dim M_i \downarrow_N = \dim M_i$, and $\dim M^k=k\dim M$, so $M_i\downarrow_N$ must be $(\dim M_i/\dim M)$ copies of $M$.
On the other hand, consider $M_i \uparrow ^G \downarrow _N$. Its dimension is $|G:I_G(M)| \dim M_i$, and the direct sum you wrote down has dimension $n_i |G:I_G(M)| \dim M$, so we have to have $n_i=\dim M_i/\dim M$.