Frobenius reciprocity I would like to ask a question on Theorem 8.6 on page 246 in this book.
There is the claim that

the multiplicity of $F$ in $E^G$ is equal to the multiplicity of $E$ in $F_H$.

Why is this just true for algebraically closed fields $k$?

$F$ is a simple $k[G]$ module, $E$ a simple $k[H]$ module, $G$ a group and $H$ a subgroup.
 A: The problem is that the dimension of a Hom (as in Theorem 8.5, and the discussion at the top of page 246) depends not only on the multiplicity, but also on the dimension of the splitting field. The splitting fields can change size, especially when the action of $G$ on $H$ induces a Galois automorphism.
For example: Let $G$ be nonabelian of order 6, $H$ normal of order 2, $k$ the rational numbers, and $E$ the 2 dimensional simple module of $H$ over $k$. Then the induced module $E^G$ is the direct sum of two copies of the 2 dimensional simple module $F$ of $G$ over $k$.
$E^G = F \oplus F$, but $F_H = E$.
The multiplicity of $F$ in $E^G$ is 2, but the multiplicity of $E$ in $F_H$ is 1.
A: Here is a more detailed answer comparing multiplicities over non-algebraically closed fields as requested in the comments.$\newcommand{\Hom}{\operatorname{Hom}}\newcommand{\Soc}{\operatorname{Soc}}\newcommand{\Head}{\operatorname{Head}}$
One version of Frobenius reciprocity is quite general and is also known as the adjoint functor theorem.

Frobenius reciprocity: Let $R$ be a commutative ring with identity, $H \leq G$ groups with $[G:H]$ finite. For any $R[H]$-module $E$ and any $R[G]$-module $F$, we have an $R$-module isomorphism: $$\Hom_{RH}(E,F_H) = \Hom_{RH}(E,\Hom_{RG}(RG,F)) \cong \Hom_{RG}( E \otimes_{RH} RG,F ) = \Hom_{RG}(E^G, F)$$

However, the clearest applications of this require significantly stronger hypothesis: we want $E$ and $F$ to be simple modules, so WLOG $R=k$ is a field (the kernel of the $R$ action on $E$ and $F$ are maximal ideals; if they are different ideals then all the Homs involved are 0). Once we have simple modules things are nicer:

Counting using Hom: If $X$ is an $R[H]$-module, and $E$ is a simple $R[H]$-module, then $$\Hom_{RH}(E,X) = \Hom_{RH}(E,\Soc_E(X)) = \Hom_{RH}(E,E^n) \cong \Hom_{RH}(E,E)^n = \Delta_E^n$$ and its dimension is $e\cdot n$ where $e=[\Delta_E:k]$ is the dimension of the endomorphism ring, and $n$ is the multiplicity of $E$ as a composition factor of the socle of the $R[H]$-module $X$. Similarly, $$\Hom_{RG}(Y,F) = \Hom_{RG}(\Head_F(Y),F) = \Hom_{RG}(F^m,F) \cong \Hom_{RG}(F,F)^m = \Delta_F^m$$ and its dimension is $f \cdot m$ where $f=[\Delta_F:k]$ is the dimension of its endomorphism ring, and $m$ is the multiplicity of $F$ as a composition factor of the head of the $R[G]$-module $Y$. Setting $X=F_H$ and $Y=E^G$, we thus have $$fm = en$$ where $e,f$ count dimensions of endomorphisms, and $m,n$ count multiplicities of composition factors.

If $k$ is a splitting field for both $H$ and $G$ then $\Delta_E=\Delta_F=k$ and $e=f=1$, so Frobenius reciprocity equates two counts of composition factors $m=n$ (one at the bottom, and one at the top). If the characteristic of $k$ does not divide the order of (the now finite group) $G$, then $E^G$ and $F_H$ are both semi-simple, so equal to their socle and heads. In this case, we get the very simple interpretation of the dimension as the count of one module inside another. This is the classical Frobenius reciprocity.
Failure of counting: If we assume $G$ is finite and the characteristic of $k$ does not divide the order of $G$, then we can have some slightly odd things happen with $e$ and $f$. In particular, it is quite possible for $f=1$ while $e>1$. For instance, $G=S_3$, $k=\mathbb{Q}$, $F=k^2$ the natural module, and $H=A_3$. Then $E=F_H$ remains irreducible over $\mathbb{Q}$, but splits over $\Delta_E=\mathbb{Q}[\zeta_3]$. Here we have $e=2$ and $f=1$, messing up the counts so that $m>n$ ($m=2$ and $n=1$).
Do we at least have an inequality? What about the other direction? Do we know $n 
\leq m$, or equivalently that $f \leq e$? Unfortunately, no. Take $G=A_3$, $H=1$, $k=\mathbb{Q}$, and $F$ to be the non-split simple module for $k[G]$. Then $F_H = E^2$ where $E=k$ is the only simple $k[H]$-module (which just so happens to be absolutely simple). $E^G = F \oplus k$ for $k$ being the trivial $k[G]$-module. Hence $e=1$, $f=2$, and $n=2$ and $m=1$.
