# Structure of induced module for subgroup $H \le G$, confused by mixing up notions of $FH$-submodule and $F$-subspace

The following is a question on a passage in the book A Course in the Theory of Groups by Derek Robinson. There he constructs the induced module $M^G$ from a $FH$-module for some subgroup $H \le G$ of finite index. The proof shows in some sense that this module is $F$-vector space isomorphic to $M^r$ with $r = |G : H|$, but it starts from a tensor product construction where $M$ and $FG$ are considered as $FH$ modules, but then suddenly speaks about vector space isomorphisms, but I do not see how this is transition is justified. I see that if $M$ and $FG$ have finite $F$-dimension, then the $FH$-dimension of both must be smaller of course, and every direct sum decomposition as $FH$-modules is also a direct sum decomposition as $F$-subspaces (as every $FH$-module obviously is an $F$-vector space) and that the $FH$-bilinearity of the tensor product implies $F$-bilinearity too. But in the argument I do not see if this might be relevant.

Suppose that $G$ is a group and $H$ is a subgroup of $G$ with finite index $r$. Let there be given an $F$-representation $\rho$ of $H$ arising from a right $FH$-module $M$. We proceed to form the tensor product over $FH$, $$M^G = M \otimes_{FH} FG,$$ wherein $FG$ is to be regarded as a left $FH$-module by means of left multiplication. So far $M^G$ is only an $F$-module. However $FG$ is also a right $FG$-module via right multiplication. Consequently $M^G$ becomes a right $FG$-module by means of the rule $$(a\times r)s = a \otimes (rs), \qquad (a \in M, r,s \in FG).$$ The $FH$-bilinearity of the map $(a,r) \mapsto a \oplus (rs)$ shows this action to be well-defined. The right $FG$-module is called the induced module of $M$, and the $F$-representation which arises from $M$ is called the induced representation $\rho^G$ of $\rho$. If $\rho$ has character $\chi$, we shall write $\chi^G$ for the induced character, that is, the character of $\rho^G$.

We plan now to investigate the nature of the module $M^G$. Choosing a right transversal $\{t_1, \ldots, t_r\}$ of $H$ in $G$, we may write each element of $FG$ uniquely in the form $\sum_{i=1}^r u_i t_i$ with $u_i \in FH$. Hence there is a decomposition of $FG$ into left $FH$-modules $FG = (FH)t_1 \oplus \ldots \oplus (FH)t_r$. By the distributive property of tensor products there is an $F$-isomorphism. $$M^G \cong M \otimes_{FH} ((FH)t_1)) \oplus \ldots \oplus M \otimes_{FH} ((FH)t_r).$$ By virtue of the equation $a \otimes ut_i = au \otimes t_i$ we can write the above as $$M^G \cong M \otimes t_1 \oplus \ldots \oplus M\otimes t_r.$$ Hence if $\{ a_1, \ldots, a_n \}$ is a basis of $M$ over $F$, the elements $a_i \otimes t_j, i = 1,2,\ldots, n, j = 1,2,\ldots, r$ form a basis for $M^G$ over $F$.

I do not understand what is happening in the last step. There suddenly everything switches from tensor products of $FH$-modules to tensor products over $F$ (guess if $FH$ is not written at the symbol $\otimes$ this is meant). But how is this possible? How can he suddently talk about $M$ as a $F$-vector space inside the tensor product, and hence claim that $a_i\otimes t_j$ forms a $F$-basis? I see by definition that $M \cong M\otimes t_j$ for all $j = 1,\ldots, r$?

## 1 Answer

A direct sum of $FH$-modules is also a direct sum of vector spaces, so it is enough to consider a single term $M \otimes t_i$ in the decomposition.

Now $M \otimes t_i = \{ m \otimes t_i : m \in M \}$. So, if $a_1,\ldots,a_n$ is a basis of $M$ as a vector space over $F$, then every element of $M$ can be written uniquely as $\sum_{i=1}^n f_i a_i$ with $f_i \in F$. Hence every element of $M \otimes t_i$ can be written uniquely as $\sum_{i=1}^n f_i(a_i \otimes t_i)$, which is saying exactly that $a_1 \otimes t_i,\ldots,a_n \otimes t_i$ is a basis of $M \otimes t_i$ as a vector space over $F$.