Induced Representation of finite groups Can induced representation be constructed without using tensor products?
Pls help me out.
 A: The easiest way to describe induction is that it is the extension of scalars functor $\mathbf{C}[G] \otimes_{\mathbf{C}[H]} -$. This is an abstract way of characterizing the following "elementary" construction (c.f. Serre, "Linear Representations of Finite Groups", 3.3).
Let $H \leqslant G$ be a subgroup. Then by Lagrange's Theorem, $G$ is partitioned into cosets of $H$. Let $S$ be a set of coset representatives. I.e. $G = \bigcup_{s \in S} sH$ and if $s, t \in S$ are such that $sH = tH$ then $s = t$. It is sometimes convenient to suppose that the representative of $H$ is the identity of the group. That is, if $s \in S$ is such that $sH = H$ then $s = e_G$.
Now let $W$ be any representation of $H$. Form the vector space
$$ W' = \operatorname{Ind}_H^G W := \bigoplus_{s \in S} sW. $$
This consists of formal expresions $sw$ with $s \in S$ and $w \in W$.
$W'$ is a representation of $G$ in the following way. Let $sw \in W'$ and $g \in G$. Then $gsH = s'H$ for a unique $s' \in S$. Thus $gs = s'h$ for a unique $s' \in S$ and $h \in H$. We define $g(sw) = s'(hw)$. The action of $H$ on $W$ means that $hw \in W$ is just another vector in $W$.
A: It can also be constructed ("co-induction") from the Hom-functor rather
than the tensor product. In the language of representations rather than modules, let $W$ be a representation of $H$ (acting on the right). Define
$V=\text{Map}_H(G,W)$ the set of $H$-equivariant maps from $G$ to $W$,
that is those $f:G\to W$ with $f(gh)=f(g)h$ for all $h\in H$. Then $G$
acts on $V$ via $(f\cdot g):g'\mapsto f(g^{-1}g')$.
When you look at infinite groups, induction and co-induction give different
results in general.
