Exercise on Induced Representations of 1 dimensional complex representation I'm having a hard time trying to solve the following problem coming from Eingof's book "Introduction to representation theory" (page 55 of the book in PDF format http://math.mit.edu/~etingof/replect.pdf):
Problem 5.8.5:

Let $ K \subset G$ be a finite Group, and let $ \chi : K \stackrel{}{\rightarrow} \mathbb{C} $ be a homomorphism, with $ \mathbb{C}_{\chi} $ be the corresponding 1-dim. representation of K, with
$$ e_{\chi} = \frac{1}{|K|} \sum_{g \in K} \chi(g)^{-1}g \in \mathbb{C}[K] $$
the idempotent corresponding to $ \chi. $ Show that the G-representation of $ Ind_{K}^{G} \mathbb{C}_{\chi} $ is naturally isomophic to $ \mathbb{C}[G]e_{\chi} $. (With G acting by left multiplication)

I think that my main problem is related to the fact that I cannot truly understand how left multiplication works in $ \mathbb{C}[G]e_{\chi} $, in order to distinguish it from $ \mathbb{C}[G] $, i.e understanding what the role of $e_{\chi}$ is.
Because then my idea, although not very elegant, was to  define my function $f:G\rightarrow \mathbb{C}$ explicitly in some natural way (e.g look at the coefficients of the g's, or something like that) from what I have on the right hand side, such that the action of left multiplication on the RHS behaves well with the one of the induced representation $g(f)(x):=f(xg)$
Recall that: $ Ind_{K}^{G} \mathbb{C}_{\chi}:=\left\{ f:G\rightarrow \mathbb{C}|f(hx)=\rho_V(h)f(x) \ \forall h\in H,x\in G \right\} $ with the just above defined action
Can someone give me a hint?
 A: Note that the vector $e_{\chi}$ acts on $\mathbb{C}[G]$ by right multiplication, i.e. the map $T: v\mapsto v e_{\chi}$ is linear. The space $\mathbb{C}[G]e_{\chi}$ that we are interested in is the subspace $\text{im}(T) \subset \mathbb{C}[G]$. When Etingof says $G$ acts by left multiplication on $\mathbb{C}[G]e_{\chi}$, this makes sense: because for any $v\in \text{im}(T)$, we know $v = Tw = we_{\chi}$ for some $w\in \mathbb{C}[G]$, and then $v \mapsto gv$ is an action since $$gv = gw e_{\chi} = T(gw) \in \text{im} (T)$$
Now, choose a set $\{g_1, \cdots, g_d\}$ of coset representatives for $K$. We can think of $\text{Ind}_{K}^{G} \mathbb{C}$ as the span of the symbols $\{v_{g_i}\}$. Here, an element $g\in G$ acts as follows. If $gg_i K = g_j K$, then there is some $k\in K$ such that $gg_i = g_j k$. This yields $g\cdot v_{g_i} = \chi(k) v_{g_j}$, giving an action of $g$ on the basis, hence on the whole space.
At last, we are in a position to give the isomorphism between $\text{Ind}_{K}^{G} \mathbb{C}$ and $\mathbb{C}[G]e_{\chi}$. Define $\phi: \text{Ind}_{K}^{G} \mathbb{C} \to \mathbb{C}[G]e_{\chi}$ as $\phi(v_{g_i}) = g_i e_{\chi}$. Extending $\phi$ linearly gives a linear map between the two spaces. First, we show $\phi$ is $G$-linear.
Suppose $gg_i = g_j k$. Then $$\phi(g\cdot v_{g_i}) = \phi(\chi(k)v_{g_j}) = \chi(k) g_j e_{\chi}$$ $$g\cdot T(v_{g_i}) = g(g_i e_{\chi}) = g_j k e_{\chi} = g_j \left(\frac{1}{|K|} \sum_{g\in K} \chi(g^{-1}) kg \right) = g_j \chi(k) \left(\frac{1}{|K|} \sum_{g\in K} \chi(g^{-1} k^{-1}) kg \right)= \chi(k) g_j e_{\chi} $$ where we used that $\chi(kg^{-1}k^{-1}) = \chi(g^{-1})$.
Next, we show $\phi$ is injective. Suppose that $\phi\left(\sum a_i v_{g_i} \right) = 0$. Since $$\phi\left(\sum a_i v_{g_i} \right) = \sum_{i=1}^{d} a_i g_i e_{\chi} = \frac{1}{|K|}\sum_{g \in G, g = g_i k \text{ where } k\in K} a_i \chi(k^{-1}) g$$ As the elements $\{g\}_{g\in G}$ are linearly independent in $\mathbb{C}[G]$, this implies all the $a_i$ are 0.
Last, we show $\phi$ is surjective. For any $v = (\sum_{g\in G} a_g g)e_{\chi}$ we can rewrite this as $$\sum_{g\in G, g = g_i k \text{ where } k\in K} a_g g_i k e_{\chi} \in \text{span}(g_1 e_{\chi}, \cdots, g_d e_{\chi})$$ since, as we saw earlier, $ke_{\chi} = \chi(k) e_{\chi}$. Thus, we can choose scalars $c_i$ such that $\phi\left(\sum c_i v_{g_i} \right) =v$.
Hence, $\phi$ is $G$-linear, injective, and surjective: it is an isomorphism of representations.
