Set-up: Consider a finite group $G$ and a subgroup $K$. Given these, we can look at the induced representation $\mathrm{Ind}_K^G 1$ (here, $1$ is the trivial $1$-dim complex representation of $K$). As a vector space, this is $$ \mathrm{Ind}_K^G 1 := \{f\colon G\to\mathbb{C}\ |\ f(gk)=f(g)\ \forall g\in G,k\in G\}. $$ We can also define the Hecke algebra $$ H(G,K) := \{f\colon G\to\mathbb{C}\ |\ f(kgk')=f(g)\ \forall g\in G,k,k'\in K\}. $$ These both clearly lie in $\mathrm{Fun}(G) := \{f\colon G\to\mathbb{C}\}$, the space of all complex-valued functions on $G$. This space clearly has a basis of delta functions $\{\delta_g\}_{g\in G}$, where each $\delta_g\colon G\to\mathbb{C}$ is given by $$ \delta_g(x) := \begin{cases} 1,\ &\text{if}\ x=g, \\ 0,\ &\text{else.}\end{cases} $$ We can endow this vector space with the convolution product $$ (f\star f')(g) := \sum_{xy=g} f(x)f'(y). $$ Then $(\mathrm{Fun}(G),\star)$ is a unital associative $\mathbb{C}$-algebra.
We also have the group algebra $\mathbb{C}[G]$ with basis elements $\{e_g\}_{g\in G}$ and multiplication $e_ge_h:=e_{gh}$.
Now there's also clearly an algebra isomorphism $$ \Phi\colon \mathrm{Fun}(G)\to\mathbb{C}[G],\quad \delta_g\mapsto e_g. $$ My question is: What is the image of $\mathrm{Ind}_K^G 1$ and $H(G,K)$ inside of $\mathbb{C}[G]$?
I believe that $\Phi(\mathrm{Ind}_K^G 1) = \mathbb{C}[G]e$ and $\Phi(H(G,K))=e\mathbb{C}[G]e$, where $$ e := \frac{1}{|K|}\sum_{k\in K} e_k, $$ which is an idempotent in $\mathbb{C}[G]$. I'm not sure why this is true. How can I see this?
My answer so far: I've shown that $\mathbb{C}[G]e\subseteq \Phi(\mathrm{Ind}_K^G 1)$.
To see this, take an arbitrary element $(\sum_{g\in G} a_ge_g)e$ in $\mathbb{C}[G]e$, where $a_g\in\mathbb{C}$. Then notice $$ \bigg(\sum_{g\in G} a_ge_g\bigg)e = \frac{1}{|K|}\bigg(\sum_{g\in G} a_ge_g\bigg)\bigg(\sum_{k\in K}e_k\bigg) = \frac{1}{|K|}\sum_{\substack{g\in G \\ k\in K}} a_ge_ge_k = \frac{1}{|K|}\sum_{\substack{g\in G \\ k\in K}} a_ge_{gk}. $$ Then we apply $\Phi^{-1}$ to see that $$ \frac{1}{|K|}\sum_{\substack{g\in G \\ k\in K}} a_ge_{gk} \mapsto \frac{1}{|K|}\sum_{\substack{g\in G \\ k\in K}} a_g\delta_{gk}. $$ We wish to show that this lies in $\mathrm{Ind}_K^G 1$, so we wish to check that this map is invariant under right-multiplication by an element of $K$. To this end, let $g'\in G$, $k'\in K$ and apply $\frac{1}{|K|}\sum_{\substack{g\in G \\ k\in K}} a_g\delta_{gk}$ to $g'k'$. Note that $\delta_{gk}(g'k')=1$ if and only if $gk=g'k'$ (and $0$ otherwise). This is equivalent to $g=g'k'k^{-1}$. Thus $$ \frac{1}{|K|}\sum_{\substack{g\in G \\ k\in K}} a_g\delta_{gk}(g'k') = \frac{1}{|K|}\sum_{k\in K} a_{g'k'k^{-1}}\delta_{g'k'}(g'k') = \frac{1}{|K|}\sum_{k\in K} a_{g'k'k^{-1}}. $$ Similarly, we apply the map $\frac{1}{|K|}\sum_{\substack{g\in G \\ k\in K}} a_g\delta_{gk}$ to $g'$. This yields $$ \frac{1}{|K|}\sum_{\substack{g\in G \\ k\in K}} a_g\delta_{gk}(g') = \frac{1}{|K|}\sum_{k\in K} a_{g'k^{-1}} $$ Since right-multiplication by an element of $K$ is an automorphism of $G$, we see that $\frac{1}{|K|}\sum_{k\in K} a_{g'k'k^{-1}}=\frac{1}{|K|}\sum_{k\in K} a_{g'k^{-1}}$ which shows that $\mathbb{C}[G]e\subseteq \Phi(\mathrm{Ind}_K^G 1)$.
I'm stuck on the reverse subset relation.
For the Hecke algebra, I believe the following result is enough: if $e\neq 0$ is an idempotent in a ring $R$, then $\mathrm{End}_G(eR)$ is ring isomorphic to $eRe$.
Note: The above is irrelevant for the Hecke algebra question. I see that the Hecke algebra image follows the same idea as the image of the induced representation.
Given your attempt so far, I'll do the other direction of the first statement in a similar style. There are of course many ways to organize it.
Claim: $\Phi(\mathrm{Ind}_K^G 1) \subseteq \mathbb{C}[G]e$.
Idea: if $x = ye$, then $xe = ye^2 = ye = x$, so $x$ itself will be a "witness" that it is in $\mathbb{C}[G]e$.
Proof: Let $f \in \mathrm{Ind}_K^G 1$, so $f \colon G \to \mathbb{C}$ where $f(gk) = f(g)$ for all $g \in G, k \in K$. Write $f = \sum_{g \in G} f(g) \delta_g$, so $\Phi(f) = \sum_{g \in G} f(g) e_g$. Then
\begin{align*} \Phi(f)e &= \frac{1}{|K|} \sum_{k \in K} \sum_{g \in G} f(g) e_{gk} \\ &= \frac{1}{|K|} \sum_{k \in K} \sum_{g \in G} f(gk^{-1}) e_g \\ &= \frac{1}{|K|} \sum_{k \in K} \sum_{g \in G} f(g) e_g \\ &= \sum_{g \in G} f(g) e_g \\ &= \Phi(f), \end{align*}
where the second equality comes from reindexing the inner sum using $g \mapsto gk^{-1}$. $\Box$