[proofplan]
Both sides are class functions of $G$, so it suffices to compare their values at every $g \in G$. The [Transversal Formula for Induced Class Functions](/theorems/2448) writes $\operatorname{Ind}^G_H 1_H(g) = \sum_{i=1}^n \mathring{1}_H(t_i^{-1} g\, t_i)$, where each summand is $1$ when $t_i^{-1} g t_i \in H$ and $0$ otherwise. The sum thus counts the indices $i$ with $t_i^{-1} g t_i \in H$. Translating $t_i^{-1} g t_i \in H \iff g \in t_i H t_i^{-1}$, and observing that $t_i H t_i^{-1}$ is the stabiliser in $G$ of the coset $t_i H$ under left multiplication, this count equals the number of cosets fixed by $g$ — which is by definition the value of the [Permutation Character](/theorems/2434) $\pi_X$ at $g$.
[/proofplan]
[step:Set up the left-multiplication action of $G$ on $X = G/H$ and its permutation character $\pi_X$]
Let $X = G/H = \{t_1 H, \ldots, t_n H\}$ be the set of left cosets of $H$ in $G$, where $t_1, \ldots, t_n$ is a left transversal of $H$ in $G$ and $n = |G : H|$. The group $G$ acts on $X$ by left multiplication:
\begin{align*}
G \times X &\to X \\
(g, tH) &\mapsto (gt) H.
\end{align*}
This is a group action: $1 \cdot tH = tH$, and $(g_1 g_2) \cdot tH = (g_1 g_2 t) H = g_1 \cdot (g_2 \cdot tH)$.
The permutation character of this action is the function
\begin{align*}
\pi_X: G &\to \mathbb{C} \\
g &\mapsto |\mathrm{fix}_X(g)| = |\{i \in \{1, \ldots, n\} : g \cdot t_i H = t_i H\}|.
\end{align*}
This $\pi_X$ is the character of the permutation representation $\mathbb{C} X = \bigoplus_{i=1}^n \mathbb{C} \cdot t_i H$ on which $G$ acts by permuting the basis $\{t_1 H, \ldots, t_n H\}$ — see [Permutation Character](/theorems/2434).
[/step]
[step:Apply the transversal formula to $\operatorname{Ind}^G_H 1_H(g)$]
The trivial character $1_H: H \to \mathbb{C}$ takes the constant value $1$, so its zero extension to $G$ is the indicator of $H$:
\begin{align*}
\mathring{1}_H: G &\to \mathbb{C} \\
y &\mapsto \mathbb{1}_H(y) = \begin{cases} 1 & y \in H, \\ 0 & y \notin H. \end{cases}
\end{align*}
By the [Transversal Formula for Induced Class Functions](/theorems/2448), with $\psi = 1_H$,
\begin{align*}
\operatorname{Ind}^G_H 1_H(g) = \sum_{i=1}^n \mathring{1}_H(t_i^{-1} g\, t_i) = \sum_{i=1}^n \mathbb{1}_H(t_i^{-1} g\, t_i) = \bigl|\bigl\{i : t_i^{-1} g\, t_i \in H\bigr\}\bigr|.
\end{align*}
Each summand contributes $1$ exactly when $t_i^{-1} g\, t_i \in H$, and $0$ otherwise; the total is the count of such indices.
[/step]
[step:Identify $\{i : t_i^{-1} g\, t_i \in H\} = \{i : g\, t_i H = t_i H\}$]
Fix $i \in \{1, \ldots, n\}$. We claim
\begin{align*}
t_i^{-1} g\, t_i \in H \quad \Longleftrightarrow \quad g\, t_i H = t_i H.
\end{align*}
**Forward direction ($\Rightarrow$).** Suppose $t_i^{-1} g\, t_i \in H$. Then $g\, t_i = t_i (t_i^{-1} g\, t_i) \in t_i H$. So $g\, t_i$ lies in the coset $t_i H$, which means $g\, t_i H = t_i H$ (cosets either coincide or are disjoint).
**Backward direction ($\Leftarrow$).** Suppose $g\, t_i H = t_i H$. Then $g\, t_i \in t_i H$, so $g\, t_i = t_i h$ for some $h \in H$, hence $t_i^{-1} g\, t_i = h \in H$.
Therefore the count in Step 2 equals
\begin{align*}
\bigl|\bigl\{i : t_i^{-1} g\, t_i \in H\bigr\}\bigr| = \bigl|\{i : g \cdot t_i H = t_i H\}\bigr| = |\mathrm{fix}_X(g)| = \pi_X(g).
\end{align*}
[guided]
The bridge $t_i^{-1} g t_i \in H \iff g \cdot t_i H = t_i H$ is the conceptual key. It identifies "how the induced character sees $g$" (via conjugation into $H$) with "how the permutation action sees $g$" (via fixed cosets).
**Why is this equivalence true?** Both statements say that $g$ stabilises the coset $t_i H$. The stabiliser of $t_i H \in X$ under the left-multiplication action is
\begin{align*}
\operatorname{Stab}_G(t_i H) = \{g \in G : g \cdot t_i H = t_i H\} = t_i H t_i^{-1}.
\end{align*}
The last equality holds because $g \cdot t_i H = t_i H \iff g \in t_i H t_i^{-1}$ (multiply by $t_i^{-1}$ on the right and by $t_i$ on the left). Equivalently, $g \in t_i H t_i^{-1} \iff t_i^{-1} g\, t_i \in H$. So the two formulations are linked by an algebraic identity for stabilisers in transitive $G$-sets.
**Coset-stabiliser identity in plain terms.** The stabiliser of the coset $tH$ in $G/H$ is the conjugate subgroup $t H t^{-1}$. This is a foundational fact about transitive $G$-sets: stabilisers of points in the same orbit are conjugate, and for the canonical orbit $G/H$ the stabilisers are precisely the conjugates of $H$.
**What the formula says structurally.** $\operatorname{Ind}^G_H 1_H(g)$ counts the cosets fixed by $g$. This is the number of fixed points of $g$ acting on $G/H$ by left multiplication — exactly what the permutation character $\pi_X$ measures.
[/guided]
[/step]
[step:Conclude $\operatorname{Ind}^G_H 1_H = \pi_X$ as class functions on $G$]
By Steps 2 and 3, for every $g \in G$,
\begin{align*}
\operatorname{Ind}^G_H 1_H(g) = |\mathrm{fix}_X(g)| = \pi_X(g).
\end{align*}
Two functions on $G$ that agree pointwise are equal, so $\operatorname{Ind}^G_H 1_H = \pi_X$ as elements of $\mathcal{C}(G)$. This completes the proof.
[/step]
