Frobenius' formula for the character of an induced representation

Choose a left transversal $T$ for $G\mathrm{/}H$. By [F2], ${\mathrm{Ind}}_H^{G}W\cong {⨁}_{t\in T}{W}_t$, where each $W_t$ is one copy of $W$ indexed by the coset representative $t$.

For $t\in T$, write $gt=t^{\mathrm{\prime }}h$ with $t^{\mathrm{\prime }}\in T$ and $h\in H$. Under the identification of step 1.1, the action of $g$ sends the $t$-summand to the $t^{\mathrm{\prime }}$-summand; if $t^{\mathrm{\prime }}=t$, so $t^{-1}gt=h\in H$, then this action on $W_t$ is exactly the action of $h=t^{-1}gt$ on $W$. Therefore the contribution of the $t$-summand to the trace is $\chi (t^{-1}gt)$ when $t^{-1}gt\in H$, and $0$ otherwise.

The trace of $g$ on the direct sum of step 1.1 is the sum of the traces on the summands fixed by the permutation it induces on $T$. Hence ${\mathrm{Ind}}_H^G\chi (g)={\sum }_{t\in T,t^{-1}gt\in H}\chi (t^{-1}gt)$.

Fix $t\in T$ with $t^{-1}gt\in H$. The elements of the left coset $tH$ are $x=th$ with $h\in H$, and then $x^{-1}gx=h^{-1}{t}^{-1}gth$. By [F3], the character value $\chi (x^{-1}gx)$ is therefore the constant $\chi (t^{-1}gt)$ on that whole coset, and every element of $tH$ contributes to the displayed sum exactly when $t^{-1}gt\in H$. So the total contribution of $tH$ to ${\sum }_{x^{-1}gx\in H}\chi (x^{-1}gx)$ is $\mathrm{\mid }H\mathrm{\mid }\chi (t^{-1}gt)$.

Summing the identity of step 4.1 over the distinct cosets indexed by $T$, and reindexing by the finite partition $G={⨆}_{t\in T}tH$, gives ${\sum }_{x\in G,x^{-1}gx\in H}\chi (x^{-1}gx)=\mathrm{\mid }H\mathrm{\mid }{\sum }_{t\in T,t^{-1}gt\in H}\chi (t^{-1}gt)$. By step 3.1 and [F4], dividing by $\mathrm{\mid }H\mathrm{\mid }$ yields the stated Frobenius formula.