Eilenberg, at NSF summer institute, Bowdoin College:

  Why mix left regular representation of G with right cosets of H?
  If you need room for a good involutory permutation τ, just make room —
add another point ∞ to the space G/H of left cosets gH of H , forming E = G/H ∪ {∞} ,
have τ ∈ E! interchange H with ∞ but leave the rest of G/H alone. Then do as before:
  Write κ_τ: E! → E! for conjugation by the involution τ — κ_τ(σ) = τ·σ·τ ; compose left-regular
representation ρ: G → (G/H)! ⊂ E! ({ρ(g)}(xH) = gxH , {ρ(g)}(∞) = ∞ ) with κ_τ: E! → E! ;
for h ∈ H and C ∈ G/H ∪ {∞} ( C = xH (x ∈ G) or C = ∞ ) , calculate {κ_τ·ρ(h)}(C) =
τ({ρ(h)}(τ(C))) = {ρ(h)}(C) , so that κ_τ·ρ and ρ agree on H .
  Now let g ∈ G , and suppose κ_τ·ρ(g) = ρ(g) . Then {κ_τ·ρ(g)}(H) = {ρ(g)}(H) = gH . But in fact
{κ_τ·ρ(g)}(H) = τ({ρ(g)}(τ(H))) = τ({ρ(g)}(∞)) = τ(∞) = H . So gH = H , and g ∈ H .
  Thus H ⊆ G is the equalizer of κ_τ·ρ and ρ . In fact, inclusion η: (G/H)! ⊂ E! and composition
κ_τ·η: (G/H)! ⊂ E! → E! have equalizer {π ∈ (G/H)! | π(H) = H} . Pf.: π ∈ (G/H)! , η(π) = {κ_τ·η}(π)
⇒ π(H) = {η(π)}(H) = {{κ_τ·η}(π)}(H) = {τ·η(π)·τ}(H) = {τ·η(π)}(∞) = τ(∞) = H , QED.

---

[(Joyal, Categories, 6/’12): H ⊆ G as equalizer of κ_τ·ρ and ρ is pullback of this along ρ_o: G → (G/H)! ]