---
rg: 2
id: pbh-graph-product-closure-proof
kind: route
title: Realize X *_C (C x K) in Aut_H(H * F_2) by transvections of one letter and a twisted conjugation of the other, then induct over vertices
target: pbh-class-closed-under-graph-products
requires:
  - boone-higman-type-a-class-closed-under-finite-extensions
  - fp-simple-highly-transitive-groups-satisfy-pbh
  - relative-automorphism-of-fp-simple-mif-group-has-type-a-action
  - type-a-action-gives-boone-higman-for-subgroups
---

Automorphisms compose right to left. `B_A` is as in the target.

## Step 0. A simple MIF envelope

Let `X, K ∈ B_A`, `C <= X` with retraction `r: X -> C`, and `L = ker r`, so
`X = L ⋊ C` internally.
- `K × X ∈ B_A` by `boone-higman-type-a-class-closed-under-finite-extensions`, so
  `K × X <= A` for some group `A` with a type (A) action.
- `A` is finitely presented with solvable word problem
  (`type-a-action-gives-boone-higman-for-subgroups`), and it satisfies (i) of
  BFFHZ Theorem C (`fp-simple-highly-transitive-groups-satisfy-pbh`). By (iv) it
  embeds in a finitely presented simple MIF group `H`.

Inside `H`, the subgroups `K` and `X` commute and meet trivially, `C <= X`, and
`L` is normal in `X`.

Put `P = H * F(x, y)` and `Ω = Aut_H(P)`, the automorphisms fixing `H`
pointwise. By BFFHZ Theorem E
(`relative-automorphism-of-fp-simple-mif-group-has-type-a-action`, `n = 2`), `Ω`
admits an action of type (A), so every subgroup of `Ω` is in `B_A`.

## Step 1. Transvections of x

For `a ∈ H * <y>` let `ρ_a` fix `H` and `y` and send `x -> x a`.
- `ρ_a ρ_b (x) = ρ_a(x b) = x a b`, since `ρ_a` fixes `b`. So `ρ_a ρ_b = ρ_(ab)`,
  and `ρ_a ∈ Ω`.
- `ρ_a = id` forces `x a = x`, that is `a = 1`.

## Step 2. Twisted conjugation

For `c ∈ H` let `γ_c` fix `H` and send `x -> x c`, `y -> c^-1 y c`.
- `γ_c γ_d (x) = γ_c(x d) = x c d` and
  `γ_c γ_d (y) = γ_c(d^-1 y d) = d^-1 c^-1 y c d`.
- So `γ_c γ_d = γ_(cd)` and `γ_1 = id`. Hence `γ_c ∈ Ω`, and `c -> γ_c` is a
  homomorphism.

## Step 3. The conjugation formula

For `a ∈ H * <y>`:

```text
γ_c ρ_a γ_c^-1 = ρ_(c γ_c(a) c^-1).                                    (GP1)
```

Check on `x`: `γ_c^-1(x) = x c^-1`, then `ρ_a` gives `x a c^-1`, then `γ_c`
gives `x c γ_c(a) c^-1`. Check on `y`: `γ_c^-1(y) = c y c^-1` is fixed by `ρ_a`,
and `γ_c(c y c^-1) = y`. `H` is fixed throughout.

Two cases, for `c ∈ C`:
- `a = k ∈ K`: `γ_c(k) = k`, and `c k c^-1 = k`. So `γ_c ρ_k γ_c^-1 = ρ_k`.
- `a = y l y^-1` with `l ∈ L`: `c γ_c(a) c^-1 = c (c^-1 y c) l (c^-1 y^-1 c) c^-1`,
  which is `y (c l c^-1) y^-1`. So `γ_c ρ_(y l y^-1) γ_c^-1 = ρ_(y (c l c^-1) y^-1)`.

## Step 4. The homomorphism

Define `Φ` by `k -> ρ_k`, `l -> ρ_(y l y^-1)` and `c -> γ_c`.
- **On `X = L ⋊ C`.** The maps on `L` and on `C` are homomorphisms, and the
  second case of (GP1) is the semidirect product relation. So `Φ` is defined on
  `X`.
- **On `C × K`.** The maps on `C` and on `K` are homomorphisms that commute by
  the first case of (GP1). So `Φ` is defined on `C × K`.

The two definitions agree on `C`, so `Φ` extends to `G = X *_C (C × K)`.

## Step 5. Injectivity

**Normal form in G.**
- The retraction `r` and the projection `C × K -> C` agree on `C`, so they give
  a retraction `R: G -> C`.
- `G` is generated by `L`, `C` and `K`, and `C` normalizes `L` and `K`. So every
  element is `n c` with `n ∈ N = <K, L>` and `c ∈ C`, and `ker R = N`. Hence
  `G = N ⋊ C`.
- `N ≅ K * L`. A reduced alternating product of nontrivial elements of `K` and
  `L` alternates between `(C × K) \ C` and `X \ C`, because `K ∩ C = 1 = L ∩ C`.
  By the normal form theorem for amalgamated products it is nontrivial.

**Φ is injective.**
- Let `w: K * L -> H * <y>` be `k -> k`, `l -> y l y^-1`. By Step 1,
  `Φ(n) = ρ_(w(n))` for `n ∈ N`.
- `w` is injective: the image of a reduced word `k_1 l_1 k_2 ...` is
  `k_1 · y · l_1 · y^-1 · k_2 · ...`, alternating between `H \ 1` and
  `<y> \ 1`, so it is reduced in `H * <y>`.
- Suppose `Φ(n c) = ρ_(w(n)) γ_c` is the identity. On `y` it sends
  `y -> c^-1 y c`, since `ρ_(w(n))` fixes `H` and `y`. In a free product a
  nontrivial element of `H` does not commute with `y`, so `c = 1`.
- Then `ρ_(w(n)) = id`, so `w(n) = 1` and `n = 1`.

Therefore `G <= Ω` and `G ∈ B_A`. This is part 1.

## Step 6. Graph products

For an induced subgraph `Λ <= Γ`:
- **Retract.** Killing `G_u` for `u ∉ Λ` defines a homomorphism `G_Γ -> G_Λ`,
  since every defining relator goes to a relator or to 1. Composed with the
  natural map `G_Λ -> G_Γ` it is the identity. So `G_Λ` embeds in `G_Γ` as a
  retract.
- **Decomposition.** Fix a vertex `v`, with link `lk(v)` and star `st(v)`. Then
  `G_(st v) = G_(lk v) × G_v`, and `G_(lk v)` embeds in both `G_(Γ \ v)` and
  `G_(st v)`. The pushout presentation of `G_(Γ \ v) *_(G_(lk v)) G_(st v)` has
  every vertex group and every commutator relator of `Γ`, so it is `G_Γ`.

So `G_Γ = X *_C (C × K)` with `X = G_(Γ \ v)`, `C = G_(lk v)` a retract of `X`,
and `K = G_v`. By induction on the number of vertices `X ∈ B_A`, so part 1 gives
`G_Γ ∈ B_A`. The base case is a single vertex. This is part 2.