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Two finite presentations define isomorphic groups if and only if a finite sequence of Tietze transformations and inverses connects them

Statement

Let $P = ⟨ X ∣ R ⟩$ and $Q = ⟨ Y ∣ S ⟩$ be finite presentations. They present isomorphic groups if and only if a finite sequence of the transformations and legal inverses of Tietze transformations: dictionary generators, redundant relators, renaming, and their inverses connects $P$ to $Q$ .

Facts & Assumptions

Given: Finite presentations $P = ⟨ X ∣ R ⟩$ and $Q = ⟨ Y ∣ S ⟩$ .

[L1]

Each Tietze transformation preserves the isomorphism type of the presented group (Each Tietze transformation preserves the isomorphism type of the presented group).

[L2]

In $⟨ Z ∣ T ⟩$ , words $u$ and $v$ represent the same element if and only if $u^{- 1} v \in ⟨ ⁣ ⟨ T ⟩ ⁣ ⟩$ (In $⟨ X ∣ R ⟩$ , the words $u$ and $v$ represent the same element if and only if $u^{- 1} v \in ⟨ ⁣ ⟨ R ⟩ ⁣ ⟩$ ).

[L3]

The canonical map from a group to a quotient group is surjective (The canonical projection $π : G \to G / N$ , $π (g) = g N$ , is a surjective group homomorphism).

[L4]

If a property $P$ satisfies $P (0)$ and $P (n) \Rightarrow P (n + 1)$ for every natural number $n$ , then $P (n)$ holds for every $n \in N$ (The principle of mathematical induction).

Proof

technique · constructive

1.1

If a finite sequence of Tietze transformations connects $P$ to $Q$ , composing the isomorphisms supplied by [L1] along that sequence gives an isomorphism between the groups they present; the zero-move case is the identity isomorphism.

L1L4

1.2

Conversely, fix an isomorphism $ϕ : G_{P} \to G_{Q}$ . If $X \cap Y \neq \emptyset$ , first apply one renaming transformation to $Q$ , replacing $Y$ by a finite set disjoint from $X$ , and compose $ϕ$ with the induced isomorphism. Write $Q = ⟨ Y ∣ S ⟩$ for this renamed presentation; after connecting $P$ to it, the inverse renaming returns to the original $Q$ . By surjectivity in [L3], for each $x \in X$ choose a word $v_{x} (Y)$ representing $ϕ ([x])$ , and for each $y \in Y$ choose a word $w_{y} (X)$ representing $ϕ^{- 1} ([y])$ ; only the finitely many choices indexed by $X \cup Y$ are made, successively by [L4].

L1L3L4givenchoose

2.1

Starting from $P$ , add every $y \in Y$ by the dictionary relation $d_{y} : = y^{- 1} w_{y} (X)$ . In the resulting presentation, $v_{x} (Y)$ and $x$ represent the same element because eliminating the new letters sends $v_{x} (Y)$ to the representative of $ϕ^{- 1} (ϕ ([x])) = [x]$ ; hence [L2] makes $d_{x} : = x^{- 1} v_{x} (Y)$ a redundant relator. Add every $d_{x}$ , and then add every $s \in S$ , which is redundant because eliminating $Y$ evaluates it as $ϕ^{- 1} ([s]) = 1$ . This is a finite legal sequence from $P$ to $C : = ⟨ X \cup Y ∣ R \cup S \cup {d_{x} : x \in X} \cup {d_{y} : y \in Y} ⟩$ .

L2step 1.2L4construct

2.2

Starting from $Q$ , add every $x \in X$ by the dictionary relation $d_{x} = x^{- 1} v_{x} (Y)$ . In that presentation, $w_{y} (X)$ and $y$ represent the same element because eliminating $X$ evaluates $w_{y} (X)$ as $ϕ (ϕ^{- 1} ([y])) = [y]$ , so [L2] licenses adding every $d_{y}$ ; each $r \in R$ is then redundant because eliminating $X$ evaluates it as $ϕ ([r]) = 1$ . Thus another finite legal sequence runs from $Q$ to the same presentation $C$ .

L2step 1.2L4construct

3.1

Reverse the sequence of step 2.2. Each relator is deleted in reverse order while the earlier relators that originally forced it remain, so the redundant-relator inverse condition is satisfied. Each dictionary generator is deleted only after every later-added relator containing it has been removed, leaving that generator in its dictionary relation alone, so the dictionary inverse condition is satisfied. Hence there is a finite legal sequence from $C$ to the renamed $Q$ . Concatenate it with step 2.1 and, when step 1.2 used a renaming, append that renaming's legal inverse. The resulting finite sequence connects the original $P$ to the original $Q$ .

step 1.2step 2.1step 2.2L4

4.1

Step 1.1 proves the forward implication and steps 1.2 through 3.1 construct the reverse implication, so the two conditions are equivalent.

step 1.1step 3.1discharge-construct∎

Remarks

The finiteness hypothesis is used to make the representative selections and the additions in steps 1.2 through 2.2 into finite sequences. No choice principle is used: each selection is from a single nonempty fibre, repeated a finite number of times.

Depends on

Used by

Solvability of the word problem does not depend on the chosen finite generating set Proposition

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Nicholas Touikan, An Introduction to Combinatorial and Geometric Group Theory, Theorem 1.6.2 (standard reference, not scraped)