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Two cyclically reduced words in a free group are conjugate if and only if one is a cyclic permutation of the other

Statement

Let $u$ and $v$ be cyclically reduced words on $X ⊔ X^{- 1}$ . In the reduced-word free group on $X$ , the elements represented by $u$ and $v$ are conjugate if and only if $v$ is a cyclic permutation of $u$ .

Through the unique generator-compatible isomorphism, the same criterion holds for elements represented by cyclically reduced words in any free group on $X$ .

Facts & Assumptions

Given: Cyclically reduced words $u$ and $v$ on $X ⊔ X^{- 1}$ .

[L1]

The reduced words on $X ⊔ X^{- 1}$ form a group under concatenation followed by free reduction, and the map sending $x \in X$ to the one-letter word $x$ has the universal property of the free group on $X$ (Reduced words form the free group on an alphabet).

[F1]

A reduced word is cyclically reduced when it is empty or its first letter is not the formal inverse of its last letter (Cyclically reduced words).

[L2]

Free groups on the same set are uniquely isomorphic compatibly with their generators (Free groups on the same set are uniquely isomorphic compatibly with their generators).

[L3]

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 · induction

1.1

If $u = p q$ literally and $v = q p$ , then $p^{- 1} u p = p^{- 1} p q p$ freely reduces to $q p = v$ , so every cyclic permutation of $u$ is conjugate to $u$ .

1.2

For the converse, suppose $v = t^{- 1} u t$ in the reduced-word group and take $t$ reduced. If $t = ε$ , then $u = v$ as elements of the underlying set of reduced words in [L1], which is a cyclic permutation obtained by taking an empty prefix.

baseL1

1.3

Assume the converse holds for conjugators shorter than a nonempty reduced word $t = a t^{'}$ , where $a$ is its first letter.

2.1

If neither seam in the literal word $t^{- 1} u t$ cancels, that word is reduced and begins with the inverse of its last letter, so [F1] says it is not cyclically reduced; but by [L1] this reduced word is the group product $t^{- 1} u t$ and hence equals the cyclically reduced word $v$ , a contradiction. Thus at least one of the two seams cancels.

F1L1step 1.3

3.1

If the right seam cancels, write $u = u^{'} a^{- 1}$ ; then $t^{- 1} u t$ freely reduces to $(t^{'})^{- 1} (a^{- 1} u^{'}) t^{'}$ , where $a^{- 1} u^{'}$ is a cyclic permutation of $u$ . If the left seam cancels, write $u = a u^{'}$ ; then it freely reduces to $(t^{'})^{- 1} (u^{'} a) t^{'}$ , where $u^{'} a$ is a cyclic permutation of $u$ . In either case the shifted word is cyclically reduced and the conjugator $t^{'}$ is shorter.

step 2.1L1

4.1

Apply [L3] to the property that the converse holds for every conjugator of length at most $n$ . The induction hypothesis makes $v$ a cyclic permutation of the shifted word in step 3.1; cyclic permutations compose, so $v$ is a cyclic permutation of $u$ .

step 1.3step 3.1L3

5.1

If one of $u, v$ is empty, conjugacy forces both to be the identity element, which is the empty word in the reduced-word group of [L1]. Combining this boundary with steps 1.1 and 4.1 proves both directions in the reduced-word model, and [L2] transports the criterion to every free group on $X$ .

step 1.1step 4.1L1L2discharge-induction∎

Depends on

Used by

Conjugacy in a free group by cyclic reduction Example

Dependency tree · two levels

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Sources

Alexei Myasnikov and Vladimir Shpilrain, Combinatorics over Free Groups, Proposition 2 (standard reference, not scraped)
Wilhelm Magnus, Abraham Karrass, and Donald Solitar, Combinatorial Group Theory (standard reference, not scraped)