Let $G$ be a free group generated by a set $S$. For $g\in G$, let $l(g)$ be the length of $g$ with respect to $S$.

Now for $a\in G$ and $g_1,\dotsc,g_n\in G$, let $$T=g_1^{-1}ag_1g_2^{-1}ag_2\dotsm g_n^{-1}ag_n.$$ We assume that $a$ is cyclically reduced; i.e., $l(a)$ is the minimum of the set $\{l(hah^{-1}): h\in G\}$.

What can we say about $l(T)$ in terms of $l(a)$? Is it known that $l(T)\geq l(a)$? (I conjectured this, but can't prove it.)


  1. The length $l(a^n)$ in terms of $l(a)$ is easily determined. Products of conjugates of $a$ is a natural generalization of powers of $a$. Thus it's natural to ask the relation of the lengths with respect to this generalization.

  2. I came up with this when studying orderability of groups.

  3. A related result of Weinbaun (1972) tells us that $T$ can't be a subword of $g$. So for example, if $g=aba^{-1}b^{-1}$ then $T$ can't be $ab$. But his result doesn't rule out the possibility that, for instance, $T=ba$.

  • $\begingroup$ Did you mean $G=F$? $\endgroup$ Sep 26, 2022 at 21:15

2 Answers 2


Let $S=\{s_1,s_2,s_3\}$.

Then $a=(s_1 s_2)^{-1}s_3(s_1s_2)$ has length $5$.

But $T=s_2as_2^{-1}=s_1^{-1}s_3s_1$ has length $3$.

Thus the conjecture about the lengths doesn't hold for the first power of $a$.

It you square $a$, you get $l(a^2)=l((s_1s_2)^{-1}s_3^2(s_1s_2))=6$.

But if $T=s_2^{-1}as_2(s_1s_2)a(s_1s_2)^{-1}=s_1^{-1}s_3s_1s_3$ has $L(T)=4$.

Thus this appears false. I guess the first observation already disproves it.

  • $\begingroup$ You are right. I forgot to mention that the $a$ should be cyclically reduced. $\endgroup$
    – user45392
    Sep 28, 2022 at 12:51

I'm sorry, I was completely wrong, please ignore this answer.
(I do find it reasonable that this post and this paper may be relevant, though).

The correct formulation is with **cyclically reduced** length, and then the answer is **yes** (otherwise it's **no** as was commented).

Definition: The cyclically reduced length of a word $w$ as $|w|_c = \min\left\{\left|uwu^{-1}\right|: u\in F \right\}$.
Geometrically, this corresponds to erasing backtraces from the $w$-path in the Cayley graph $\textrm{Cay}\left(F/\langle\langle w \rangle\rangle, S\right)$ and then counting edges (instead of just counting edges).
A word $w$ is cyclically reduced if $|w|_c=|w|$.

Claim: $\left|\prod_{i=1}^n g_i^{-1} a g_i\right|_c \ge n\cdot |a|_c$.
(This answers your question positively in the case $|a|_c = |a|$).

Proof: Let's see both the geometric and algebraic interpretations.
There is a unique decomposition of the word $g_i = h_i g'_i$ such that $h_i^{-1} a h_i$ is cyclically reduced and $h_i$ is maximal among such words.
Geometrically, this means we decompose the traceback-erased $\rho$-shaped path $g_i^{-1}a g_i$ to its tail $g_i'^{\pm1}$ and cycle $h_i^{-1} a h_i$.
We may assume $h_i=1, g_i'=g_i$ (and in particular $a$ is cyclically reduced). Then all cancellations between $g_i^{-1}ag_i$ must occure between $g_i, g_{i+1}^{-1}$ only. Geometrically, the path $\prod_{i=1}^n g_i^{-1} a g_i $ is a tree with an $a$-cycle attached to every leaf. This finishes the proof.

  • $\begingroup$ I do not see why this argument is correct: for example, if $F=\langle x,y \rangle$ and $a=x$, then the first copy of $a$ cancels in the reduction of $T=a \cdot x^{-1}y a y^{-1} x$. The correct proof would have to use the fact that $T$ is the product of conjugates of $a$ only (and no conjugates of $a^{-1}$ are involved). Indeed, otherwise we can take $a=xy^{-n}$, which has length $n+1$, and $T=xyx^{-1}y^{-1}=a \cdot y a^{-1} y^{-1}$, which has length $4$. $\endgroup$ Sep 27, 2022 at 16:15
  • 1
    $\begingroup$ Your claimed formula is not correct. Consider in the free group with generators $x$ and $y$. Consider $a=xyx^{-1}y^{-1}$ which is cyclically reduced, the length of $a$ is 4. Let $a'=yx^{-1}y^{-1}x$, it's a conjugate of $a$. But the length of $aa'$ is 6, cyclically reduced length of $aa'$ is 4$. $\endgroup$
    – user45392
    Sep 28, 2022 at 12:40
  • $\begingroup$ You may want to modify your claimed inequality to $|\prod_{i=1}^ng_i^{-1}ag_i|_c\geq|a|_c$ (by deleting the $n$ multiplier on the RHS), which looks stronger than my conjecture, but equivalent. But I don't see how your argument proves this either. After you wrinting $g_i=h_ig_i'$ such that $a_i:=h_i^{-1}ah_i$ is still cyclically reduced and $l(h_i)$ is maximal, you write the product to the form $T=\prod_{i=1}^n g_i'^{-1}a_ig_i$. Now the difficulty is that there might be cancellations among $a_i$ and $a_{i+1}$. (It's relatively easy to control the number of such cancellations when $n=2$.) $\endgroup$
    – user45392
    Sep 28, 2022 at 13:09
  • $\begingroup$ Sorry, in my example above, the reduced length of $aa'$ is still 6, not 4. (In fact, because conjugate is allowed, to prove $|T|_c\geq |a|_c$ is equivalent to prove $|T|\geq|a|$, if $a$ is cyclically reduced. ) $\endgroup$
    – user45392
    Sep 29, 2022 at 2:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.