# Length of a product of conjugates of an element in a free group

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.)

Background:

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, $$T=ba$$.

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

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.

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

(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.

• 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 $4$. Sep 27, 2022 at 16:15
• 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$. Sep 28, 2022 at 12:40 • You may want to modify your claimed inequality to$|\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$.) Sep 28, 2022 at 13:09 • Sorry, in my example above, the reduced length of$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. ) Sep 29, 2022 at 2:08