# Proof that finite symmetrized relator sets, which are $C'(1/6)$, with equal normal closures are unique

The following statement is made in the Wikipedia article on small cancellation theory without reference or proof. Can anyone either provide a proof or point me to a reference with a proof?

The statement: "If $R$ and $S$ are finite symmetrized subsets of $F(X)$ with equal normal closures in $F(X)$ such that both presentations $\langle X | R \rangle$ and $\langle X | S \rangle$ satisfy the $C'(1/6)$ condition then $R = S$."

Dehn's algorithm/Greendlinger's lemma give this: By Dehn's algorithm, in a $C'(1/6)$ the shortest trivial words of length greater than $0$ are exactly the shortest relations, so $R$ and $S$ have the same shortest relations. Consider $R_1,S_1$ which are the relation sets with the shortest relations removed.

Claim: $R_1,S_1$ have the same shortest relations. Let $r_0$ be the length of the shortest relations in $R/S$, and let $r_1,s_1$ be the length of the shortest relation in $R_1,S_1$. We can assume without loss of generality that $r_0< s_1 \leq r_1$. Let $w$ be a shortest relation in $S_1$, and apply a step of Dehn's algorithm with respect to the presentation $R$, and note that we can not shorten it with respect to a shortest word in $R$ (that would mean $w$ breaks $C'(1/6)$ in $S$) so we actually use a relation in $R_1$, which means we shorten with a word that is at least as large as $w$. This new reduced word is $v \bar v$ where $\bar v$ is a subword of $w$ with less than half its length, and $v$ is less than half the length of some word in $R_1$.

$|v \bar v|<s_1$, and we can suppose towards a contradiction that $0< |v \bar v|$, hence $r_0\leq|v \bar v|$ since the word is still trivial with respect to $R$. In fact $r_0< |v \bar v|$ since if it were equal, then $v \bar v$ would be a shortest word in $R$, but that would mean one of the factors would be more than $\frac{1}{6} r_0$, contradiction both $S$ and $R$ being $C'(1/6)$. For similar reasons you can not use such a word to reduce by a shortest word in $R$. So we must reduce by a word in $R_1$, and so the reduction can not lead to the trivial word. During the reduction at most $\frac{1}{6} r_0$ can effect the $v$, and more than $\frac{2}{6} r_0$ must cancel out with $\bar v$, so after cancelling with $\bar v$ there is less than $\frac{1}{6}r_0$ of the original $w$ in this new word. We apply Dehn's algorithm again, and we get that we still can't reduce by something of length $r_0$, but if we reduce by something in $R_1$ more than $\frac{1}{6}r_0$ will affect some piece that came from $R_1$ contradicting $R_1$ is $C'(1/6)$.

The above argument can be continued for $R_2,S_2$ and so on, till you exhaust $R$ and $S$. A quick sketch of this is you will have $0<r_0<...<r_n<s_{n+1} \leq r_{n+1}$, you will get a $v \bar v$ which can't be one of lower $r_n$ for the same reasons as the claim, and similarly the rest is pretty much the same, except you need to argue for all lower $r_i$ instead of just $r_0$.

Two standard sources which prove Greendlinger's lemma and Dehn's algorithm are Combinatorial Group Theory by Lyndon and Schupp and another standard source is Geometry of Defining Relations in Groups by Olʹshanskii.

• This is an interesting idea but why do the two normal closures remain equal after the first step? Jan 13, 2017 at 1:30
• @MoisheCohen Good catch. I edited, and it came out less easy... I feel like there should be a pretty nice van Kampen diagram argument though. ( I think I have an easier argument that just uses that the product of these relations and their conjugates shouldn't really get smaller so something of length $s_1$ should could only be made of $r_0$'s or one $r_1$. )
– user29123
Jan 13, 2017 at 10:48