Let $M(X)$ be a free monoid with involution $^{-1}$ on $X$ . A word $w\in M(X)$ is said to be cyclically reduced if every cyclic conjugate of it is reduced. In other words, $w$ is cyclically reduced iff $w$ is a reduced word and that if $w=uvu^{-1}$ for some words $u$ and $v$ , then $w=v$ .

For example, if $X=\lbrace a,b,c\rbrace$ , then words such as $$c^{-1}bc^2a\qquad\mbox{and}\qquad abac^2ba^2$$ are cyclically reduced, where as words $$a^2bca^{-1}\qquad\mbox{and}\qquad cb^2b^3c$$ are not, the former is reduced, but of the form $a(abc)a^{-1}$ , while the later is not even a reduced word.

Remarks. The concept of cyclically reduced words carries over to words in groups. We consider words in a group $G$ .

• If a word is cyclically reduced, so is its inverse and all of its cyclic conjugates.
• A word $v$ is a cyclic reduction of a word $w$ if $w=uvu^{-1}$ for some word $u$ , and $v$ is cyclically reduced. Clearly, every word and its cyclic reduction are conjugates of each other. Furthermore, any word has a unique cyclic reduction.
• Every group $G$ has a presentation $\langle S| R\rangle$ such that
  1. $R$ is cyclically reduced (meaning every element of $R$ is cyclically reduced),
  2. closed under inverses (meaning if $u\in R$ , then $u^{-1}\in R$ ), and
  3. closed under cyclic conjugation (meaning any cyclic conjugate of an element in $R$ is in $R$ ).
  Furthermore, if $G$ is finitely presented, $R$ above can be chosen to be finite.
  Proof. Every group $G$ has a presentation $\langle S| R'\rangle$ . There is an isomorphism from $F(S)/N(R')$ to $G$ , where $F(S)$ is the free group freely generated by $S$ , and $N(R')$ is the normalizer of the subset $R'\subseteq F(S)$ in $F(S)$ . Let $R''$ be the set of all cyclic reductions of words in $R'$ . Then $N(R'')=N(R')$ , since any word not cyclically reduced in $R'$ is conjugate to its cyclic reduction in $R''$ , and hence in $N(R'')$ . Next, for each $u\in R''$ , toss in its inverse and all of its cyclic conjugates. The resulting set $R$ is still cyclically reduced. Furthermore, $R$ satisfies the remaining conditions above. Finally, $N(R)=N(R'')$ , as any cyclic conjugate $v$ of a word $w$ is clearly a conjugate of $w$ . Therefore, $G$ has presentation $\langle S| R\rangle$ .

  If $G$ is finitely presented, then $S$ and $R'$ above can be chosen to be finite sets. Therefore, $R''$ and $R$ are both finite. $R$ is finite because the number of cyclic conjugates of a word is at most the length of the word, and hence finite.