Prove that the relation $\leq$ on the braid group $B_n$ is anti symmetric.

Question

For $A,B$ in the braid group $B_n$, we define $A \leq B$ if $B=CAC'$ for some $C$ and $C'$ in $B_n^{+}$, i.e., $C$ and $C'$ are positive words. We need to show that this relation is anti symmetric. This has been directly stated in papers regarding braid groups but I wanted to prove it. To show $\leq$ is anti symmetric, we need to prove that $A \leq B$ and $B \leq A$ implies $A=B$. Now, if $A \leq B$, then $B= CAC'$ and if $B \leq A$ we get $A=DBD'$ for positive words $C,C',D,D'$. This gives us $A=DCAC'D'$. I directly concluded that $DC$ and $C'D'$ would be identity braids but it is not very clear why, even though $DC$ and $C'D'$ are positive braids. Please help, I am really stuck here.

Answer 1

Let $\ell: B_n \rightarrow \mathbb Z$ be the abelianization. Note that $\ell$ is a homomorphism. More concretely: generating elements $\sigma_i$ get mapped to $1$ by $\ell$ (and their inverses to $-1$). So $$\ell (DCAC'D') = \ell (D) + \ell (C) + \ell (A) + \ell (C') + \ell (D').$$ Since positive braids have non-negative length, it follows that if $\ell (DCAC'D') = \ell (A)$, that $$\ell (C) = \ell (C') = \ell (D) = \ell (D') = 0,$$ in other words, that the positive braids $C, C', D$, and $D'$ are all equal to the identity braid.