# The canonical form of elements in the plactic monoid of rank $3$

I am reading the paper 'Plactic algebra of rank 3'. You can visit http://link.springer.com/article/10.1007/s00233-011-9337-3.

For an integer $n\geq1$ we consider the finitely presented monoid $M_{n}=\langle a_{1},\ldots,a_{n}\rangle$ defined by the relations \begin{align} a_{i}a_{k}a_{j}=a_{k}a_{i}a_{j}\ for\ i\leq j<k,\\ a_{j}a_{i}a_{k}=a_{j}a_{k}a_{i}\ for\ i\leq j<k. \end{align} It is called the plactic monoid of rank $n$. It is know that the elements of $M_{n}$ can be written in a canonical form,from which it follows that they are in a one to one correspondence with Young tableaux of certain type.

Considering the case $n=3$. So $M=\langle a,b,c\rangle$ with the convention that $a<b<c$, where \begin{align} aba=baa,\ bab=bba,\ aca=caa,\ cac=cca,\\ cbb=bcb,\ cbc=ccb,\ bac=bca,\ acb=cab. \end{align} The canonical form of an element $w\in M$ looks in this case as follows: \begin{align} w=(cba)^{k_{1}}(ba)^{k_{2}}(ca)^{k_{3}}(cb)^{k_{4}}(a)^{k_{5}}(b)^{k_{6}}(c)^{k_{7}}, \end{align} where $k_{i}\geq0$ such that either $k_{4}=0$ or $k_{5}=0$.

My qusetion is how to get $k_{4}=0$ or $k_{5}=0$? Any help will be appreciated.

The normal form requires that each row of the tableau is non-decreasing and each column is strictly increasing. Thus if the normal form looks like $$w=(cba)^{k_{1}}(ba)^{k_{2}}(ca)^{k_{3}}(cb)^{k_{4}}(a)^{k_{5}}(b)^{k_{6}}(c)^{k_{7}},$$ then the bottom row will be $$a^{k_{1}}a^{k_{2}}a^{k_{3}}b^{k_{4}}a^{k_{5}}b^{k_{6}}c^{k_{7}}$$ Thus in order to have a non-decreasing row, you must have either $k_4 = 0$ or $k_5 = 0$.