A subfactor $N \subset M $ is **irreducible**  if $N' \cap M = \mathbb{C} $.   
It's **maximal** if it admits no non-trivial intermediate subfactors $N \subset P \subset M $.  
It's **cyclic** if its lattice of intermediate subfactors is distributive (see [here][1]).    
For "**index**" and "**depth**", see [Jones-Sunder (1997)][2] pages 29 and 91.  
 
$\begin{array}{c|c|c|c|c|c}
\text{cyclic}^{1} & \text{maximal} & \text{finite depth}& \text{finite index}&\text{irreducible}& \text{examples}^{23} \newline
 \hline
 &     \times    &   &         & \times& L(S_{\mathbb{N}^{*}}) \subset L(S_{\mathbb{N}}) \newline
 \hline
 &   \times       &   & \times & \times & A_{\infty}\text{-subfactors}  \newline
  \hline
 &   \times       & \times  & \times& \times& \mathbb{Z}/p\mathbb{Z} , \  S_{n} \subset S_{n+1}  ,  \text{Haagerup}\newline
 \hline
 \times  &   & \times  & \times& \times& \mathbb{Z}/m\mathbb{Z}, A_{n}\otimes A_{n+1}\text{-subfactor} \newline
 \hline
      \times   &   & \times  &         & \times& \mathbb{Z}  \newline
 \hline
&   \times       &  \times &         & \times& \textbf{?} 
\end{array}$     
$^{1}$ cyclic and non-maximal (obviously, maximal $\Rightarrow$ cyclic)   
$^{2}$  $p \in \mathbb{P}$ , $n \ge 2$ and $ m\notin \mathbb{P}$  
$^{3}$ the groups $G$ or $G_{1} \subset G_{2}$ correspond to the subfactors $R^{G} \subset R$ or $R^{G_{2}} \subset R^{G_{1}}$

> **Question**: is there a maximal finite depth infinite index irreducible subfactor ?    


  [1]: http://mathoverflow.net/questions/133050/the-cyclic-subfactors-theory-a-quantum-arithmetic
  [2]: http://www.cambridge.org/us/knowledge/isbn/item1155032/?site_locale=en_US