A subfactor $N\subset M$ is essentially the same thing as an $N$-$M$-bimodule. I'll recall the basic definitions in the language of bimodules, and I hope that subfactor people will excuse me.

Explanation:To go from a subfactor $N\subset M$ to a bimodule, consider the actions of $N$ amd $M$ on the standard form $L^2(M)$. To go the other way around, given a bimodule ${_N}H_M$, you get an inclusion $N\hookrightarrow M'$.

A subfactor $N\subset M$ is said to have ** finite index** if the corresponding bimodule ${_N}H_M$ is dualizable. This means that there is a dual bimodule ${_M}K_N$, a unit map ${}_NL^2(N)_N\to {_N}H\ \boxtimes_M K_N$ and a counit map ${_M}K\ \boxtimes_N H_M \to {}_ML^2(M)_M$ that satisfy the usual zigzag identities $(H\to H\boxtimes K\boxtimes H \to H) = 1_H$ and $(K\to K\boxtimes H\boxtimes K \to K) = 1_K$.

A subfactor is said to have ** finite rank** if the irreducible summands of

$H\boxtimes K\boxtimes H\boxtimes K\boxtimes H\ldots \boxtimes K$,

the irreducible direct summands of

$H\boxtimes K\boxtimes H\boxtimes K\boxtimes H\ldots \boxtimes H$, ($*$)

and the irreducible direct summands of

$K\boxtimes H\boxtimes K\boxtimes H\boxtimes K\ldots \boxtimes K$ lie in finitely many isomorphism classes.

(see my last comment to Makoto's answer for a disambiguation)

**Question 1:** Does there exist a subfactor that is of finite rank but of infinite index?

**Question 2:** If I furthermore assume that all the branching multiplicites are finite, (i.e. that every ($*$) splits as a *finite* direct sum of irreducible bimodules), is it still possible?