*-isomorphism between subfactor and $M_n(\mathbb{C})$ I am not getting anywhere with the following problem. Do you have any tips?
Let $M$ be a $II_1$ (finite with no non-zero abelian subprojections) factor. Show that for any $n\in \mathbb{N}$, there is a subfactor $N$, such that $N$ is *-isomorphic to $M_n(\mathbb{C})$
Thanks for your help
 A: They key observation is that $M_n(\mathbb C)$ is spanned by the matrix units $E_{kj}$, and that the matrix units also carry the information about multiplication. A set of matrix units is characterized by
$$\tag1
E_{kj}E_{rs}=\delta_{j,r}\,E_{ks},\qquad\qquad E_{kj}^*=E_{jk},\qquad\qquad k,j,r,s=1,\ldots,n. 
$$
So we need to find such a set inside $N$.

*

*You begin with $n$ pairwise orthogonal projections $f_{11},\ldots,f_{nn}$ that add to the identiy (these exist because $N$ has projections of all traces, in particular $1/n$, and you can always subdivide the orthogonal complement).


*Because these projections have equal trace, they are equivalent. It is tempting to simply take $f_{kj}$ to be a partial isometry between $f_{jj}$ and $f_{kk}$, but this is too naive; such partial isometries may not satisfy the relations $(*)$.


*What you do, instead, is to go through a fixed projection. Since each $f_{kk}$ is equivalent to $f_{11}$, there exist partial isometries $f_{12},\ldots,f_{1n}$ with
$$
f_{1k}^*f_{1k}=f_{kk},\qquad\qquad f_{1k}f_{1k}^*=f_{11},\qquad\qquad k=1,\ldots,n. 
$$


*Next you define
$$
f_{k1}=f_{1k}^*,\qquad\qquad f_{kj}=f_{k1}f_{1j},\qquad\qquad k,j=1,\ldots,n.
$$
It is now easy to check that the $\{f_{kj}\}$ satisfy $(1)$.


*Finally you define $\theta:M_n(\mathbb C)\to N$ by
$$
\theta\Big(\sum_{k,j} a_{kj}E_{kj}\Big)=\sum_{k,j} a_{kj}f_{kj}.
$$
The relations $(*)$ guarantee that $\theta$ is a $*$-homomorphism. It is easy to check that $\theta$ is injective, but not necessary because it comes for free from $M_n(\mathbb C)$ being simple.
