What is the isomorphism function in $M_m(M_n(\mathbb R))\cong M_{mn}(\mathbb R)$? What is the isomorphism function in $M_m(M_n(\mathbb R))\cong M_{mn}(\mathbb R)$.
I tried this $[[a_{ij}]_{kl}]\mapsto[a_{ijkl}]$ , but I couldn't prove all steps.
 A: Hint: Starting with an element $[a_{ij}]_{kl}$ in $M_m(M_n(\Bbb R))$, the "obvious" homomorphism is to just "erase all the squares around $n\times n$ submatrices" and let all the contents form an $mn\times mn$ matrix $b$.
For example, the elements of $[a_{i,j}]_{1,1}$ would wind up being $b_{i,j}$ in the "big" matrix $b$ with components $b_{\cdot,\cdot}$. Then $[a_{i,j}]_{1,2}$ would wind up being $b_{i,n+j}$. 
Then also $[a_{i,j}]_{2,1}$ would be  $b_{n+i,j}$.
And $[a_{i,j}]_{4,5}$ would be  $b_{3n+i,4n+j}$.
Can you see the pattern?

Once you grasp the indexing, you can take a look at why the multiplication works. Here is the relevant stuff (with no explanation, because I want you to think about it.) Below, $c$ is another $mn\times mn$ matrix like $b$ is for a matrix with entries $[d_{i,j}]_{k,l}$.
$$([b][c])_{pn+i,qn+j}:=\sum_{k=1}^{mn}b_{pn+i,k}c_{k,qn+j}=\sum_{r=0}^{m-1}\sum_{s=1}^{n}b_{pn+i,rn+s}c_{rn+s,qn+j}$$
Can you see $([a]_{i,s}[d]_{s,j})_{p+1,q+1}$ in that expression? This is basically expressing "block multiplication."

Alternatively, if you are comfortable with the fact that matrix rings can be interpreted as linear transformations on vector spaces, you could restate the entire problem in terms of linear transformations of an $n$ dimensional vector space $V$, and the tranformations of $V^m$. This would ameliorate the need for matrix multiplication, but you would have to be comfortable with the translation.
