I'm solving problem 3.D in H. Georgi *Lie Algebra etc* for fun where one is to compute the matrix elements of the direct product $\sigma_2\otimes\eta_1$ where $[\sigma_2]_{ij}\text{ and }[\eta_1]_{xy}$ are two different Pauli matrices in two different two dimensional spaces. 

Defining the basis in our four dimensional tensor product space 
$$\tag{1}\left|1\right\rangle = \left|i=1\right\rangle\left|x=1\right\rangle\\
\left|2\right\rangle = \left|i=1\right\rangle\left|x=2\right\rangle\\
\left|3\right\rangle = \left|i=2\right\rangle\left|x=1\right\rangle\\
\left|4\right\rangle = \left|i=2\right\rangle\left|x=2\right\rangle$$

Now we know that when we multiply representations, the generators add in the sense of 

$$\tag{2}[J_a^{1\otimes2}(g)]_{jyix} = [J_a^1]_{ji}\delta_{yx} +\delta_{ji}[J_a^2]_{yx}, $$ where the $J$s are the generators corresponding to the different representations $D_1$ and $D_2$ ($g$ stands for the group elements). 

Using all of this I find that in the basis of $(1)$ the matrix representation of the tensor product is given by 

$$\tag{3}\sigma_2\otimes\eta_1 = \begin{pmatrix} 
0 & \mathbf{1} & -i & 0 \\
1 & 0 & 0 & -i \\ 
i & 0 & 0 & 1 \\
0 & i & 1 & 0 \end{pmatrix}$$

(The bold $\mathbf{1}$ is just notation, see below!)
I am not asking you to redo the calculations for me but does $(3)$ make sense? 

Appendix. 
My calculations were done in the following fashion [using equation $(2)$]: 
$$\tag{4}\langle 1| \sigma_2\otimes \eta_1 |1\rangle  = 
\\
\langle j=1,y=1| \sigma_2\otimes \eta_1 |i=1,x=1\rangle
\\
=
[\sigma_2]_{11}\delta_{11}+\delta_{11}[\eta_1]_{11}
\\
= 0.$$
Similarly for eg 
$$\tag{5}
\langle 1| \sigma_2\otimes \eta_1 |2\rangle  = 
\\
\langle j=1,y=1| \sigma_2\otimes \eta_1 |i=1,x=2\rangle
\\
=
[\sigma_2]_{11}\delta_{12}+\delta_{11}[\eta_1]_{12}
\\
= 1. 
$$
This is how the bold $\mathbf{1}$ was obtained. 

**So are my calculations $(4), (5)$ totally wrong?** 

The Pauli matrices
$$\begin{align}
  \sigma_1 &=
    \begin{pmatrix}
      0&1\\
      1&0
    \end{pmatrix} \\
  \sigma_2  &=
    \begin{pmatrix}
      0&-i\\
      i&0
    \end{pmatrix} \\
  \sigma_3 &=
    \begin{pmatrix}
      1&0\\
      0&-1
    \end{pmatrix} \,.
\end{align}
$$