Tensor product of two different Pauli matrices $\sigma_2\otimes\eta_1 $ 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
$$\begin{align}
  \sigma_1 &=
    \begin{pmatrix}
      0&1\\
      1&0
    \end{pmatrix} \\
  \sigma_2  &=
    \begin{pmatrix}
      0&-i\\
      i&0
    \end{pmatrix} \\
  \sigma_3 &=
    \begin{pmatrix}
      1&0\\
      0&-1
    \end{pmatrix} \,.
\end{align}
$$ 
 A: I think it is easier to compute direct products when you write the matrices in component form; basically, you just have to multiply each element of the first matrix by the whole second matrix: 
$$ \mathbf{A}\otimes\mathbf{B} = \begin{bmatrix} A_{11} \mathbf{B} & \cdots & A_{1n} \mathbf{B} \\ \vdots & \ddots & \vdots \\ A_{n1} \mathbf{B} & \cdots & A_{nn} \mathbf{B}\end{bmatrix} $$
In your case, using the Pauli matrices $\boldsymbol{\sigma}_2$ and $\boldsymbol{\eta}_1$, we get:
$$ \boldsymbol{\sigma}_2 \otimes \boldsymbol{\eta}_1 = \begin{bmatrix} 0 & -i\boldsymbol{\eta}_1 \\ i\boldsymbol{\eta}_1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & -i \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ i & 0 & 0 & 0\end{bmatrix} $$
A: Your equation (2) is right, in principle: it is the standard coproduct of Lie algebras,
but it is irrelevant, and should have never been used for anything here. The language confused you. It should read 
$$ \boldsymbol{J^a} = \boldsymbol{j^a} \otimes 1\!\!1 +1\!\!1\otimes \boldsymbol{j^a} .$$
If you wished to apply it to two doublet reps, you should have used the same Pauli matrix $\sigma^a$ for both $j^a$s, and multiplying by the same angle and exponentiating you would have seen how nicely the group elements tensor-factor in the respective subspaces 1 and 2: $\exp (i\theta^a \boldsymbol{J}^a)=$ $\exp(i\theta^a(\boldsymbol{j^a} \otimes 1\!\!1 +1\!\!1\otimes \boldsymbol{j^a}))=\exp(i\theta^a(\boldsymbol{j^a} \otimes 1\!\!1)) \exp(i\theta^a (1\!\!1\otimes \boldsymbol{j^a}))= \exp(i\theta^a \boldsymbol{j^a}) \otimes\exp(i\theta^a\boldsymbol{j^a} )$. 
But, instead, your assignment asked you to simply mechanically evaluate the tensor product of two different matrices, to see if you understand the rules @jabirali correctly applied to get the correct answer you were meant to find. So, your equation (3) is magnificently wrong: you evaluated $\boldsymbol{\sigma_2} \otimes 1\!\!1 +1\!\!1\otimes \boldsymbol{\sigma_1} $. jabirali is actually using your conventions, basis (1), exactly. 
As a further exploratory excursion, you might use his and your rules, "right matrix into entries of left matrix", "left-coarse, right-fine" to compute (2) for a common matrix, e.g. $\sigma^2$, and then C-G rotate/reduce the 4x4 matrix to find $J_2$ in the triplet representation (3x3 block) and a singlet (0! in the remaining 1x1 block).
A: Each Pauli matrix has two non-zero elements. Therefore, direct product of Pauli matrices will have four non-zero elements. Your answer, unfortunately, has eight.
