Is this a simple Lie algebra? This question comes from Georgi, Lie Alegbras in Particle Physics. Consider the algebra generated by $\sigma_a\otimes1$ and $\sigma_a\otimes \eta_1$ where $\sigma_a$ and $\eta_1$ are Pauli matrices (so $a=1,2,3$). He claims this is "semisimple, but not simple". To me, that means we should look for an invariant subalgebra (a two-sided ideal). The multiplication table is pretty easy to figure out:
$[\sigma_a,\sigma_b]=i\epsilon_{abc}\sigma_c,$
$[\sigma_a,\sigma_b\otimes\eta_1]=i\epsilon_{abc}\sigma_c\otimes\eta_1$
$[\sigma_a\otimes\eta_1,\sigma_b\otimes\eta_1]=i\epsilon_{abc}\sigma_c\otimes1$
I'm dropping off the identity in all the places where it looks like it should be. So the only subalgebra is the $\mathfrak{su}(2)$ generated by $\sigma_a\otimes 1$, and that is not invariant from the second line above. So this looks like a simple algebra to me. Is there a typo somewhere I do not see?
 A: Unless I am mistaken, your algebra is $\mathfrak{so}(4)=\mathfrak{su}(2)\oplus\mathfrak{su}(2)$. The generators are $\sigma_a\otimes 1 \pm \sigma_a\otimes\eta_1$.
A: In this relatively simple example, one can observe that the subalgebras $\{\sigma_a \otimes \frac{1\mp\eta_1}{2}\}$ are the two commuting copies of $su(2)$.
For more complicated situations, one actually has an algorithm to veify the simplicity of a Lie algebra. This is because (the root systems of) simple Lie algebras are classified by Cartan, thus one just needs to verify if the root system of the given Lie algebra coincides with one of the types. Actually, an algebra which is semisimple but not simple will necessarily have orthogonal simple roots.
Taking the given example for illustration. One can choose $\sigma_3\otimes 1$ and $\sigma_3\otimes \eta_1$ as the generators of the Cartan's subalgebra. The corresponding root generators can be found as: $\sigma_{\pm} \otimes \frac{1\mp\eta_1}{2}$. The positive roots can be chosen as: $\alpha = [1,1 ]$, and $\beta  = [1,-1 ]$. Now we notice that on one hand they are linearly independent thus they are simple roots.
On the other hand we have $\alpha.\beta=0$ in contradiction to the property of a simple root system requiring that for any distinct simple roots $\alpha.\beta<0$.
