Are the Pauli matrices closed under commutation? I tried to make a group multiplication table for the Pauli matrices, but I keep getting multiples in front of the elements. What am I doing wrong? I thought the Pauli matrices formed a group that was closed under commutation? Here's what I have:
$\begin{array}{c|ccc} [,] & \sigma_1 & \sigma_2 & \sigma_3 \\ \hline \sigma_1 & 0& 2i\sigma_3 & -2i\sigma_2 \\ \sigma_2 & -2i\sigma_3 & 0 & 2i\sigma_1 \\ \sigma_3 & 2i\sigma_2 & 2i\sigma_1 & 0  \end{array}$
 A: Lie algebras are not a group w.r.t. to the commutator (the Lie bracket).
The first reason is that the commutator is not associative.
Another is that they almost always lack an identity element, since the identity matrix is, for example, not in $\mathfrak{su}(2)$, and Schur's lemma would, in the fundamental representation, guarantee that only multiples of the identity can be commuting with all elements of the algebra. From the lack of the identity, it follows that also the existence of an inverse is ill-defined, so they can't be a group.
They are closed under the Lie bracket operation though, but your table doesn't contradict that.
It is not possible to make the commutator into a group operation on finite-dimensional Lie algebras - just look at the trace of $[A,B] = 1$ to see that this cannot be true as long as the trace is defined for both sides of the equation.
The closest thing to providing a more "familiar" associative structure with a neutral element on a Lie algebra (aside from its vector addition) is to pass to the universal enveloping algebra. Even there, you still lack inverses for the algebra operation (which is good, since otherwise we'd have a (skewed) field, which is boring).
