If $G$ is some matrix group of which $A$ is a subgroup, show $L(A)$ is a subalgebra for $L(G)$.
We only consider matrix Lie algebras in the very brief lecture series I am doing. Since no information is given at all on what the criteria are for an element to be part of $S$, how do I do this?
There are several ways to describe the Lie algebra of a Lie group. For this problem the most useful description is $$L(H) = \{\alpha'(0): \; \alpha : \mathbb{R} \rightarrow H \; \mathrm{differentiable}\},$$ the set of all derivatives-at-$0$ of functions from $\mathbb{R}$ to $H$.
(It turns out that you can restrict to looking at curves of the form $\alpha(t) = e^{tA}$, which implies that $$L(H) = \{A: \; e^{tA} \in H \; \mathrm{for} \; \mathrm{all} \; t\}.)$$
The hard part is showing that it's closed under commutators. Here's a sketch: if $A = \alpha'(0)$ and $B = \beta'(0)$ are two derivatives at zero, then look at the functions $$F_h(t) = \alpha(h) \beta(t) \alpha(h)^{-1}.$$ You will find that $$[A,B] = \lim_{h \rightarrow 0} \frac{F_h'(0) - \beta'(0)}{h}$$ is a limit of elements of $L(H)$, so it also lies in $L(H).$