My question comes from the observation of two examples.
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The quaternion product $\otimes$ makes $\mathbb R^4$ a unitary associative algebra (over $\mathbb R$), and with that product $\mathbb R^4/0$ becomes a Lie group whose Lie algebra is $\mathbb R^4$. The interesting fact is the Lie bracket on the Lie algebra is exactly the commutator of $\otimes$: $[q_1, q_2] = q_1q_2-q_2q_1$ (for readability the $\otimes$ is omitted).
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The matrix product makes $L(n)\equiv\mathbb R^{n\times n}$ a unitary associative algebra, and with that product $GL(n)$ becomes a Lie group whose Lie algebra is $\mathbb R^{n\times n}$. The Lie bracket on the Lie algebra is exactly the commutator of the matrix product: $[M_1, M_2] = M_1M_2-M_2M_1$.
So my question is, are the two examples above just special cases? Or the general guess below is true?
A general guess:
Let $V$ be a finite dimonsional vector space over $\mathbb R$, and $\otimes: V\times V\rightarrow V$ a bilinear operator that makes $V$ a unitary associative algebra. $G\subset V$ is a submanifold embedded in $V$, and with the product $\otimes$ it becomes a Lie group (with Lie algebra $\mathfrak g$). Since $\mathfrak g$ is just the tangent space of $G$ at $I$ (the identity element of the unitary algebra $V$ shared by $G$) and $G$ is embedded in the vector space $V$, $\mathfrak g$ can be identified with a subspace of $V$.
Then, the Lie bracket on the Lie algebra $\mathfrak g$ associated with $G$ , which by definition is determined by the Poisson's bracket on the left-invariant vector fields on $G$, will necessarily coincide with the commutator of the associative algebra $V$? i.e. $[v_1,v_2] = v_1v_2-v_2v_1$ ? (the $\otimes$ is omitted)
If the guess is not always the case, then what extra conditions are needed to make it true?
Update (a detailed proof):
Qiaochu Yuan's answer and comments pointed out that the "general guess" above could be even augmented. Actually all the invertible elements of $V$ form a maximal Lie group $V^\times$ (i.e. the group of units) whose Lie algebra is exactly the one on $V$ with the commutator as its bracket. Every Lie group embedded in $V$ is just a subgroup of the maximal.
Proof
$V^\times$ is a group : trivial
$V^\times$ is an open set of $V$ (thus a manifold and a Lie group):
For any $g\in V$, $\lambda_g: h\mapsto gh$ is a linear operator on $V$. We first show that $g$ is an invertible element iff $\lambda_g$ is invertible as a linear operator.
If $g$ is invertible, $\lambda_g$ is surjective since for any $h\in V$, we have $\lambda_g(g^{-1}h) = gg^{-1}h = h$; it's injective since
the kernel of $\lambda_g$ contains a single element $0$ ($\lambda_g(h) = 0 \Rightarrow g^{-1}\lambda_g(h) = g^{-1}0=0 \Rightarrow h=0$). So $\lambda_g$ is bijective and as a linear operator it's invertible.
If $\lambda_g$ is invertible, then it's surjective so there exists $h$ such that $\lambda_g(h)=1$, but such $h$ is exactly the inverse of $g$.
Then we construct a map $(\det\circ \lambda): V\rightarrow \mathbb R$, where $\lambda: g\mapsto \lambda_g$ is from $V$ to $L(V,V)$ (the space of linear operators from $V$ to itself). Note $L(V,V)$ itself is a vector space and the map $\lambda$ is linear, thus continuous. To show $\lambda$ is linear, choose a basis $\{e_1,e_2,…e_n\}$ for $V$, then $\lambda(g)$ is just a linear combination of $\{\lambda(e_1),\lambda(e_2),…\lambda(e_n)\}$ for any $g\in V$.
So now we know the combined map $(\det\circ \lambda)$ is continuous (with in mind that $\det: L(V,V) \rightarrow \mathbb R$ is continuous). It's clear that $(\det\circ \lambda)(g)\ne 0$ iff $\lambda_g$, thus $g$, is invertible. So $V^\times$ is open as it's the preimage of the open set $\mathbb R/0$.
The bracket on the Lie algebra of $V^\times$ is exactly the commutator
Let $\mathfrak g$ be the Lie algebra of $V^\times$, and the left invariant vector field on $V^\times$ induced by $v\in \mathfrak g$ is denoted by $\zeta_v$.
We first show $\zeta_v(g)=gv$ (note the Lie group $V^\times$ and its Lie algebra $\mathfrak g$ are both subsets of $V$ and points from the two spaces can be multiplied directly).
By the definition of left invariant vector field we have $\zeta_v (g) = (T\lambda_g)_I(v)$, where $(T\lambda_g)_I$ is the tangent map of $\lambda_g$ at identity $I$. Benefit from the fact that $V^\times$ is an open set of a vector space (where $+/-$ is allowed) we can directly compute $(T\lambda_g)_I(v)$ as below:
$$(T\lambda_g)_I(v) = \lim_{t\rightarrow 0}\frac{\lambda_g(I+vt) – \lambda_g(I)}{t}= \lim_{t\rightarrow 0}\frac{g(I+vt) – g}{t}=\lim_{t\rightarrow 0}\frac{gvt}{t}=gv$$
Let $\zeta_w$ be another left invariant vector field induced by $w\in \mathfrak g$. The value of the poisson bracket $[,]_p$ of the two vector field at $I$ can be computed as:
$[\zeta_w, \zeta_v ]_p(I) = (\zeta_w(I)) \zeta_v – (\zeta_v(I)) \zeta_w$
where $(\zeta_w(I)) \zeta_v $ denotes the directional derivative of $ \zeta_v$ at $I$ along $ \zeta_w(I) $, and $ (\zeta_v(I)) \zeta_w $ the opposite. This formula makes sense since we're working on a vector space.
Choose a basis $\{e_1,e_2,…,e_n\}$ for $V$， we have
$$\zeta_w(g) = gw = (g^ie_i)w = x^i(g)(e_iw)$$
(Einstein summation convention is used here), where $x^i$ is a coordinate function and $(e_iw)$, as a vector, can be also viewed as a constant vector field. To be more compact, the vector field $\zeta_w$ can be written as:
$$\zeta_w = (e_iw)\cdot x^i$$
Similarly, $$\zeta_v =(e_jv) \cdot x^j $$
Thus
$$\begin{aligned} {[\zeta_w, \zeta_v ]_p(I)} & = (\zeta_w(I)) \zeta_v – (\zeta_v(I)) \zeta_w \\
& = (w) ((e_jv)\cdot x^j) – (v) ((e_iw)\cdot x^i) \\
& = ((w)x^j) e_jv – ((v)x^i) e_iw \end{aligned}$$
Since $(w)x^j = w^j, (v)x^i = v^i$, so finally we have
$$[w, v]_{\mathfrak g} = [\zeta_w, \zeta_v ]_p(I) = w^je_jv – v^i e_iw = wv – vw$$
The Lie group $V^\times$ is maximal: trivial
Every Lie group embedded in $V$ is a Lie subgroup of $V^\times$ and its Lie algebra is a Lie subalgebra of $\mathfrak g$: trivial
Best Answer
You don't specify what $G$ is, and if you take $G$ to be as large as possible a stronger statement is true: if $A$ is a finite-dimensional associative algebra over $\mathbb{R}$, then its group of units
$$GL_1(A) \cong A^{\times} \cong \{ a \in A : \exists b \in A : ab = ba = 1 \}$$
is a Lie group, and its Lie algebra $\mathfrak{gl}_1(A)$ can be canonically identified with $A$ (not just a subspace of $A$ but the whole thing), and furthermore the bracket is the commutator bracket. Moreover the exponential map $\exp : \mathfrak{gl}_1(A) \to GL_1(A)$ is given by $a \mapsto \sum_{n \ge 0} \frac{a^n}{n!}$ (convergence of this series should be understood to be with respect to any norm, all of which are equivalent).