§ Lie bracket as linearization of conjugation

Let us have $Y = GXG^{-1}$ with all of these as matrices. Let's say that $G$ is very close to the identity: $G = I + E$ with $E^2 = 0$ ( $E$ for epsilon). Note that now, $G^{-1} = (I + E)^{-1}$, which by abuse of notation can be written as $1/(I + E)$, which by taylor expansion is equal to $I - E + E^2 - E^3 + \dots$. Since $E$ is nilpotent, we truncate at $E^2$ leaving us with $(I - E)$ as the inverse of $(I+E)$. We can check that this is correct, by computing:

$$\begin{aligned} &(I+E)(I - E) = \\ &= I - E + E - E^2 \\ &= I - E^2 = \\ &= I - 0 = I \end{aligned}$$

This lets us expand out $Y$ as:

$$\begin{aligned} &Y = GXG^{-1} \\ &Y = (I + E)X(I + E)^{-1} \\ &Y = (I + E)X(I - E) \\ &Y = IXI -IXE + EXI - EXE \\ &Y = X - XE + EX - EXE \end{aligned}$$

Now we assert that because $E$ is small, $EXE$ is of order $E^2$ and will therefore vanish. This leaves us with:

$$GXG^{-1} = Y = X + [E, X]$$

and so the lie bracket is the Lie algebra's way of recording the effect of the group's conjugacy structure.