§ Lie bracket as linearization of conjugation

Let us have Y=GXG1Y = GXG^{-1} with all of these as matrices. Let's say that GG is very close to the identity: G=I+EG = I + E with E2=0E^2 = 0 (EE for epsilon). Note that now, G1=(I+E)1G^{-1} = (I + E)^{-1}, which by abuse of notation can be written as 1/(I+E)1/(I + E), which by taylor expansion is equal to IE+E2E3+I - E + E^2 - E^3 + \dots. Since EE is nilpotent, we truncate at E2E^2 leaving us with (IE)(I - E) as the inverse of (I+E)(I+E). We can check that this is correct, by computing:
(I+E)(IE)==IE+EE2=IE2==I0=I \begin{aligned} &(I+E)(I - E) = \\ &= I - E + E - E^2 \\ &= I - E^2 = \\ &= I - 0 = I \end{aligned}
This lets us expand out YY as:
Y=GXG1Y=(I+E)X(I+E)1Y=(I+E)X(IE)Y=IXIIXE+EXIEXEY=XXE+EXEXE \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 EE is small, EXEEXE is of order E2E^2 and will therefore vanish. This leaves us with:
GXG1=Y=X+[E,X] 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.