Deducing the formula for group commutator in terms of Lie bracket from the Baker-Campbell-Hausdorff formula

Method
We denote by $$*$$ the operation for which the Baker-Campbell-Hausdorff formula:

$$X * Y = \log(\exp(X)\exp(Y))$$

The goal is to find an explicit formula for:

$$\log([\exp(X),\exp(Y)] = \log(\exp(X)\exp(Y)\exp(-X)\exp(-Y))$$

This can be described as:

$$\log((\exp(X)\exp(Y))(\exp(Y)\exp(X))^{-1})$$

It can be written as:

$$\! (X * Y) * (-(Y * X))$$

Abelian case
In this case, we have:

$$X * Y = \log(\exp(X)\exp(Y)) = X + Y$$

We also have:

$$Y * X = Y + X$$

Thus, we get:

$$-(Y * X) = -(Y + X)$$

We thus have:

$$(X * Y) * (-(Y * X)) = (X + Y) + (-(Y + X)) = X + Y - Y - X = 0$$

Thus, we get:

$$\log([\exp(X),\exp(Y)]) = 0$$

Case of class two
In this case, we have:

$$X * Y = \log(\exp(X)\exp(Y)) = X + Y + \frac{1}{2}[X,Y]$$

We thus get:

$$Y * X = \log(\exp(Y)\exp(X)) = Y + X + \frac{1}{2}[Y,X]$$

Thus:

$$-(Y * X) = -(Y + X + \frac{1}{2}[Y,X])$$

We thus get:

$$(X * Y) * (-(Y * X)) = X + Y + \frac{1}{2}[X,Y] + (-(Y + X + \frac{1}{2}[Y,X])) + \frac{1}{2}[X + Y + \frac{1}{2}[X,Y], -(Y + X + \frac{1}{2}[Y,X])]$$

Simplify the right side. The linear terms cancel, and we get:

$$(X * Y) * (-(Y * X)) = \frac{1}{2}[X,Y] - \frac{1}{2}[Y,X] - \frac{1}{2}[X + Y + \frac{1}{2}[X,Y], Y + X + \frac{1}{2}[Y,X]]$$

Use $$[Y,X]= -[X,Y]$$ and obtain:

$$(X * Y) * (-(Y * X)) = \frac{1}{2}[X,Y] + \frac{1}{2}[X,Y] - \frac{1}{2}[X + Y + \frac{1}{2}[X,Y], Y + X + \frac{1}{2}[Y,X]]$$

We now expand the later bracket:

$$[X + Y + \frac{1}{2}[X,Y], Y + X + \frac{1}{2}[Y,X]] = [X+ Y, Y+X] + \frac{1}{2}X,Y],Y + X] +\frac{1}{2} [X + Y,[Y,X + \frac{1}{4}X,Y],[Y,X$$

All the terms on the right side are zero. The first one is zero because it is the bracket of something with itself, and the remaining are zero because of the class two constraint.

Plugging these back in, we get:

$$(X * Y) * (-(Y * X)) = \frac{1}{2}[X,Y] + \frac{1}{2}[X,Y] = [X,Y]$$

In other words, we obtain that the group commutator is the same as the Lie bracket.

Case of class three
In this case:

$$X * Y = \log(\exp(X)\exp(Y)) = X + Y + \frac{1}{2}[X,Y] + \frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]]$$

We thus get:

$$Y * X = Y + X + \frac{1}{2}[Y,X] + \frac{1}{12}[Y,[Y,X]] - \frac{1}{12}[X,[Y,X]]$$

Simplifying, we obtain that:

$$-(Y * X) = -(Y + X + \frac{1}{2}[Y,X] + \frac{1}{12}[Y,[Y,X]] - \frac{1}{12}[X,[Y,X]])$$

Taking the sum, we obtain:

$$(X * Y) + (-(Y * X)) = \frac{1}{2}[X,Y] - \frac{1}{2}[Y,X]$$

The degree three terms cancel each other when we add, using that $$[X,[X,Y]] = -[X,[Y,X]]$$ and $$[Y,[X,Y]] = -[Y,[Y,X]]$$.

This simplifies to:

$$(X * Y) + (-(Y * X)) = [X,Y]$$

We can now compute:

$$(X * Y) * (-(Y * X)) = (X * Y) + (-(Y * X)) + \frac{1}{2}[(X * Y),-(Y * X)] + \frac{1}{12}[(X * Y),[(X * Y),-(Y * X)]] - \frac{1}{12}[-(Y * X),[(X * Y),-(Y * X)]]$$

We can show that the degree three terms are zero, because of the class three condition and the fact that $$[X+Y,[X + Y,X+ Y]]$$. We also obtain that $$(X * Y) + (-(Y * X)) = [X,Y]$$:

$$(X * Y) * (-(Y * X)) = [X,Y] + \frac{1}{2}[X + Y + \frac{1}{2}[X,Y] + \frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]],-(Y + X + \frac{1}{2}[Y,X] + \frac{1}{12}[Y,[Y,X]] - \frac{1}{12}[X,[Y,X]])]$$

Expanding and big bracket and using the class three condition, we obtain:

$$(X * Y) * (-(Y * X)) = [X,Y] + \frac{1}{2}[X + Y,\frac{-1}{2}[Y,X]] + \frac{1}{2}[\frac{1}{2}[X,Y],-(Y + X)]$$

Simplifying, we get:

$$(X * Y) * (-(Y * X)) = [X,Y] + \frac{1}{2}[X,[X,Y]] + \frac{1}{2}[Y,[X,Y]]$$