= Solution
Expand the three <matrix exponentials> and retain terms through degree two. Their product is
$$
E(t)=I+t(A+B)+\frac{t^2}{2}(A^2+AB+BA+B^2)+O(t^3),
$$
which agrees with $e^{t(A+B)}$ to that order. Keeping the cubic terms gives the more informative defect
$$
E(t)-e^{t(A+B)}=t^3\left(-\frac1{24}[A,[A,B]]+\frac1{12}[B,[B,A]]\right)+O(t^4),
$$
where $[A,B]=AB-BA$ is the <commutator>. For example, the coefficients of $A^2B,ABA,BA^2$ in the product are $1/8,1/4,1/8$, instead of the exact $1/6,1/6,1/6$. The cubic defect is generically nonzero. Therefore \b[<Strang splitting> has order $p=2$]. The symmetry $E(-t)=E(t)^{-1}$ is consistent with this cancellation of the quadratic defect; for commuting $A,B$ the composition is exact rather than merely second order.
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