 Solution

Expand the three matrix exponentials and retain terms through degree two. Their product iswhich agrees with $e^{t(A+B)}$ to that order. Keeping the cubic terms gives the more informative defectwhere $[A,B]=AB-BA$ is the commutator. For example, the coefficients of $A^{2}B,ABA,B{A}^2$ 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 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.