The relation of time-evolution operators from Schwartz's textbook

Question

In the section 7.2.2 of Schwartz's QFT textbook, it says:

define the generation definition of time-evolution operators:

$$U_{21}\equiv U(t_2,t_1)=T{\exp[-i\int^{t_2}_{t_1} dt'V_I(t')]}\tag{7.46}$$

where $V_I$ is the interaction part of Hamiltonian in the interaction picture.

Then it has:

$$U_{21}U_{12}=1\tag{7.47}.$$

I don't understand how this relation is arrived. I tried to expand the definition directly, but to the second order of integral, it seems it cannot be canceled:

$$\int^{t_2}_{t_1} dt' dt'' V_I(t')V_I(t'')-\int^{t_2}_{t_1} dt' dt'' T[V_I(t')V_I(t'')]$$

which is obviously not zero.

Answer 1

Schwartz forgot to mention that the time-evolution operator $U(t_2,t_1)$ is antitime-ordered for $t_2<t_1$, cf. e.g. my Phys.SE answer here. Then eq. (7.47) becomes essentially just a telescoping product when discretized.

Answer 2

He must be tacitly using anti-time ordering when $t_2<t_1$. Then the cancellation works.

He should really use the path-ordering symbol "P" rather than "T".