# Tensor product of Pauli strings?

We define $$$$\sum_{i=1}^{4^l} P_i \otimes P_i, \sum_{i=1}^{4^m} Q_i \otimes Q_i,$$$$ where $$P_l$$ is the $$n$$ qubit Pauli string and $$Q_m$$ is the $$m$$ qubit Pauli string.

Does the following equality hold? $$$$\sum_{i=1}^{4^l} P_i \otimes P_i \otimes \sum_{i=1}^{4^m} Q_i \otimes Q_i = \sum_{i=1}^{4^{l+m}} O_i \otimes O_i,$$$$ where $$O_i$$ is the $$l+m$$ qubit Pauli string.

Would the summation be the same although the elements of the summation are different? In fact, they should be the same.

It doesn't hold. You can consider the simple example where $$l=1$$ and $$m=1$$ as a counter-example.
$$\sum_i P_i \otimes P_i = II + XX + YY + ZZ$$ $$\sum_i Q_i \otimes Q_i = II + XX + YY + ZZ$$
\begin{align} \sum_i P_i \otimes P_i \sum_i Q_i \otimes Q_i &= IIII + IIXX + IIYY + IIZZ \\ &+ XXII + XXXX + XXYY + XXZZ \\ &+ YYII + YYXX + YYYY + YYZZ \\ &+ ZZII + ZZXX + ZZYY + ZZZZ \\ &\neq \sum_i O_i \otimes O_i \end{align}
What you have in the RHS are $$4^{l+m}$$ Pauli strings of a set of $$4^{2(l+m)}$$ elements, since they are defined as $$O_i \otimes O_i$$, then their length is $$2(l+m)$$. Notice that these Pauli strings are symmetric with respect to the middle. However, from the LHS you don't have this symmetry.