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Define $o = A \otimes B$. Compute the results of $o^{\otimes N}$. We have \begin{align} o^{\otimes N} &= (A \otimes B) \otimes (A \otimes B) \otimes ... \otimes (A \otimes B). \end{align} Can we groups the terms A and terms B and have \begin{align} o^{\otimes N} &= (A \otimes A \otimes ... \otimes A) \otimes (B \otimes B \otimes ... \otimes B)? \end{align}
I think this is wrong, but the eq. (2) in this question used this.
Viewed rigorously, it is wrong. Consider a simple example $o = X\otimes Y, A = X, B = Y$. $o^3 = X\otimes Y\otimes X\otimes Y\otimes X\otimes Y \neq X\otimes X\otimes X\otimes Y\otimes Y\otimes Y$
It is often assumed without explicitly stating that when the operators are shuffled, the qubits are correspondingly permuted to reflect this reordering. When is this done? Often when dealing with algorithms that utilize entanglement, like QKD and teleportation that go from single bits/bell states and generalize to binary strings or sets of Bell pairs to represent binary strings, it is convenient to reorder and denote via bit strings.
$|\psi_{00}\rangle^{\otimes n} = \left(\frac{|00\rangle + |11\rangle}{\sqrt{2}}\right)^{\otimes n} = \left(\frac{|00\rangle + |11\rangle}{\sqrt{2}}\right)\otimes \left(\frac{|00\rangle + |11\rangle}{\sqrt{2}}\right)\otimes \cdots \otimes \left(\frac{|00\rangle + |11\rangle}{\sqrt{2}}\right)$
When two parties are sharing this entanglement, say Alexa and Bard, it is convenient to take the first qubits of each pair that Alexa holds to one side and the rest to the right and write as
$|\psi_{00}\rangle^{\otimes n} \equiv \frac{1}{2^{n/2}}\left(|\underbrace{0\cdots 0}_{n}\rangle|\underbrace{0\cdots 0}_{n}\rangle + |0\cdots 1\rangle|0\cdots 1\rangle + \dots |1\cdots 1\rangle|1\cdots 1\rangle\right)$
This way, it is easy to see if Alexa measures a bit string and obtains $\mathbf{b}$, it can be easily seen that Bard will also have $\mathbf{b}$ on his qubits.
