Is there a way to factor out the middle tensor product? Given a state $(|1\rangle \otimes |0\rangle \otimes |1\rangle) + (|0\rangle \otimes |0\rangle \otimes |1\rangle)$, is it possible to factorise out the $|0\rangle$ in the middle of both of them?
Alternatively written as $|101\rangle + |001\rangle$ for convenience.
For a state such as the following it is possible to factorise out:
$|101\rangle + |100\rangle = |1\rangle (|01\rangle + |00\rangle)$
as shown above. Is there any way to factorise out the middle tensor in such a form?
 A: $(|1\rangle \otimes |0\rangle \otimes |1\rangle) + (|0\rangle \otimes |0\rangle \otimes |1\rangle)$ is clearly a product state. It's tricky to find good notation to denote this and I don't think there's an established standard way. Here's one option:
Let $\sigma_{ij}$ be an automorphism that flips the $i$th and $j$th position of a tensor product $\bigotimes_{k=1}^n V$. Then
$$(|1\rangle \otimes |0\rangle \otimes |1\rangle) + (|0\rangle \otimes |0\rangle \otimes |1\rangle) = \sigma_{12} \Big( \ |0\rangle \ \otimes \ \big( |1\rangle \otimes |1\rangle + |0\rangle \otimes |1\rangle \big) \ \Big) $$
A: If you consider the definition of tensor product by universal property, which is unavoidably equal to one you have in mind, then you have two natural isomorphisms $V\otimes W\cong W\otimes V$ and $(V_1\otimes V_2)\otimes V_3\cong V_1\otimes (V_2\otimes V_3).$ This implies that you can simply say that the position on which we write particular factor of tensor is just a matter of convinience. I bet you start with three qubits. The order of them depends on you. If someone fix some order, then these isomorphisms tells you that you can make a permutation. But after you done a permutation and you want put these cubits in some gate then you also have to change the order in this gate (in imput and output). The point is that the tensor product gives you a flexibility in making permutations, in sucha a way, that if you permute the tensors and perumute the gates (i.e. imputs and outputs) then result will be the same. Which in fact is all we care about.
