In this video by Numberphile, the Dehn invariant is computed by noting that there are 12 edges all having dihedral angle $\frac{\pi}{2}$
We compute the invariant using linearity in first slow
$$ 1 \otimes \frac{\pi}{2} + 1 \otimes \frac{\pi}{2} ... \text{12 times} = 12 \otimes \frac{\pi}{2 } \tag{1}$$
But, suppose we had added things in the second slot, and then use the fact that the second slot is mod $2\pi$:
$$ 1 \otimes \frac{\pi}{2} + 1 \otimes \frac{\pi}{2} ... \text{12 times} = 1 \otimes (12 \frac{\pi}{2}) = 1 \otimes 6 \pi = 1 \otimes 0 \tag{2}$$
How would I show (1)=(2)?
How you’d show that depends on your definition of the tensor product. Under the definition of the tensor product as a quotient module, the tensors $(na)\otimes b$ and $a\otimes(nb)$ are equivalent by definition.