9:00 I like that you are computing things explicitly, but I think in the case of [swap 1 & 2] [swap 3 & 4] it’s visually intuitive that the operations don’t interfere with each other, and therefore order doesn’t matter (=they commute). More interesting to me is that (1↔2) ∘ (1↔3) ∘ (1↔2) = (2↔3). Thanks for sharing these nice videos, Matt.
Hi Dr Matt In the proof that the set generated by an element g of G is in fact a subgroup of G, you did not mention the associativity. This property comes from the associativity in Z , that's easy to see but I think we must mention it in this proof. Don't you think so or is it only a forgetting ?
For negative k, g^k means (g^{-1})^k, the k-th power of the inverse of g. By definition of a group, every element g has an inverse - so this is just the k-th power of that.
In another video the author responded "Indeed, I should make a video on cyclic notation. I made a couple mistakes inn this video that suggest that I didn't understand it myself at the time!", so all is cool.
