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I know that for the one object {1,2,3}⊗{1,2,3} = {1,2,3}. But what is (1,2) ⊗ (2,3)?

hom(A⊗B,C) ≅ hom(A,hom(B,C)) so for A = B = C

hom(C⊗C,C) ≅ hom(C,hom(C,C)), so if the group is S3 = hom(C,C), then (1,2) ⊗ (2,3) is somehow not a permutation, but instead a morphism into the group itself?

If I write the permutations as permutation matrices, and the tensor product as the kronecker product of matrices, I end up with (2,3)⊗(1,2) being a 9x9 matrix which corresponds to (12)(48)(57)(69) but that sounds totally wrong.

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    $\begingroup$ Why do you expect there to be such a bifunctor? If there is a bifunctor as you write, the group is forced to be abelian. $\endgroup$
    – David Roberts ♦
    Commented Sep 15, 2019 at 10:57
  • $\begingroup$ Can a non-abelian group be a monoidal category? $\endgroup$
    – user145873
    Commented Sep 15, 2019 at 11:17
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    $\begingroup$ Yes, if you instead take the objects to be group elements, and the underlying category to be discrete. David's comment is closely related to the Eckmann-Hilton argument: ncatlab.org/nlab/show/Eckmann-Hilton+argument $\endgroup$
    – Todd Trimble
    Commented Sep 15, 2019 at 12:46

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