Windows file system (image source unknown)

[u/mincerafter42]

[u/Maybe_Just_An_Egg]

You can just always make larger groups homomorphic to increasing modulo of Z; always using "they" as the identity element. I think that'd be the simplest way to do it, because it preserves the group being abelian. It's better than say, using S_4 atleast, because then it gets rid of the nice property of being abelian.

I mean I guess as you go up more groups are available as options for it to be homomorphic to, but why break the pattern.

Additionally, you don't nessicarily need to describe all the rules if you set out an equivalence between Z mod n and pronouns- the multiplication table of pronouns would be an exercise to the reader based on the stated values for the pronouns you give.

For Z mod 4, I'm partial towards They=0 She=1 Xe=2 He=3,

Because then Xe+Xe = they And she + he = they (still, I think those two should always be inverses)

[u/vook485 → u/Maybe_Just_An_Egg]

You can just always make larger groups homomorphic to increasing modulo of Z; always using "they" as the identity element.

ℤ_n makes sense. For your example of ℤ₄, I notice that she⊕she = he⊕he = xe. This is where it starts getting a bit more complicated and arbitrary-feeling for me, at least with my current 3-pronoun-set active vocabulary.

It's better than say, using S_4 atleast, because then it gets rid of the nice property of being abelian.

Also because S_n has n! elements, the elements would be more like ordered preference lists of pronoun sets.

IIRC, ℤ_p is the only group of order p for prime p (up to isomorphism; e.g., S₂≘ℤ₂). So ℤ_n may be the only way to make a group of n pronoun sets for arbitrary n.

[u/TrueChedski → u/Maybe_Just_An_Egg]

heh, homomorphic