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Stream: community: general

Topic: the group with no elements

John Baez (Aug 11 2020 at 01:16):

I blogged about something simple and fun:

• The group with no elements.

There's a puzzle about Lawvere theories at the end.

Alexander Campbell (Aug 11 2020 at 02:02):

John Baez said:

There's a puzzle about Lawvere theories at the end.

I've left a comment with a solution to your puzzle.

John Baez (Aug 11 2020 at 02:14):

I'll check it out!

Morgan Rogers (he/him) (Aug 11 2020 at 11:08):

I posted a reply about models of theories in other contexts that I hope makes the ideas in your post seem even more fun :grinning_face_with_smiling_eyes:

Morgan Rogers (he/him) (Aug 11 2020 at 11:23):

Incidentally, I've had an idea on a back-burner for a while that is tangentially related to my reply, and which I intend to get back to towards the end of the year (sorry for hijacking this topic, I may move this to #theory: topos theory later). We can view any commutative ring as special abelian group in the topos of actions of its multiplicative monoid, and the category of modules of a ring is a sub-abelian category of the category of abelian groups over this topos. I'm hoping that these observations might be exploited to get a new perspective on homological algebra. If anyone is interested in this, I would be very happy to collaborate on it.

Eric M Downes (Apr 02 2024 at 04:09):

There's a nice symmetry to the inverse $\zeta$ and constant identity $e$ equations for the possibly-empty group;
$\mu:GG\to G, ~\zeta:G\to G, ~e:G\to G,~\Delta:G\to GG$

$\mu\circ(\zeta\times id_G)\circ\Delta=\mu\circ(id_G\times\zeta)\circ\Delta=e$
$\mu\circ(e\times id_G)\circ\Delta=\mu\circ(id_G\times e)\circ\Delta=id_G$

Note how the first eq. becomes the second upon substitution $e\mapsto id_G, \zeta\mapsto e$. We can (sort of) extend this sequence one further by pre-composing the associative rule with $(1\times\Delta)\circ\Delta$ obtaining an equation of the same form
$\mu\circ(\mu\circ\Delta\times id_G)\circ\Delta=\mu\circ(id_G\times\mu\circ\Delta)\circ\Delta$

So the full analogy is $\zeta\,:\,e\,:\,id_G\,:\,\mu\circ\Delta~::~-1:0:1:2$, where the latter correspond to the resulting exponent of a group element passed to the former. (Though the analogy is tightest with the first three.)