Timeline for Functional equation $f(x*y) = f(f(x)*f(y))$
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 6, 2022 at 23:14 | comment | added | Jérôme JEAN-CHARLES | @Steinberg Sorry you are right. | |
| Oct 6, 2022 at 15:04 | comment | added | Benjamin Steinberg | I only claimed the idempotent ones arise this way. Not that they are all idempotents | |
| Oct 6, 2022 at 14:34 | comment | added | Jérôme JEAN-CHARLES | @Steinberg On (Z/2nZ,+) f(x) = (x+n) mod 2n is not idempotent yet it a momorphism. | |
| Sep 25, 2022 at 19:09 | history | edited | Jérôme JEAN-CHARLES | CC BY-SA 4.0 |
minor spelling.
|
| Sep 23, 2022 at 21:18 | comment | added | Benjamin Steinberg | I guess the taking representatives for a congruence gets all the idempotent ones. | |
| Sep 23, 2022 at 21:00 | comment | added | Benjamin Steinberg | It seems to me like the way to get these is take a congruence on your semigroup and a set of representatives and let f be the map taking an element to its representative. | |
| Sep 23, 2022 at 20:53 | comment | added | Joseph Van Name | The operation $*$ does not satisfy the identity f(xyz)=f(f(x)f(y)f(z)) in the case when $\mathbb{S}=\mathbb{Z}_2$, $*$ is addition modulo $2$ (XOR), and f(x)=x+1 mod 2 (NOT). If we added the assumption that $f=f^2$, then we would satisfy the identity f(xyz)=f(f(x)f(y)f(z)) though. | |
| Sep 23, 2022 at 20:44 | history | edited | Jérôme JEAN-CHARLES | CC BY-SA 4.0 |
Added an example of interest
|
| Sep 23, 2022 at 19:57 | comment | added | Jérôme JEAN-CHARLES | @Terry Ok for associativity , for Z/kZ,: I have the solution for the additive but not yet for the multiplicative case your observation help clarifying. | |
| Sep 23, 2022 at 19:48 | comment | added | Jérôme JEAN-CHARLES | @Joseph at start I had the involutive axiom in (N,+) a lot but not all functions obey the idempotency. | |
| Sep 23, 2022 at 17:45 | history | edited | Sam Hopkins | CC BY-SA 4.0 |
deleted 12 characters in body
|
| Sep 23, 2022 at 16:59 | comment | added | Terry Tao | Any such function induces a commutative semigroup structure $\tilde *$ on $f({\mathbb S})$ by $f(x) \tilde * f(y) := f(x*y) = f(f(x) * f(y))$, and the restriction of $f$ to $f({\mathbb S})$ obeys the simpler law $x \tilde * y = f(x) \tilde * f(y)$; conversely, these axioms imply the original property. When ${\mathbb S}$ is a group, the latter condition is equivalent to $f(x) = x \tilde * a$ for some order two element $a$ of $(f({\mathbb S}), \tilde *)$. | |
| Sep 23, 2022 at 15:32 | history | edited | YCor | CC BY-SA 4.0 |
formatting, changed tag
|
| Sep 23, 2022 at 15:14 | history | asked | Jérôme JEAN-CHARLES | CC BY-SA 4.0 |