Timeline for Adding a formal inverse of an element to a free monoid
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Nov 16, 2010 at 11:04 | comment | added | user6976 | @Chad: Actually, one can embed $M$ into a genuine semidirect product if you replace $\mathbb N$ by $\mathbb Z$ with operation $\max$; $\mathbb Z$ acts on itself by shift. That operation preserves $\max$. Perhaps you could define your $M$ this way. (I am not sure this approach works, though.) | |
| Nov 16, 2010 at 11:00 | comment | added | user6976 | @Chad: Are you sure that $M$ is a monoid? The subsemigroup ${\mathbb N}$ is a semilattice, and $\mathbb Z$ acts on it (so $M$ looks like a semi-direct product of a semilattice and a group) but the transformations induced by elements of $\mathbb Z$ are not invertible. | |
| Nov 16, 2010 at 4:16 | comment | added | Chad Groft | My guess is that adjoining $z^{-1}$ does not get you all the way to $F_2$, but it's not a very well-informed guess. | |
| Nov 16, 2010 at 3:11 | comment | added | user6976 | What is your guess for $z=a^2b^3a^2bab^2$? | |
| Nov 16, 2010 at 2:45 | comment | added | Chad Groft | Note also that the criteria above don't cover the whole space of possibilities. $z=a^2b^3a^2bab^2$ is the simplest word I can find which isn't covered. | |
| Nov 16, 2010 at 2:20 | comment | added | Chad Groft | Good point. I'll edit the answer to note this. | |
| Nov 16, 2010 at 2:19 | history | edited | Chad Groft | CC BY-SA 2.5 |
Conjecture was false, see below
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| Nov 16, 2010 at 2:12 | comment | added | user6976 | @Chad: Your suspicion seems wrong. Indeed, if $w=abaab$ is invertible, then $ab$ is invertible ($w$ starts and ends with $ab$), hence $a$ is invertible (since $a$ is the product of $(ab)^{-1}$, $w$ and $(ab)^{-1}$, hence $b$ is invertible. Thus inverting $w$ turns the monoid into a group. | |
| Nov 16, 2010 at 1:35 | history | answered | Chad Groft | CC BY-SA 2.5 |