Timeline for Group of positive rationals under multiplication not isomorphic to group of rationals
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 11, 2020 at 10:18 | comment | added | Bertrand Haskell | The generalization of the result mentioned by Martin Sleziak can be found in Kuratowski's and Mostowski's "Set theory" (p.85 of the 2-nd edition). Given an isomorphism f between two algebraic systems (not necessarily groups), the result allows to "build" the whole spectrum of formulas preserved by f. | |
| May 16, 2011 at 19:26 | vote | accept | pberlijn | ||
| Apr 18, 2011 at 10:19 | comment | added | joriki | To clarify that slightly (since "being able to divide" is a property of humans and computers, not of groups :-): A group is called divisible if for each element $x$ and each natural number $n$ there is an element $y$ such that $ny$, defined as the $n$-fold sum of $y$, is $x$. Then $(\mathbb{Q},+)$ is divisible and $(\mathbb{Q}_{>0},\times)$ isn't. | |
| Apr 18, 2011 at 9:45 | comment | added | Tobias Kildetoft | It might be added that the property mentioned here (being avle to divide by any natural number n, which would then correspond to being able to take n'th roots in the multiplicative case) is called being divisible. | |
| Apr 18, 2011 at 9:28 | comment | added | joriki | Martin's comment referred to an earlier version of the answer that only had the first couple of sentences. | |
| Apr 18, 2011 at 9:27 | history | edited | joriki | CC BY-SA 3.0 |
added 325 characters in body; added 5 characters in body
|
| Apr 18, 2011 at 9:20 | history | edited | joriki | CC BY-SA 3.0 |
deleted 60 characters in body
|
| Apr 18, 2011 at 9:12 | history | edited | joriki | CC BY-SA 3.0 |
added 358 characters in body; added 1 characters in body
|
| Apr 18, 2011 at 9:11 | comment | added | Martin Sleziak | More-or-less, joriki's answer can be rephrased like this. If we find a property, which is preserved by isomorphisms and which is fulfilled only for one of the groups, then the groups are not isomorphic. In this case, the property of the group $(G,\circ)$ is $(\forall x\in G)(\exists y\in G) y\circ y=x$. | |
| Apr 18, 2011 at 9:04 | history | answered | joriki | CC BY-SA 3.0 |