Group of arbitrary order

Question

Can we construct a group of order $n$ for any $n \in \mathbb{Z}^+$ i.e set of positive integers? Are there theorems which characterize the order of any finite group? What is the smallest possible restriction you can have on a group such that there does not exist such a group of particular order.

http://en.wikipedia.org/wiki/Cyclic_group – Harald Hanche-Olsen Jun 22 '12 at 05:55 — Harald Hanche-Olsen, Jun 22 '12 at 05:55

score 4 · Answer 1 · answered Jun 22 '12 at 05:55

4

For every $n$, there is a cyclic group of order $n$.

In fact ( you can use Upward-lowenheim skolem or compactness theorem from logic to show), there are groups of every infinite cardinality.

answered Jun 22 '12 at 05:55

William

19,935

May I ask inform me any refrences about what you claimed above (you can...). Thanks – Mikasa Jun 22 '12 at 07:19
This result is far more general and holds for arbitrary structures with certain properties. To find these theorems and learn model theory, I would recommend Model Theory: an Introduction by David Marker. – William Jun 22 '12 at 07:41

score 1 · Accepted Answer · answered Jun 22 '12 at 06:19

The last part of the question is not entirely clear to me, but if you impose the restriction that the group be nonabelian then there are no qualifying groups of orders 1, 2, 3, 4, 5, 7, 9, 11, 13, 15, 17, 19, 23, 25, 29, .... These numbers are tabulated at the Online Encyclopedia of Integers Sequences.

score 0 · Answer 3 · answered Jun 22 '12 at 06:06

0

For a given n, take the cyclic permutation group of n elements having a permutation which sends 1 to 2, 2 to 3, 3 to 4, and so on... This group of permutations is of order n.

answered Jun 22 '12 at 06:06

Aang

14,672

thank you for explicit response. – DurgaDatta Jun 22 '12 at 06:08

Group of arbitrary order

3 Answers3