characterization of ordered groups of rank one
Lemma An ordered group has rank one if and only if, for every two elements and such that , there exists an integer such that .
Proof Suppose that the Archimedean property is satisfied and that is an isolated subgroup of . We shall show that if contains any element other than the identity, then . First note that there must exist an such that . By assumption, there must exist an element such that . By conclusion 1 of the basic theorem on ordered groups, either , or (since we assumed that the case is excluded). If , set . If not, by conclusion 5, if , then we will have and therefore will set when .
Let be any element of . There are five possibilities:
We shall show that in each of these cases, .
Trivial — 1 is an element of every group.
Trivial — is assumed to belong to
Since is an isolated subgroup, .
By the Archimedean property,there exists an integer such that . Since and is isolated (http://planetmath.org/IsolatedSubgroup), it follows that .
By conclusion 5 of the basic theorem on ordered groups, . By conclusion 1 of the same theorem, either or or . In each of these three cases, it follows that from what we have already shown. Since is a group, implies .
This shows that the only isolated subgroups of are the two trivial subgroups (i.e. the group and itself), and hence has rank one.
Next, suppose that does not enjoy the Archimedean property. Then there must exist and such that for all integers . Define the sets as
and define .
We shall show that is a subgroup of . First, note that, by a corollary of the basic theorem on ordered groups, , so for all , hence . Second, suppose that . Then . By conclusion 5 of the basic theorem, implies and implies . Thus, , so . Hence, if , then . Third, suppose that and . Then there must exist integers and such that and , so
Using conclusion 4 of the main theorem repeatedly, we conclude that
so . Hence, if and , then . this the proof that is a subgroup of .
Not only is a subgroup of , it is an isolated subgroup. Suppose that and and . Since , there must exist an such that , hence . By conclusion 2 of the basic theorem on ordered groups, and imply . Combining this with the facts that and , we conclude that , so . Hence .
Note that is not trivial since . The reason for this is that for any because we assumed that for all . Hence, the order of the group must be at least 2 because and are two examples of isolated subgroups of .
|Title||characterization of ordered groups of rank one|
|Date of creation||2013-03-22 14:55:15|
|Last modified on||2013-03-22 14:55:15|
|Last modified by||rspuzio (6075)|