Homepage Solution manuals Ivan Niven An Introduction to the Theory of Numbers Exercise 5.7.22* (Density of $\langle P \rangle$ if $P$ is on the unbounded convex component)

Exercise 5.7.22* (Density of $\langle P \rangle$ if $P$ is on the unbounded convex component)

Let 𝒞 f ( ) be defined as in the preceding problem. Show that if P is a point of infinite order, P 𝒞 0 , then the points nP form a dense subset of 𝒞 0 . Show that if P is of infinite order, P 𝒞 1 , then the points nP are dense on 𝒞 f ( ) .

Answers

Proof.

(a)
Suppose that P 𝒞 0 .

By Problem 21, we know that 𝒞 0 is a subgroup of 𝒞 f ( ) . We can do the same reasoning as in Problem 20: the group 𝒞 0 is a compact connected real Lie group of dimension 1 , so is isomorphic to S 1 = . The subgroups of S 1 are finite, or dense on S 1 . If P has infinite order in 𝒞 f ( ) , then P = { nP , n } 𝒞 0 is an infinite subgroup of 𝒞 0 , thus its image in S 1 is dense on S 1 . This shows that P is dense on 𝒞 0 , because an isomorphism of Lie groups is also an homeomorphism.

(b)
Suppose that P 𝒞 1 .

By Problem 21, 2 P 𝒞 0 , and by induction, for all k , 2 kP 𝒞 0 and ( 2 k + 1 ) P 𝒞 1 . Therefore, for all k , 2 kP 𝒞 0 and ( 2 k + 1 ) P 𝒞 1 .

Consider the subgroup H = P = { nP , n } of 𝒞 f ( ) . Then H 𝒞 0 = 2 P is an infinite subgroup of 𝒞 0 . By part (a), H 𝒞 0 is dense on 𝒞 0 .

Consider the map

η { 𝒞 1 𝒞 0 M P + M .

Then η is well defined: since P in 𝒞 1 , if M 𝒞 , then P + M 𝒞 0 by Problem 21.

Moreover η is continuous, because the addition is continuous on 𝒞 0 by the explicit formulas.

η is bijective, with reciprocal η 1 : N N P . Therefore η 1 is also continuous, so η is an homeomorphism.

By our previous remarks, η ( 𝒞 1 H ) = 𝒞 0 H . Since 𝒞 0 H = 2 P is dense on 𝒞 0 , then 𝒞 1 H = η 1 ( 𝒞 0 H ) is dense on 𝒞 1 .

Therefore the subgroup { nP , n } is dense on 𝒞 f ( ) .

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2025-06-05 09:55
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