21-373: Algebraic Structures Extra Credit Assignment

Peter Battaglino 18. February, 2005

Let a group G be given with the following property:

∀a, b ∈ G, ⟨a⟩ ⊂ ⟨b⟩ or ⟨b⟩ ⊂ ⟨a⟩; (∗)

in other words, the subgroups of G generated by single elements of G are linearly ordered by inclusion.

Problem 1: If G is a finite group, show that G is cyclic (and, hence, that every subgroup of G is cyclic). Establish also that there exist positive integers m and n such that

{|H| | H ⩽ G} = {n^{k} | k = 1, ⋯, m}.

Hint: Show first that for each positive integer q there is at most one subgroup of the form ⟨a⟩ of order q. Then choose an element of G (other than e) of lowest order.

\underline{Solution:} Suppose that the group G is finite. Then we can represent G as the set

G = {g_{1}, g_{2}, ⋯, g_{n}},  (1)

where

⟨g_{1}⟩ ⊂ ⟨g_{2}⟩ ⊂ ⋯ ⊂ ⟨g_{n}⟩.  (2)

Since g_{i} ∈ ⟨g_{i}⟩ for all i ∈ {1, ⋯, n}, we see that the group ⟨g_{n}⟩ contains every element of G and is generated by the element g_{n} ∈ G. Thus we see that G is cyclic. All the subgroups of G are cyclic, so each one is generated by an element of G. In addition, the orders of subgroups of G form a bounded set of positive integers. This bounded set must have a smallest element n besides 1 (1 corresponding to the identity e of G). In addition, according to theorem 4.3 of our text, the element n corresponds to a unique subgroup ⟨a⟩ ∈ G with order n. It is also a direct consequence of this theorem that every subgroup of G has an order that divides the order of G. We see immediately that there is an element b ∈ G with order |G|/n. Now, since the element a generates the unique subgroup of G with smallest order, and by property (*), we know that there are no elements of G with order between |b| and |G|. That is, the order of b is the next-largest order, besides |G| of all possible orders of subgroups of G. In addition, since a is the unique element with smallest order, ⟨a⟩ ⊂ ⟨b⟩. This means, once again by theorem 4.3, that |a| | |b|, so we can find an element c ∈ G such that |a||c| = |b|. But |b| = |G|/n and |a| = n, so this means that |c| = |G/n^{2}. This order is the third-largest order of any subgroup of G. We can see that if we continue this process, we will exhaust the subgroups of G from largest order to smallest. We will eventually arrive at an element of G whose order is as small as possible without being as small as |a| = n. If we call this element f, then we still satisfy ⟨a⟩ ⊂ ⟨f⟩, so we have n | |f|. By construction, there exists a positive integer m such that |f| = |G|/n^{m−2}. But since n | |f|, there exists a positive integer q such that nq = |f|, so there must exist yet a smaller subgroup of G with order q. But since there are no subgroups contained in ⟨f⟩ and containing ⟨a⟩ by supposition, we must have that q = n, so |f| = n^{2} = |G|/n^{m−2}, which means that |G| = n^{m}. Since we have exhausted all of the subgroups of G, and each of them has order of the form n^{k}, where k is an integer between 1 and m inclusive, we have shown what we set out to prove.

Problem 2: If G is an infinite group, show that G has no proper infinite subgroup and that every finite subgroup of G is cyclic. Hint: It helps first to show that every element of G has finite order.

\underline{Solution:} Problem 3: Find an example of an infinite group G with the property (*). Is G cyclic?

\underline{Solution:} Consider the group G of rotations about angles of the form mπ/2^{n}, where m, n ∈ ℤ. It is simple to show that this is a group. Upon successive rotations of mπ/2^{n} and pπ/2^{q}, the total rotation is (2^{q}m+2^{n}p)π/2^{n+q}, which is a rotation in our group. The identity is the zero rotation, and every rotation has an inverse. We see that each subgroup of G has an order of the form 2^{k}, where k ∈ {1, ⋯, m}, and m is a positive integer, and there is only one unique subgroup of any given order. The (*) property is exhibited by the following fact: any rotation of mπ/2^{n} is in the subgroup generated by the rotation of π/2^{n}, and is also in the subgroup generated by π/2^{n+1}, etc. Every element of the group has finite order (as would any rotation of a rational multiple of π), so it follows that none of the elements of the group is capable of generating the entire (infinite) group. Thus, G cannot be cyclic.