Connect powers, divisibility and generators

Try

Repeatedly add four modulo twenty. The residues are zero, four, eight, twelve, sixteen, then zero again. The group has twenty elements, but the chosen element has order five. Group order means the number of elements in a finite group; element order means the least positive power that returns to identity.

Write the corresponding multiplicative statement: if a has order twenty, then a⁴ has order five. The operation changes notation, but the divisibility mechanism is the same.

Predict

Predict which powers of a generator still generate the entire cyclic group. An exponent sharing a factor with the group order should return too early. Test this intuition for a group of order ten, then state a condition involving a greatest common divisor rather than a list of lucky exponents.

Experiment

Let a generate a group of order ten. Calculate the least positive m for which 10 divides km.

Exponent k	gcd(10,k)	Order of aᵏ
1	1	10
2	2	5
4	2	5
5	5	2

The generators are a,a³,a⁷,a⁹. Draw their power traces or list residues until each first returns to zero. Reaching every residue is a stronger fact than merely having a nonzero starting residue.

Understand

If a has finite order n, then aʲ=e exactly when n divides j. Divide j by n: j=qn+r with 0≤r<n. Since aʲ=aʳ, a nonzero remainder producing identity would contradict minimality of n. This proves the divisibility criterion for all integer exponents, including negative ones.

To find the order of aᵏ, write d=gcd(n,k), n=dn₀ and k=dk₀. The condition n dividing km becomes n₀ dividing k₀m. Coprimality gives n₀ dividing m, so the least positive choice is n₀=n/d. In particular, aᵏ generates precisely when gcd(n,k)=1.

Every subgroup of a cyclic group is cyclic. For a nontrivial subgroup H of an infinite cyclic group, choose the least positive d such that aᵈ∈H. For any aᵏ∈H, divide k=qd+r. Then aʳ=aᵏ(aᵈ)⁻q belongs to H, so minimality forces r=0. Hence H=⟨aᵈ⟩. The trivial subgroup is handled separately. In a finite cyclic group the same argument can choose d among positive exponents with aᵈ∈H, including n if H is trivial; division of n by d shows d divides n.

Consequently a cyclic group of order n has exactly one subgroup of order d for each positive divisor d of n, namely ⟨a^(n/d)⟩. This converse to divisibility is special to cyclic groups; later examples show it cannot be transferred to every finite group.

Apply

List the subgroups of a cyclic group of order ten and draw containment arrows. Translate each containment into divisibility of generating exponents. Explain why the subgroup with more elements is generated by a smaller divisor in this presentation.

Then compare C₂×C₂ with C₄. Both have four elements and are abelian, but only the latter is cyclic. Every nonidentity element of the product has order two. This is a structural counterexample to “abelian implies cyclic,” not merely an unfamiliar multiplication table.

Continue

Permutation groups supply noncommutative examples and a universal representation of groups. Keep element order and composition order distinct: one counts repetitions, while the other specifies which map acts first.