Recipe

Find the cyclic subgroup generated by an element

Use when: you're given a specific element g of a group and asked for the entire cyclic subgroup ⟨g⟩ it generates

Goal: either the complete finite list of ⟨g⟩'s elements, or a correct description of its infinite structure

Taught in Cyclic subgroups, revisited

1. 1

   Compute the smallest positive n with g^n=e, or confirm no such n exists at all.

   Example

   In \mathbb{Z}_{12}, find 5's order: the smallest n with 5n\equiv0\pmod{12} is n=12, since \gcd(5,12)=1.

   • g has finite order n — continue to Step 2.
   • g has infinite order — skip ahead to Step 3.

   Why: the equality |\langle g\rangle|=\operatorname{ord}(g) means this one number already answers how big the subgroup is, before listing a single element.

2. 2

   Write out g^0, g^1, \ldots, g^{n-1}; this list is exactly \langle g\rangle, with n distinct elements.

   Example

   \operatorname{ord}(5)=12 in \mathbb{Z}_{12}, so \langle5\rangle has all 12 elements, since 5\cdot0,5\cdot1,\ldots,5\cdot11 \pmod{12} hit every residue.

   Why: each further power just repeats one of these n, since g^n=e resets the cycle back to g^0.

3. 3

   Write \langle g\rangle as \{\ldots,g^{-2},g^{-1},g^0,g^1,g^2,\ldots\}: every integer power, all distinct, in either direction.

   Example

   In (\mathbb{Z},+), 2 has infinite order, so \langle2\rangle=\{\ldots,-4,-2,0,2,4,\ldots\}=2\mathbb{Z}, running forever both ways.

   Why: an infinite-order element never returns to the identity, so no finite list of powers can ever close back into a repeating cycle.

4. 4

   Confirm the finite list has exactly \operatorname{ord}(g) entries, or confirm the infinite description genuinely never repeats, before moving on.

   Example

   \mathbb{Z}_{12}'s \langle4\rangle: \operatorname{ord}(4)=3, so listing 4^0=0,4^1=4,4^2=8 gives exactly the 3 elements \{0,4,8\}.

   Why: this catches a miscounted order, or a skipped or duplicated power, before it's used in later work such as Lagrange's theorem.