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Recipe

Find the subgroups of a finite cyclic group

Use when: you have a cyclic group G=⟨g⟩ of known finite order n and need one or all of its subgroups

Goal: the exact subgroup(s) of G, each one correctly matched to a divisor of n, with no missing or extra orders

Taught in Subgroups of cyclic groups

Quick recipe

  1. List the divisors of n
  2. Compute the generator g^{n/d} for each divisor d
  3. List that subgroup's elements as powers of the generator
  4. Repeat across every divisor, if listing all subgroups, and check the total count
  1. 1

    Write down every positive divisor of n=|G|; no other number can ever be a subgroup's order.

    Example

    For n=12: the divisors are 1,2,3,4,6,12.

    • you need every subgroup of G — repeat Step 2 for each divisor found here.
    • you need only the subgroup of a specific order d, and d divides n — use just that one divisor in Step 2.
    • your target order d does not divide n — stop here; no subgroup of that order exists.

    Why: this section's theorem guarantees exactly one subgroup per divisor of n, and none for any non-divisor.

  2. 2

    For a divisor d, the unique subgroup of order d is generated by g^{n/d}: divide n by d, and raise g to that power.

    Example

    For n=12, d=6: the generator is g^{12/6}=g^2, or in \mathbb{Z}_{12}'s additive notation, the multiples of 2.

    Why: this section's theorem shows m=n/d is exactly the smallest exponent landing a power of g in a subgroup of order d.

  3. 3

    Write out d elements: the generator raised to the 0th through (d-1)th power, or, additively, its multiples, stopping before it repeats.

    Example

    \langle2\rangle in \mathbb{Z}_{12} with d=6: 0,2,4,6,8,10, six elements, then back to 0.

    Why: a generator of order d produces exactly d powers before returning to the identity, the same equality proved for cyclic subgroups generally.

  4. 4

    If Step 1's first case applies, run Steps 2-3 for every divisor of n; then confirm the number of subgroups found equals the number of divisors of n.

    Example

    n=12 has 6 divisors, and running Steps 2-3 on each produces exactly 6 subgroups: \{0\}, \{0,6\}, \{0,4,8\}, \{0,3,6,9\}, \{0,2,4,6,8,10\}, and \mathbb{Z}_{12} itself.

    Why: matching the divisor count confirms no divisor's subgroup was skipped and no order was duplicated or invented.

Common mistakes

  • Assuming a subgroup exists for every divisor of |G| even when G is not cyclic, unlike A_4's missing order-6 subgroup at order 12.
  • Trying to build a generator before checking whether d actually divides n.
  • Using g^d instead of g^{n/d} as the generator for the subgroup of order d.
  • Stopping after finding one subgroup when the task actually asked for every subgroup of G.