In ℤ/12ℤ (integers mod 12 under addition), what is the order of the element 4?
A4, because 4 is the element itself
B12, because the group has 12 elements and every element must traverse the whole group
C3, because 4 + 4 + 4 = 12 ≡ 0 (mod 12) — three steps reach the identity
D6, because the order of an element is always |G| divided by the element
The order of an element is the smallest positive integer n such that n copies of the element (under the group operation) equal the identity. In ℤ/12ℤ, we add: 4+4 = 8, 4+4+4 = 12 ≡ 0. Three steps reach 0, so ord(4) = 3. Note that option D reflects a common error — |G|/a is not a formula for element order. The correct relationship is that gcd(a, |G|) determines the order in cyclic groups: ord(4) = 12/gcd(4,12) = 12/4 = 3.
Question 2 Multiple Choice
A group G has order 21. Which of the following element orders is IMPOSSIBLE in G?
A1
B3
C6
D7
By Lagrange's theorem, the order of any element must divide the order of the group. Since 21 = 3 × 7, the divisors of 21 are 1, 3, 7, and 21. The number 6 does not divide 21 (21/6 is not an integer), so no element of order 6 can exist in G. This is one of the most powerful applications of element order: without constructing G explicitly, we can rule out entire families of element orders just by factoring |G|.
Question 3 True / False
If an element a has order n in a group G, then the set {e, a, a², ..., aⁿ⁻¹} forms a cyclic subgroup of G with exactly n elements.
TTrue
FFalse
Answer: True
This is the fundamental structural consequence of element order. The powers of a are all distinct (if aⁱ = aʲ for i < j, then a^(j−i) = e, contradicting the minimality of n), so there are exactly n of them. They form a subgroup — closed under the group operation and containing inverses — isomorphic to ℤ/nℤ. So ord(a) simultaneously tells you how many steps before you cycle back and the size of the smallest subgroup containing a. These are the same thing.
Question 4 True / False
In a group of order 20, it is possible for an element to have order 6.
TTrue
FFalse
Answer: False
By Lagrange's theorem, the order of an element must divide the order of the group. Since 6 does not divide 20 (20 = 4 × 5; divisors are 1, 2, 4, 5, 10, 20), no element of order 6 can exist in any group of order 20. This constraint is absolute — it requires no knowledge of the group's specific structure, only its size.
Question 5 Short Answer
Why does the order of an element in a finite group necessarily divide the order of the group? Trace the argument through subgroup theory.
Think about your answer, then reveal below.
Model answer: The powers of an element a of order n form a cyclic subgroup H = {e, a, a², ..., aⁿ⁻¹} of size n. By Lagrange's theorem, the size of any subgroup of a finite group G must divide |G|. Since |H| = n = ord(a), it follows that ord(a) divides |G|.
The chain of reasoning is: element order → size of generated cyclic subgroup → Lagrange's theorem → divisibility. Each step is tight: ord(a) = n exactly because the n powers are all distinct and aⁿ = e; those n elements form a genuine subgroup; and Lagrange proves any subgroup's size divides |G| by the coset partition argument. This means knowing |G| immediately constrains which orders are possible, making element order one of the primary tools for deducing group structure.