The set of non-zero elements of ℤ₆ under multiplication modulo 6 is:

Not a group, because 2 has no inverse

Not a group, because it is not closed

Not a group, because 2 has no inverse

Not a group, because multiplication modulo 6 is not associative

The set of odd integers under addition is:

Not closed under the operation, so not even a semigroup

Which of the following statements are true for every finite group G? (More than one option may be correct.)

The order of every element of G divides |G|

If |G| is prime then G is cyclic

The identity element of G is unique

The order of every element of G divides |G|

If |G| is prime then G is cyclic

The identity element of G is unique

Monoids and Groups

Almost every question in this topic is "is this a group?", and it is answered by walking four conditions in a fixed order — closure, associativity, identity, inverses — and stopping at the first failure. Doing them in that order matters, because closure is the cheapest to test and the most often broken by the examiner: a set that looks fine can fail simply because the operation takes two members outside it. The names above the axioms are worth keeping straight, since a question will say "monoid" when it means something weaker than a group: a semigroup has closure and associativity, a monoid adds an identity, and a group adds inverses for every element. Add commutativity and a group is abelian. Two results then answer most of the rest. Lagrange's theorem: the order of a subgroup divides the order of the group, and so does the order of every element — so in a group of order 12 no element can have order 5, and a question offering 5 as an option is testing exactly that. And for the standard examples: (ℤₙ, +) is always a group, cyclic, generated by any element coprime to n, so it has φ(n) generators — four for ℤ₁₂, four for ℤ₁₀, six for ℤ₇. Multiplication is the fussier case: (ℤₙ \ {0}, ×) is a group only when n is prime, because otherwise a proper divisor of n has no inverse — in ℤ₆ nothing multiplies 2 to give 1. That single distinction accounts for a large share of the questions set on this bullet, and it comes straight from the definition rather than from anything to memorise.

Four Axioms, in the Order Worth Testing Them

The structures and what each one requires
Structure	Requires	A worked example
Semigroup	Closure, associativity	The positive integers under addition — no identity, since 0 is not in the set.
Monoid	+ an identity element	The natural numbers including 0 under addition — 0 is the identity, but 3 has no inverse.
Group	+ an inverse for every element	The integers under addition; (ℤₙ, +) for any n.
Abelian group	+ commutativity	All of the above examples. Matrix multiplication is the standard non-abelian case.

🧠 Test closure first — it is the cheapest and the most often broken

Associativity is usually inherited from ordinary arithmetic and rarely the thing an examiner breaks. Closure is broken constantly, and it is broken quietly: "the set of odd integers under addition" fails because 3 + 5 = 8, and nothing about the set looks suspicious until you try it. Two members and one operation settle it, so try the two most awkward members you can find before checking anything else. The other cheap failure is a missing identity caused by the set boundary rather than the operation — the positive integers under addition are associative and closed and still not a monoid, because 0 was excluded.

Lagrange, Orders and Cyclic Groups

The order of a subgroup divides the order of the group. In a group of order 12 the only possible subgroup orders are 1, 2, 3, 4, 6 and 12 — the divisors of 12. An option offering a subgroup of order 5 can be discarded without any construction.
The order of an element divides the order of the group too, since the element generates a cyclic subgroup of exactly that order. In (ℤ₁₂, +) the element 3 has order 4, because 3 + 3 + 3 + 3 = 12 ≡ 0 and no smaller multiple vanishes; the element 8 has order 3.
(ℤₙ, +) is cyclic, and its generators are the elements coprime to n — so there are φ(n) of them. ℤ₁₂ has φ(12) = 4 generators (1, 5, 7, 11); ℤ₁₀ has φ(10) = 4; ℤ₇, being of prime order, has φ(7) = 6, which is every element except 0.
A group of prime order is cyclic and has no proper non-trivial subgroup, straight from Lagrange: the only divisors of a prime p are 1 and p. This is why questions about groups of order 5, 7 or 11 have short answers.

⚠️ Multiplication mod n is a group only for prime n

(ℤₙ, +) is a group for every n. (ℤₙ \ {0}, ×) is a group only when n is prime. In ℤ₅ every non-zero element has a multiplicative inverse — 2 × 3 = 6 ≡ 1, 4 × 4 = 16 ≡ 1 — so it is a group of order 4. In ℤ₆ the element 2 has none: 2 × 1, 2 × 2, …, 2 × 5 give 2, 4, 0, 2, 4 and never 1, because 2 shares a factor with 6. The general rule is that an element is invertible mod n exactly when it is coprime to n, which is why the invertible elements always number φ(n) and form a group of that size whatever n is. A question that says "ℤₙ under multiplication" is almost always testing whether you checked the primality of n first.

Key takeaways

Test the axioms in order — closure, associativity, identity, inverses — and stop at the first failure. Closure is the cheapest and the most often broken.
Semigroup = closure + associativity; monoid adds an identity; group adds inverses; abelian adds commutativity. A question saying "monoid" means something weaker than a group.
Lagrange: the order of every subgroup and of every element divides the order of the group. In a group of order 12 no element has order 5.
(ℤₙ, +) is cyclic with φ(n) generators — the elements coprime to n. ℤ₁₂ has four: 1, 5, 7 and 11.
(ℤₙ \ {0}, ×) is a group only when n is prime. In ℤ₆ the element 2 has no inverse, because it shares a factor with 6.
An element is invertible mod n exactly when it is coprime to n, so the invertible elements always number φ(n) and form a group of that size.
A group of prime order is cyclic and has no proper non-trivial subgroup, straight from Lagrange.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The number of generators of the cyclic group (ℤ₁₂, +) is:
1. 1
2. 4
3. 6
4. 12
Show answer
Answer: B — 4
An element k generates (ℤₙ, +) exactly when gcd(k, n) = 1, so the generators number φ(n). Here φ(12) = 12 × (1 − 1/2) × (1 − 1/3) = 4, and they are 1, 5, 7 and 11. Option A is what you get by assuming a cyclic group has one generator, which is true only for ℤ₁ and ℤ₂; option D counts every element, and 0 generates nothing at all. Option C is φ(7), a common answer to a neighbouring question — for prime p, φ(p) = p − 1, so ℤ₇ has six generators and every non-zero element is one. Compute the totient rather than testing twelve elements: 4 arrives in one line and the enumeration takes twelve.
The set of non-zero elements of ℤ₆ under multiplication modulo 6 is:
1. A group
2. Not a group, because it is not closed
3. Not a group, because 2 has no inverse
4. Not a group, because multiplication modulo 6 is not associative
Show answer
Answer: C — Not a group, because 2 has no inverse
The set {1, 2, 3, 4, 5} is not a group, and the reason matters because three of the four options give a reason. Closure fails as well — 2 × 3 = 6 ≡ 0, which is outside the set — but the reason the option list is testing is the missing inverse: 2 × 1, 2 × 2, 2 × 3, 2 × 4, 2 × 5 give 2, 4, 0, 2, 4 and never 1, so 2 has no inverse. Option B is not false, which makes this a question about reading: with two true reasons available an examiner will usually name the deeper one, and invertibility is the axiom that separates the composite case from the prime one. Option D is simply wrong — multiplication mod n is always associative, inherited from the integers. In ℤ₅ the same construction is a group, because 5 is prime.
In a group of order 12, which of the following cannot be the order of a subgroup?
1. 2
2. 3
3. 5
4. 6
Show answer
Answer: C — 5
By Lagrange, a subgroup order must divide the group order. The divisors of 12 are 1, 2, 3, 4, 6 and 12, so 5 is impossible — no construction needed, and no knowledge of which group of order 12 is meant. That is the value of the theorem in an exam: it rules out options from arithmetic alone. Note what Lagrange does not promise, which is the neighbouring question: a divisor of the group order need not actually occur as a subgroup order in every group. It happens to for order 12, but the converse of Lagrange is false in general, and a question asking whether a subgroup of a given order must exist is a different and harder question from this one.
The set of natural numbers including 0, under addition, is:
1. A group
2. A monoid but not a group
3. A semigroup but not a monoid
4. Not even a semigroup
Show answer
Answer: B — A monoid but not a group
Walk the axioms. Closure: the sum of two naturals is a natural. Associativity: inherited from the integers. Identity: 0 is present and a + 0 = a. So it is a monoid. Inverses: 3 would need −3, which is not a natural number — so it is not a group, and the answer is B. Option C is the answer for the positive integers, which exclude 0 and therefore have no identity: a one-element difference in the set changes the answer, and that is the distinction the question is set on. Option A is what a candidate answers by checking three axioms and stopping, which is exactly why the order of the checks should end at inverses rather than begin there.
The set of odd integers under addition is:
1. An abelian group
2. A monoid
3. Not closed under the operation, so not even a semigroup
4. A group but not abelian
Show answer
Answer: C — Not closed under the operation, so not even a semigroup
Closure fails at the first attempt: 3 + 5 = 8, which is even and outside the set. So it is not even a semigroup, and the remaining three axioms never need testing. This is why closure is worth checking first — nothing about "the odd integers" looks suspicious, and a candidate who begins with associativity and identity will do three correct pieces of work before discovering that the structure does not exist. Option D can also be dismissed on principle: addition is commutative wherever it is defined, so "a group but not abelian" is impossible for any set under ordinary addition, whatever the set is.
Which of the following statements are true for every finite group G? (More than one option may be correct.)
1. The order of every element of G divides |G|
2. G is abelian
3. If |G| is prime then G is cyclic
4. The identity element of G is unique
Show answer
Answer: A — The order of every element of G divides |G|; C — If |G| is prime then G is cyclic; D — The identity element of G is unique
Three of the four hold for every finite group. A is Lagrange applied to the cyclic subgroup an element generates. C follows from it: if |G| is prime p, a non-identity element has order dividing p and greater than 1, hence exactly p, so it generates the whole group. D is true in any group at all, finite or not — if e and f are both identities then e = e·f = f. Only B fails, and it fails on the smallest counterexample available: the symmetric group on three letters has order 6 and is not commutative. Assuming commutativity is the single most common unwarranted step in this topic, which is why "abelian" appears in the axiom list as an extra condition rather than part of the definition of a group.
In the group (ℤ₁₂, +), the order of the element 3 is _____.
Numerical answer — type the value.
Show answer
Answer: 4
The order of an element is the smallest positive k with k·3 ≡ 0 (mod 12). Adding: 3, 6, 9, 12 ≡ 0 — so k = 4. Equivalently, the order of a in (ℤₙ, +) is n / gcd(a, n) = 12 / gcd(3, 12) = 12 / 3 = 4, which is the formula worth carrying because it answers the whole family at once: the element 8 has order 12 / gcd(8, 12) = 12 / 4 = 3, and the element 5 has order 12 / 1 = 12, making it a generator. Note that 4 divides 12, as Lagrange requires — a quick check on any answer here, since an order that failed to divide the group order would be wrong whatever the arithmetic said.
The number of elements of ℤ₁₀ that have a multiplicative inverse modulo 10 is _____.
Numerical answer — type the value.
Show answer
Answer: 4
An element is invertible mod n exactly when it is coprime to n, so the count is φ(10) = 10 × (1 − 1/2) × (1 − 1/5) = 4 — the elements 1, 3, 7 and 9. Check one: 3 × 7 = 21 ≡ 1 (mod 10), and 9 × 9 = 81 ≡ 1. The other six share a factor with 10 and have no inverse: 2 × k is always even, so it never reaches 1, and 5 × k is always 0 or 5. Those four elements form a group of order 4 under multiplication mod 10, which is the general fact worth taking away — the invertible elements mod n always form a group, of size φ(n), whether or not n is prime; primality is only what makes that group all the non-zero elements.