Cyclic groups

At a Glance

• Frequency: 8 sub-parts across 8 of 14 years (2013, 2015, 2016, 2017, 2018, 2020, 2022, 2025)
• Priority tier: T1
• Marks (count): 10 (3), 15 (3), 20 (1), 5 (1)
• Average solve time: ~7 min
• Difficulty mix: easy 6, medium 2
• Section: A | Dominant type: proof

Why This Chapter Matters

Cyclic groups appear in 8 of the last 14 years — one of the most reliable Section A atoms in Paper 2. The questions are almost uniformly easy to medium: six of the eight questions in the corpus are rated easy. The core theory is compact: understand the order formula $\operatorname{ord}(a^k)=n/\gcd(n,k)$, know that a cyclic group has exactly one subgroup per divisor and $\phi(n)$ generators, and be comfortable with the CRT isomorphism $\mathbb Z_m\times\mathbb Z_n\cong\mathbb Z_{mn}$ when $\gcd(m,n)=1$. These three facts cover every question that has ever appeared; the harder 15-mark questions (2016, 2017, 2020) simply require writing them out as proofs rather than quoting them.

Minimum Theory

Cyclic groups and generators. A group $G=\langle a\rangle$ is cyclic if every element is a power of a single element $a$ (the generator). In an additive cyclic group, “power” means “multiple.” The group $\mathbb Z_n=\{0,1,\ldots,n-1\}$ under addition mod $n$ is the standard finite cyclic group of order $n$; the group $\mathbb Z$ of integers under addition is the standard infinite cyclic group. Any two cyclic groups of the same finite order $n$ are isomorphic to $\mathbb Z_n$.

Order of elements. In a cyclic group $G=\langle a\rangle$ of order $n$, the order of $a^k$ is $\operatorname{ord}(a^k)=\frac{n}{\gcd(n,k)}.$ Proof: $(a^k)^m=e\iff n\mid km\iff n/\gcd(n,k)\mid m$ (after cancelling $\gcd$). It follows that $a^k$ is a generator (has order $n$) iff $\gcd(k,n)=1$; the number of generators is $\phi(n)$ (Euler’s totient).

Subgroups. A cyclic group of order $n$ has exactly one subgroup of order $d$ for each divisor $d\mid n$, namely $\langle a^{n/d}\rangle$. There are $\tau(n)$ subgroups in total and $\tau(n)-1$ proper subgroups (excluding $G$ itself). Quotients: $G/H$ for $H=\langle a^{n/d}\rangle$ is cyclic of order $d$, isomorphic to $\mathbb Z_d$.

Product and CRT. $\mathbb Z_m\times\mathbb Z_n\cong\mathbb Z_{mn}$ iff $\gcd(m,n)=1$. Proof: the element $(1,1)$ has order $\operatorname{lcm}(m,n)=mn$; equal to $|G|$ makes it a generator. The explicit isomorphism is the Chinese Remainder map $x\bmod mn\;\mapsto\;(x\bmod m,\,x\bmod n)$.

Multiplicative group of $\mathbb Z_p$. For prime $p$, $\mathbb Z_p^\times=\{1,2,\ldots,p-1\}$ under multiplication mod $p$ is cyclic of order $p-1$. Its subgroups are in bijection with divisors of $p-1$; the unique subgroup of order $d\mid(p-1)$ is generated by $g^{(p-1)/d}$ where $g$ is a primitive root mod $p$.

Question Archetypes

Six patterns cover every cyclic-groups question in the corpus.

count-generators (2 question(s); 2015, 2020)

Recognition Cues

• “How many generators does the cyclic group of order $n$ have?” — answer $\phi(n)$.
• “Prove that a cyclic group of order $n$ has exactly $\phi(n)$ generators.”
• Explicit: list the generators of $\mathbb Z_8$, or $\mathbb Z_n$ for given $n$.

Solution Template

Count only:

1. State: generator of $\langle a\rangle$ of order $n$ means $\operatorname{ord}(a^k)=n$ iff $\gcd(k,n)=1$.
2. Count integers $k\in\{1,\ldots,n\}$ with $\gcd(k,n)=1$: this is $\phi(n)$ by definition.

Proof (15-mark):

1. Prove the order formula $\operatorname{ord}(a^k)=n/\gcd(n,k)$ by showing $(a^k)^m=e\iff n\mid km\iff n/d\mid m$ where $d=\gcd(n,k)$.
2. Deduce $a^k$ generates $G$ iff $\gcd(k,n)=1$.
3. The count is $\phi(n)$ by definition of Euler’s totient.

Worked Example(s)

2015 Paper 2, 2015-P2-Q1a-i (5 marks)

How many generators of cyclic group of order 8? Explain.

$\phi(8)=\phi(2^3)=8-4=4$. The generators are $a^k$ for $\gcd(k,8)=1$, i.e. $k\in\{1,3,5,7\}$: $\boxed{\;\text{4 generators: }a,\ a^3,\ a^5,\ a^7.\;}$

Check: $a^2$ has order $8/\gcd(2,8)=4$, not $8$ — it is not a generator.

2020 Paper 2, 2020-P2-Q2a (15 marks)

$G$ finite cyclic of order $n$. Prove $G$ has $\phi(n)$ generators.

Let $G=\langle a\rangle$, $|G|=n$.

Lemma (order of $a^k$). Set $d=\gcd(n,k)$. Then $(a^k)^m=e\iff n\mid km$. Write $n=dn'$, $k=dk'$ with $\gcd(n',k')=1$; then $n\mid km\iff n'\mid k'm\iff n'\mid m$ (since $\gcd(n',k')=1$ and Euclid’s lemma). The smallest positive $m$ is $n'=n/d$: $\operatorname{ord}(a^k)=\frac{n}{\gcd(n,k)}.$

Generator condition. $a^k$ generates $G$ iff $\operatorname{ord}(a^k)=n$ iff $n/\gcd(n,k)=n$ iff $\gcd(k,n)=1$.

Count. The generators are $\{a^k:1\le k\le n,\ \gcd(k,n)=1\}$; there are exactly $\phi(n)$ such $k$ by definition. $\blacksquare$

Common Traps

• A generator must have order $n$, not just be a “non-identity” element. $a^2$ in $\langle a\rangle$ of order 8 has order 4, not a generator.
• The lemma must be proved in the 15-mark version — the examiner expects the $n\mid km\iff n/d\mid m$ chain.
• Distinct exponents $0\le k<n$ give distinct group elements, so counting exponents equals counting generators.

cyclic-isomorphism (2 question(s); 2017, 2022)

Recognition Cues

• “Show $G\cong H$” where both groups are cyclic of the same finite order.
• Or “show $\mathbb Z_m\times\mathbb Z_n\cong\mathbb Z_{mn}$” — the CRT argument.
• The question may give one group multiplicatively and another additively.

Solution Template

Via finding a generator (for products):

1. Find an element of order $|G|$ in the product. For $(1,1)\in\mathbb Z_m\times\mathbb Z_n$: its order is $\operatorname{lcm}(m,n)=mn$ iff $\gcd(m,n)=1$.
2. Element of order $|G|$ generates $G$ → $G$ cyclic.
3. Both groups cyclic of same order → isomorphic to $\mathbb Z_{mn}$.

Via explicit isomorphism:

1. Both groups have the same finite order and are cyclic; map generator to generator.
2. Show the map is a bijective homomorphism ($\phi(ab)=\phi(a)\phi(b)$ via the generator relation).

Worked Example(s)

2017 Paper 2, 2017-P2-Q3a (15 marks)

Show $\mathbb Z_5\times\mathbb Z_7\cong\mathbb Z_{35}$.

Order of $(1,1)$. $k(1,1)=(0,0)$ iff $5\mid k$ and $7\mid k$ iff $35\mid k$ (since $\gcd(5,7)=1$). So $\operatorname{ord}(1,1)=\operatorname{lcm}(5,7)=35=|\mathbb Z_5\times\mathbb Z_7|$. Hence $\mathbb Z_5\times\mathbb Z_7=\langle(1,1)\rangle$ is cyclic of order 35.

Conclusion. Any two cyclic groups of order 35 are isomorphic, so $\mathbb Z_5\times\mathbb Z_7\cong\mathbb Z_{35}$. The CRT isomorphism is $x\bmod35\mapsto(x\bmod5,x\bmod7)$.

Key fact. $\operatorname{lcm}(m,n)=mn/\gcd(m,n)$; this equals $mn$ iff $\gcd(m,n)=1$. Coprimality is necessary: $\mathbb Z_2\times\mathbb Z_2\not\cong\mathbb Z_4$ (max element order 2 vs. 4).

2022 Paper 2, 2022-P2-Q1a (10 marks)

Show $G=\{1,-1,i,-i\}$ (multiplication) $\cong$ $G'=(\{0,1,2,3\},+_4)$.

$G=\langle i\rangle$ (order of $i$ is 4) and $G'=\langle 1\rangle$ (order of $1$ under $+_4$ is 4); both cyclic of order 4, hence both $\cong\mathbb Z_4$. The explicit isomorphism: $\phi:G\to G',\quad\phi(i^k)=k\bmod4.$ Homomorphism: $\phi(i^a\cdot i^b)=\phi(i^{a+b})=(a+b)\bmod4=\phi(i^a)+_4\phi(i^b)$ ✓. Bijection: explicit 4-element correspondence. $\boxed{\;\phi(1)=0,\ \phi(i)=1,\ \phi(-1)=2,\ \phi(-i)=3.\;}$

Common Traps

• “Same order” alone does not imply isomorphic for non-cyclic groups; for cyclic groups it does (since they are all isomorphic to $\mathbb Z_n$).
• The coprimality condition for products: $\mathbb Z_m\times\mathbb Z_n\cong\mathbb Z_{mn}$ requires $\gcd(m,n)=1$. With $\gcd=2$: $\mathbb Z_2\times\mathbb Z_2$ and $\mathbb Z_4$ are non-isomorphic (different element orders). State the coprimality before concluding.

cyclic-generation-proof (1 question(s); 2016)

Recognition Cues

• “Show every non-zero element of the additive group $\mathbb Z_p$ generates $\mathbb Z_p$.”
• The phrase “prime order” or “group of prime order” — Lagrange + primality forces everything.

Solution Template

1. Let $a\ne0$ in $\mathbb Z_p$ (prime). The cyclic subgroup $\langle a\rangle\subseteq\mathbb Z_p$ has order dividing $|\mathbb Z_p|=p$ (Lagrange).
2. Since $p$ is prime, the only divisors are $1$ and $p$. The order of $\langle a\rangle$ cannot be 1 (that would require $a=0$).
3. Hence $|\langle a\rangle|=p$, so $\langle a\rangle=\mathbb Z_p$.

Alternatively (without Lagrange): $\gcd(a,p)=1$ (since $p$ prime and $p\nmid a$). The multiples $0,a,2a,\ldots,(p-1)a$ are all distinct mod $p$ (if $ia\equiv ja$ then $p\mid(i-j)a$, and Euclid forces $p\mid(i-j)$, so $i=j$ for $0\le i,j<p$). So $\langle a\rangle$ contains all $p$ residues.

Worked Example(s)

2016 Paper 2, 2016-P2-Q2b (15 marks)

Show every non-zero element of $\mathbb Z_p$ ($p$ prime) generates $\mathbb Z_p$.

Let $1\le a\le p-1$. Since $p$ is prime and $p\nmid a$, we have $\gcd(a,p)=1$.

By Lagrange: $|\langle a\rangle|$ divides $p$. Since $p$ is prime, $|\langle a\rangle|\in\{1,p\}$. Since $a\ne0$, the subgroup $\langle a\rangle\ne\{0\}$, so $|\langle a\rangle|=p$, giving $\langle a\rangle=\mathbb Z_p$. $\blacksquare$

Self-contained alternative: The $p$ elements $\{0,a,2a,\ldots,(p-1)a\}\pmod p$ are distinct (if $ia\equiv ja\pmod p$ then $p\mid(i-j)a$; since $\gcd(a,p)=1$, Euclid gives $p\mid i-j$; for $0\le i,j<p$ this forces $i=j$). So $\langle a\rangle$ exhausts $\mathbb Z_p$.

$\boxed{\;\text{Every non-zero element of }\mathbb Z_p\text{ generates }\mathbb Z_p.\;\blacksquare\;}$

Common Traps

• This result is special to prime order. In $\mathbb Z_6$ (composite), $a=2$ only generates $\{0,2,4\}$ because $\gcd(2,6)=2\ne1$.
• In $\mathbb Z_p$, “generates” is additive: $\langle a\rangle=\{ka\bmod p:k\in\mathbb Z\}$, not powers $a^k$.
• When citing Lagrange, the theorem requires $\langle a\rangle$ to be a subgroup of $\mathbb Z_p$ — it is, since subsets closed under the group operation and containing inverses are subgroups.

group-example (1 question(s); 2013)

Recognition Cues

• “Give an example of an infinite group in which every element has finite order.”
• Or: “an example of a torsion group that is infinite.”

Solution Template

1. State the example: $G=\mathbb Q/\mathbb Z$ (additive group of rationals modulo integers).
2. Show $G$ is infinite: contains $1/n+\mathbb Z$ for every $n\ge2$, giving infinitely many distinct cosets.
3. Show every element has finite order: for $g=p/q+\mathbb Z$ (in lowest terms), $q\cdot g=p+\mathbb Z=0_G$ since $p\in\mathbb Z$. So $\operatorname{ord}(g)$ divides $q$ — finite.

Alternatively: $\mu_\infty=\{z\in\mathbb C^\times:z^n=1\text{ for some }n\ge1\}$ (group of all roots of unity) — infinite because it contains all primitive $n$-th roots for every $n$; every element has finite order by definition.

Worked Example(s)

2013 Paper 2, 2013-P2-Q1b (10 marks)

Give an example of an infinite group in which every element has finite order.

Example: $G=\mathbb Q/\mathbb Z$. Elements are cosets $r+\mathbb Z$, $r\in[0,1)\cap\mathbb Q$. The set is infinite (contains $1/n+\mathbb Z$ for each $n\ge2$, all distinct). For $g=p/q+\mathbb Z$ (lowest terms): $q\cdot g=p+\mathbb Z=0_G\quad(\text{since }p\in\mathbb Z),$ so $\operatorname{ord}(g)\le q<\infty$. $\boxed{\;\mathbb Q/\mathbb Z\text{ is infinite, with every element of finite order.}\;}$

Alternative: the group $\mu_\infty=\{e^{2\pi ir}:r\in\mathbb Q\}$ under multiplication. In fact $\mu_\infty\cong\mathbb Q/\mathbb Z$ via $r+\mathbb Z\mapsto e^{2\pi ir}$.

Common Traps

• Do not propose $\mathbb Q$ itself: the only finite-order element of $(\mathbb Q,+)$ is $0$, so $\mathbb Q$ is not a torsion group.
• In $\mathbb Q/\mathbb Z$, the coset $r+\mathbb Z$ has order exactly $q$ (denominator in lowest terms), not always 1 or $\infty$.

group-relation-proof (1 question(s); 2025)

Recognition Cues

• A small non-abelian group $G=\{e,x,x^2,y,yx,yx^2\}$ (order 6) with $o(x)=3$, $o(y)=2$.
• “Show $xy=yx^2$” (or equivalent conjugate relation $yxy^{-1}=x^2$).
• This is the dihedral group $D_3\cong S_3$.

Solution Template

1. Note $o(y)=2$ so $y^{-1}=y$.
2. Conjugation preserves order: $o(yxy)=o(x)=3$. The only elements of $G$ with order 3 are $x$ and $x^2$.
3. If $yxy=x$: then $yx=xy$, so $x,y$ commute, making $G$ abelian. Contradiction.
4. Therefore $yxy=x^2$. Left-multiply by $y$: $xy=yx^2$. $\square$

Worked Example(s)

2025 Paper 2, 2025-P2-Q1b (10 marks)

$G=\{e,x,x^2,y,yx,yx^2\}$, $o(x)=3$, $o(y)=2$, $G$ non-abelian. Show $xy=yx^2$.

Since $y^2=e$, $y^{-1}=y$. Consider $yxy\in G$: conjugation preserves order, so $o(yxy)=o(x)=3$. The elements of order 3 in $G$ are $x$ and $x^2$ (all others have order 1 or 2).

Case $yxy=x$: then $yx=xy$, so $x,y$ commute. Every product of $x,y$ commutes, making $G$ abelian — contradicting the hypothesis.

Therefore $yxy=x^2$. Left-multiply by $y$: $(y^2)xy=yx^2$, i.e.: $\boxed{\;xy=yx^2.\;}\qquad\blacksquare$

Common Traps

• The proof hinges on two facts: conjugation preserves order (so $yxy\in\{x,x^2\}$) and $G$ is non-abelian (so $yxy=x$ is impossible). Both must be stated.
• Don’t assume $xy=yx^2$ at the start and “verify” it — that’s circular. The argument must derive it.
• The elements $y, yx, yx^2$ each have order 2 (they are “reflections” in $D_3$); the only order-3 elements are $x$ and $x^2$.

subgroups-of-cyclic (1 question(s); 2018)

Recognition Cues

• “Find all (proper) subgroups of $\mathbb Z_p^\times$” or “of the multiplicative group of the field $\mathbb Z_p$.”
• Equivalently: find all subgroups of a cyclic group of order $n$ (one per divisor of $n$).

Solution Template

1. Show the multiplicative group is cyclic by finding a primitive root $g$ (verify $g$ has order $p-1$ by computing its powers mod $p$).
2. List divisors $d\mid(p-1)$ (including $d=1$ and $d=p-1$; exclude $d=p-1$ for proper subgroups).
3. For each proper divisor $d$: the subgroup of order $d$ is $\langle g^{(p-1)/d}\rangle$; list its elements explicitly.

Worked Example(s)

2018 Paper 2, 2018-P2-Q3a (20 marks)

Find all proper subgroups of $\mathbb Z_{13}^\times$.

$|\mathbb Z_{13}^\times|=12$. Divisors of 12: $\{1,2,3,4,6,12\}$. Proper divisors (exclude 12): $\{1,2,3,4,6\}$.

Primitive root. Compute powers of 2 mod 13: $2^1=2$, $2^2=4$, $2^3=8$, $2^4=3$, $2^6=12$, $2^{12}=1$. No smaller positive exponent gives 1 → $\operatorname{ord}(2)=12$, so $g=2$ is a primitive root.

Proper subgroups (using $H_d=\langle 2^{12/d}\rangle$ of order $d$):

$\boxed{\;\{1\},\;\{1,12\},\;\{1,3,9\},\;\{1,5,8,12\},\;\{1,3,4,9,10,12\}.\;}$

Common Traps

• “Proper” includes the trivial subgroup $\{1\}$ and excludes $G$ itself. List five subgroups for the five proper divisors of 12.
• Before using $2^{12/d}$ as the generator, verify that $g=2$ is actually a primitive root (order 12). Not every small integer is a generator — $\operatorname{ord}(3)=3$, not 12.
• The subgroup $H_6=\{1,3,4,9,10,12\}$ is the set of quadratic residues mod 13 (the unique index-2 subgroup). Use it as a sanity check.

Marks-Aware Writing

5-mark questions (2015-Q1a-i): One sentence stating $\phi(8)=4$, one sentence listing the generators $a,a^3,a^5,a^7$. Include the criterion $\gcd(k,8)=1$ to show understanding.

10-mark questions (2013-Q1b, 2022-Q1a, 2025-Q1b): Two to three paragraphs. For examples: state the example, show it is infinite, show every element has finite order (two separate bullet-points of reasoning). For isomorphisms: define $\phi$ explicitly, verify homomorphism, verify bijection. For the relation proof: three steps (order preservation → two cases → eliminate non-abelian case).

15-mark questions (2016-Q2b, 2017-Q3a, 2020-Q2a): Full proof format. For the generator count proof: explicitly prove the order lemma $\operatorname{ord}(a^k)=n/\gcd(n,k)$ — it is worth about 5 marks on its own. For the isomorphism: both “exhibit a generator” and “explicit map” routes are accepted; the CRT route is faster.

20-mark questions (2018-Q3a): Structure the answer clearly: (1) show $\mathbb Z_{13}^\times$ is cyclic (primitive root 2 computed explicitly), (2) state the divisor-subgroup correspondence (one subgroup per divisor, no more), (3) list each of the five proper subgroups with elements verified. The table format is the clearest presentation.

Practice Set