Set Theory & Elementary Group Theory

Subgroup of cyclic group

Every subgroup of a cyclic group is itself cyclic, and for a finite cyclic group of order $n$, there exists a unique subgroup of order $d$ for each positive divisor $d$ of $n$.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Fundamental Theorem of Cyclic Subgroups & Divisor Structure

Let $G = \langle a \rangle$ be a cyclic group. (1) Every subgroup of $G$ is cyclic. (2) If $G$ is a finite cyclic group of order $n$, then for every positive divisor $d$ of $n$, $G$ contains a unique subgroup of order $d$, generated by $a^{n/d}$. (3) If $G$ is an infinite cyclic group, its non-trivial subgroups are of the form $\langle a^m \rangle$ for unique positive integers $m$, all of which are infinite cyclic.

📐 Geometric Interpretation

Visualized as symmetries of regular polygons (dihedral or cyclic rotations), where subgroups correspond to sub-rotations that close upon themselves after $d$ steps, partitioning the full circular symmetry into equal angular sectors.

⚙️ Algebraic & Structural Meaning

Establishes a complete isomorphism and lattice classification of subgroups of cyclic groups, mapping group-theoretic inclusion directly to the divisibility lattice of integers.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Cyclicity of the parent group $G$	`Entire group $G$`	Strictly Mandatory / Necessary	If $G$ is non-cyclic (e.g., Klein 4-group $V_4$ or symmetric group $S_3$), not all subgroups are cyclic (e.g., $S_3$ contains non-cyclic subgroups or subgroups that fail uniqueness properties).
Divisibility condition $d \mid n$	`Divisors of group order $n$`	Strictly Mandatory for existence and uniqueness	If $d$ does not divide $n$ ($d \nmid n$), a finite cyclic group of order $n$ contains NO subgroup of order $d$.
Generator power relation $k = n/d$	`Exponent integer $k \in \mathbb{Z}$`	Essential for explicit construction	Choosing an arbitrary element whose order does not divide $n$ will generate a subgroup whose order fails to divide $n$, contradicting Lagrange's Theorem.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Number of Subgroups of $\mathbb{Z}_n$

$$ \text{Number of subgroups of } \mathbb{Z}_n = \tau(n) $$

Rule: The total number of subgroups of a finite cyclic group $\mathbb{Z}_n$ equals the total number of positive divisors of $n$, denoted by $\tau(n) = \sum_{d \mid n} 1$.

Application: undefined

Number of Elements of Order $d$ in $\mathbb{Z}_n$

$$ \text{Number of elements of order } d \text{ in } \mathbb{Z}_n = \phi(d) \text{ if } d \mid n \text{ (else 0)} $$

Rule: If $d$ divides $n$, the number of elements of exact order $d$ in any cyclic group of order $n$ is given by Euler's totient function $\phi(d)$.

Application: Frequently asked in KVS PGT and DSSSB: 'How many elements of order 5 exist in $\mathbb{Z}_{50}$?' Answer: $\phi(5) = 4$.

Number of Generators of $\mathbb{Z}_n$

$$ \text{Number of generators of } \mathbb{Z}_n = \phi(n) $$

Rule: A cyclic group of order $n$ has exactly $\phi(n)$ distinct generators, corresponding to elements coprime to $n$.

Application: Standard UP TGT question asking for the number of generators of $\mathbb{Z}_{12}$.

Intersection of Subgroups

$$ \langle a^r \rangle \cap \langle a^s \rangle = \langle a^{\text{lcm}(r, s)} \rangle $$

Rule: The intersection of two cyclic subgroups generated by $a^r$ and $a^s$ is generated by $a$ raised to the least common multiple of $r$ and $s$ (appropriately scaled by group order if finite).

Application: Advanced TGT/PGT matching questions on subgroup lattices.

Order of Subgroup Generated by $a^k$

$$ |\langle a^k \rangle| = \frac{n}{\gcd(n, k)} $$

Rule: To find the exact order of the subgroup generated by any element $a^k$ in $\mathbb{Z}_n$, divide $n$ by the greatest common divisor of $n$ and $k$.

Rapid Option Elimination via Divisibility

$$ d \nmid n \implies \text{No subgroup of order } d $$

Rule: Instantly eliminate any option claiming a subgroup of order $d$ exists in $\mathbb{Z}_n$ if $d$ does not divide $n$.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Assuming Non-Cyclic Groups Behave Like Cyclic Groups

Common Pitfall: Candidates assume that if every proper subgroup of a group is cyclic, the group itself must be cyclic, or vice versa.

Counterexample: The Klein 4-group $V_4 = \{e, a, b, ab\}$

Why it fails: $V_4$ is not cyclic (every non-identity element has order 2), yet all its proper subgroups ($\langle a \rangle, \langle b \rangle, \langle ab \rangle$) are cyclic of order 2.

🛡️ Exam Defense: Never assume group cyclicity from subgroup cyclicity alone unless specific order conditions (like prime power orders) are met.

⚠️ COMMON ERROR #02

Confusing 'Number of Subgroups' with 'Number of Elements of a Given Order'

Common Pitfall: Applying $\phi(d)$ when asked for the number of subgroups, or applying divisor counting $\tau(d)$ when asked for elements of order $d$.

Counterexample: In $\mathbb{Z}_{12}$, divisors of 12 are 1, 2, 3, 4, 6, 12 (6 divisors $\implies$ 6 subgroups). But elements of order 4 are $\phi(4) = 2$ ($3$ and $9$).

Why it fails: Subgroups of order $d$ are unique in cyclic groups, so number of subgroups of order $d$ is 1 (if $d \mid n$), whereas elements of order $d$ number $\phi(d)$.

🛡️ Exam Defense: Read carefully: 'subgroups of order $d$' $\to$ exactly 1 subgroup. 'Elements of order $d$' $\to$ exactly $\phi(d)$ elements.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Subgroups of $\mathbb{Z}_n$	$n \in \mathbb{Z}^+$	1 subgroup for each $d \mid n$	Sub-rotations of regular $n$-gon	Keywords: 'Number of subgroups of $\mathbb{Z}_n$'
Generators of $\mathbb{Z}_n$	$n \in \mathbb{Z}^+$	Total generators = $\phi(n)$	Primitive roots of unity rotation	Keywords: 'How many generators'
Elements of Order $d$	$d \mid n$	Total elements = $\phi(d)$	Cycles of exact length $d$	Keywords: 'Elements of order $k$'
Infinite Cyclic Group $\mathbb{Z}$	$G = \langle 1 \rceil$	Subgroups are $m\mathbb{Z}$ ($m \ge 0$)	Infinite integer lattice scaling	Keywords: 'Subgroups of $\mathbb{Z}$'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the number of subgroups of the cyclic group $\mathbb{Z}_{360}$.

(A) 16(B) 24(C) 12(D) 8

📚 Method A: Formal Textbook Derivation

Step 1: Write the prime factorization of $360$. $360 = 36 \times 10 = 2^3 \times 3^2 \times 5^1$. Step 2: Use the divisor function formula $\tau(360) = (3+1)(2+1)(1+1)$. Step 3: Compute $\tau(360) = 4 \times 3 \times 2 = 24$.

⚡ Method B: 25-Second Exam Shortcut

Recognize that the number of subgroups of $\mathbb{Z}_n$ equals $\tau(n)$. Prime factorization $\to$ add 1 to each exponent $\to$ multiply.

💡 Tip: Do not list divisors manually for large numbers; always prime factorize and use $\tau(n) = \prod (k_i + 1)$.

CANONICAL PROBLEM 02 Dual-Method Breakdown

How many elements of order 10 are present in a cyclic group of order 30?

(A) 1(B) 2(C) 4(D) 6

📚 Method A: Formal Textbook Derivation

Step 1: Check if $10 \mid 30$ (Yes). Step 2: The number of elements of order $d$ in a cyclic group of order $n$ (where $d \mid n$) is $\phi(d)$. Step 3: Calculate $\phi(10)$. $\phi(10) = \phi(2) \phi(5) = 1 \times 4 = 4$.

⚡ Method B: 25-Second Exam Shortcut

For cyclic groups, number of elements of order $d$ is $\phi(d)$ if $d \mid n$. Directly compute $\phi(10) = 4$.

💡 Tip: Ensure $d$ divides $n$ first. If $d$ did not divide $n$, the answer would instantly be 0.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: Every subgroup of a cyclic group is:

A Not necessarily a group

B Always cyclic

C Always finite

D Always abelian but not cyclic

Derivation & Analysis:

By the Fundamental Theorem of Cyclic Groups, every subgroup of a cyclic group is cyclic.

MCQ 2: The number of subgroups of $\mathbb{Z}_{50}$ is:

A 3

B 4

C 5

D 6

Derivation & Analysis:

$50 = 2^1 \times 5^2$. The number of divisors is $(1+1)(2+1) = 2 \times 3 = 6$. Hence, there are 6 subgroups.

MCQ 3: In the additive group $\mathbb{Z}$, the number of distinct subgroups is:

A Finite

B Countably infinite

C Uncountably infinite

D Zero

Derivation & Analysis:

Subgroups of $\mathbb{Z}$ are of the form $m\mathbb{Z}$ for $m \ge 0$. Since there is one subgroup for each non-negative integer $m$, the set of subgroups is countably infinite.

MCQ 4: If $G = \langle a \rangle$ is a cyclic group of order 12, the order of the subgroup $\langle a^4 \rangle$ is:

A 3

B 4

C 6

D 12

Derivation & Analysis:

The order of $\langle a^k \rangle$ in $\mathbb{Z}_n$ is $\frac{n}{\gcd(n, k)}$. Here $\frac{12}{\gcd(12, 4)} = \frac{12}{4} = 3$.

MCQ 5: The number of generators of a cyclic group of order 20 is:

A 4

B 8

C 10

D 20

Derivation & Analysis:

The number of generators of $\mathbb{Z}_{20}$ is $\phi(20) = \phi(4) \phi(5) = 2 \times 4 = 8$.

MCQ 6: Which of the following groups has a unique subgroup of order 7?

A $S_7$

B $\mathbb{Z}_{42}$

C The Klein 4-group

D $\mathbb{Z}_{13}$

Derivation & Analysis:

$\mathbb{Z}_{42}$ is a finite cyclic group of order 42. Since $7 \mid 42$, by the Fundamental Theorem of Cyclic Groups, $\mathbb{Z}_{42}$ contains a unique subgroup of order 7.

MCQ 7: If $H$ and $K$ are subgroups of a cyclic group $G$ of orders 4 and 6 respectively, then $H \cap K$ must have order:

A 1

B 2

C 4

D 12

Derivation & Analysis:

The order of $H \cap K$ must divide both $|H| = 4$ and $|K| = 6$. The greatest common divisor is $\gcd(4, 6) = 2$. Thus, $|H \cap K| = 2$.

MCQ 8: Assertion (A): Every subgroup of $\mathbb{Z}_n$ is normal. Reason (R): Every subgroup of an abelian group is normal.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation of (A)

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Cyclic groups are abelian. Every subgroup of an abelian group is normal. Since $\mathbb{Z}_n$ is cyclic, it is abelian, so all its subgroups are normal. Reason correctly explains Assertion.

MCQ 9: The number of elements of order 3 in $\mathbb{Z}_{18}$ is:

A 1

B 2

C 3

D 6

Derivation & Analysis:

Since $3 \mid 18$, the number of elements of order 3 in $\mathbb{Z}_{18}$ is $\phi(3) = 2$.

Q10

MCQ 10: If a finite group $G$ has exactly one subgroup for each divisor of its order, then $G$ is necessarily:

A An abelian group

B A cyclic group

C A symmetric group

D A dihedral group

Derivation & Analysis:

A finite group $G$ is cyclic if and only if it has at most one subgroup of order $d$ for every divisor $d$ of $|G|$. Having exactly one subgroup for each divisor characterizes finite cyclic groups.

Common questions

Yes. This is a foundational theorem in abstract algebra (part of the Fundamental Theorem of Cyclic Groups). Both finite and infinite cyclic groups possess this property.

Candidates frequently confuse 'number of subgroups of order $d$' (which is always 1 if $d \mid n$ in a cyclic group) with 'number of elements of order $d$' (which is $\phi(d)$).

Master prime factorization shortcuts. Memorize totient values up to $\phi(12)$ and use $\tau(n) = \prod(k_i + 1)$ to bypass manual counting.

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