Chapter 03

Medium

Subgroups & Cyclic Groups

00 · Symbol Glossary

$H \le G$H is a subgroup of G

Read " $H$ is a subgroup of $G$ ." Means $H \subseteq G$ and $H$ is itself a group under $G$ 's operation. The symbol deliberately echoes $\le$ for numbers — subgroups really do behave like an ordering on the "sub-structures" of $G$ .

$\{e\}$Trivial subgroup

The subgroup containing only the identity. Every group has this as its smallest possible subgroup — never empty, since $e$ is always present.

$\langle g \rangle$Cyclic subgroup generated by g

The set $\{g^n : n\in\mathbb{Z}\}$ — every power (positive, negative, and zero) of a single element $g$ . Read "the subgroup generated by $g$ ."

$\langle g \rangle = G$G is cyclic with generator g

Means every element of $G$ is some power of the single element $g$ . A group satisfying this for some $g$ is called cyclic.

$\gcd(a,b)$Greatest common divisor

The largest positive integer dividing both $a$ and $b$ . Central to determining generators of $\mathbb{Z}_n$ and orders of powers, via the identity $|g^k| = |g|/\gcd(|g|,k)$ developed in this chapter.

$d \mid n$d divides n

Recalled from Chapter 01: $n = dk$ for some integer $k$ . Governs exactly which subgroups a cyclic group of order $n$ has — one for each positive divisor of $n$ , and no others.

$A_n$Alternating group

The subgroup of $S_n$ consisting of even permutations (those expressible as a product of an even number of transpositions). $|A_n| = n!/2$ . Previewed here as an example of a subgroup found by isolating a property closed under composition.

01 · What Is a Subgroup?

A subset can sit inside a group without inheriting group structure — for instance, $\{1,2\}\subset\mathbb{Z}$ has no identity for $+$ once you leave $0$ out. A subgroup is a subset that manages to be a self-contained group in its own right, using the ambient operation unchanged.

Definition — Subgroup

Let $(G,\ast)$ be a group. A subset $H\subseteq G$ is a subgroup of $G$ , written $H\le G$ , if $H$ is itself a group under $\ast$ (restricted to $H$ ).

Checking all four group axioms (G1–G4) from scratch every time would be wasteful — associativity is automatically inherited from $G$ (it held for all triples in $G$ , so certainly for the triples that happen to live in $H$ ). This leaves only three things to check, exactly mirroring the subspace test from linear algebra (Chapter 01 of Linear Algebra) and the abstract pattern set by Theorem 1.3's partition argument.

Theorem 3.1 — The Subgroup Test

A nonempty subset $H \subseteq G$ is a subgroup if and only if:

H1. $H$ is closed under $\ast$ : $a,b\in H \implies ab \in H$ .

H2. $H$ is closed under inverses: $a \in H \implies a^{-1} \in H$ .

Proof

( $\Rightarrow$ ) If $H\le G$ , it satisfies its own G1 and G4, which are exactly H1 and H2.

( $\Leftarrow$ ) Suppose $H$ satisfies H1, H2, and is nonempty. Closure (G1) is H1 directly. Associativity (G2) holds for all elements of $G$ , hence for the subset $H$ . Identity (G3): since $H\neq\varnothing$ , pick any $a\in H$ ; by H2, $a^{-1}\in H$ ; by H1, $aa^{-1}=e\in H$ . Inverses (G4) is H2 directly. All four axioms hold, so $H$ is a group under $\ast$ , i.e. $H\le G$ .

The identity is never assumed — it's derived

Notice the proof did not need to separately assume $e\in H$ ; nonemptiness plus H1 and H2 forces $e\in H$ automatically. This mirrors the subspace test's insistence on checking $\mathbf{0}\in W$ first, except here it's a consequence rather than a hypothesis — a small but genuine difference between the group and vector-space versions of this pattern.

Example — Every group has (at least) two trivial subgroups

$\{e\}$ : closed under the operation ( $ee=e$ ), closed under inverses ( $e^{-1}=e$ ), nonempty. Always a subgroup, called the trivial subgroup.

$G$ itself: trivially satisfies both conditions. Always a subgroup of itself.

Any subgroup strictly between these two extremes is called a proper nontrivial subgroup — the interesting case.

Step-by-step — Is $H=\{0,2,4\}$ a subgroup of $(\mathbb{Z}_6,+)$?

Check nonempty: yes, contains

0

Check closure (H1):

0+2=2\in H

2+2=4\in H

2+4=6\equiv0\in H

4+4=8\equiv2\in H

. Every sum of two elements of

H

lands back in

H

Check inverses (H2): additive inverse of

0

0\in H

. Inverse of

2

-2\equiv4\in H

. Inverse of

4

-4\equiv2\in H

Conclusion: both conditions hold.

H=\{0,2,4\}\le\mathbb{Z}_6

Common mistake — A subset that is closed but not a subgroup

Let $H=\{0,1,2\}\subset\mathbb{Z}_6$ . Closure fails immediately: $1+2=3\notin H$ . Not even closed, so certainly not a subgroup — but notice this is a weaker failure than being closed-but-missing-inverses. Always check H1 before H2; if H1 fails, there's no need to check inverses at all.

02 · More Subgroup Examples

Example — Multiples of an integer, revisited

$n\mathbb{Z} = \{0,\pm n,\pm2n,\ldots\}$ inside $(\mathbb{Z},+)$ : closed (sum of two multiples of $n$ is a multiple of $n$ ), closed under inverses (negative of a multiple of $n$ is a multiple of $n$ ), nonempty. $n\mathbb{Z}\le\mathbb{Z}$ for every positive integer $n$ — infinitely many subgroups of a single infinite group, one per choice of $n$ (plus $\mathbb{Z}$ itself, the case $n=1$ ).

Example — Rotations inside $D_n$

Let $R = \{e,r,r^2,\ldots,r^{n-1}\}\subset D_n$ (the rotations alone, no reflections). Closed: a rotation composed with a rotation is a rotation. Inverses: the inverse of a rotation is a rotation (rotating the opposite way). $R \le D_n$ , and $|R|=n$ exactly half of $|D_n|=2n$ .

Example — Even permutations form a subgroup of $S_n$

A transposition (2-cycle) like $(1\,2)$ swaps exactly two elements. Every permutation can be written as a product of transpositions, and — although a given permutation has many such expressions — the parity (even or odd number of transpositions used) is always the same. Call a permutation even if it needs an even number of transpositions.

$A_n = \{\text{even permutations}\}$ : composing two even permutations uses an even-plus-even (even) total count of transpositions, so $A_n$ is closed. The inverse of an even permutation is even (reversing a product of transpositions uses the same count). $A_n \le S_n$ , with $|A_n|=n!/2$ for $n\geq2$ — this specific subgroup will resurface heavily once Chapter 09's Sylow theory and Chapter 21's solvability-by-radicals both need it.

03 · A Shortcut for Finite Subgroups

For a finite group (or a finite candidate subset), checking inverses separately turns out to be unnecessary.

Theorem 3.2 — Finite Subgroup Test

If $G$ is finite and $H\subseteq G$ is nonempty and closed under $\ast$ (H1 only), then $H\le G$ automatically.

Proof

Let $a\in H$ . Since $H$ is closed under $\ast$ , all powers $a,a^2,a^3,\ldots$ lie in $H$ . Because $H\subseteq G$ is finite, this list cannot contain infinitely many distinct elements, so eventually two powers coincide: $a^i=a^j$ for some $i<j$ . Cancelling $a^i$ (valid by Theorem 2.1, since $G$ is a group) gives $a^{j-i}=e$ , so some positive power of $a$ equals $e$ — meaning $a$ has finite order, say $|a|=k$ , and $a^k=e\in H$ by closure.

Now if $k=1$ , $a=e$ and $a^{-1}=e=a\in H$ trivially. If $k>1$ , then $a^{-1}=a^{k-1}$ (since $a\cdot a^{k-1}=a^k=e$ ), and $a^{k-1}\in H$ by closure (it's a power of $a$ , and closure gives every power). Either way, $a^{-1}\in H$ . Since $a$ was arbitrary, H2 holds automatically, and Theorem 3.1 finishes the job.

Common mistake — This shortcut requires finiteness

Consider $H=\{0,1,2,3,\ldots\}$ (the nonnegative integers) inside $(\mathbb{Z},+)$ . Closed under addition: sum of two nonnegative integers is nonnegative. But $H$ is infinite, and indeed H2 fails: $1\in H$ but $-1\notin H$ . Theorem 3.2 does not apply here — finiteness is doing real work in the proof (guaranteeing powers eventually repeat), and dropping it breaks the conclusion.

04 · Cyclic Groups

Definition — Cyclic Subgroup Generated by an Element

For $g\in G$ , define $\langle g\rangle = \{g^n : n\in\mathbb{Z}\} = \{\ldots,g^{-2},g^{-1},e,g,g^2,\ldots\}$ .

Theorem 3.3 — $\langle g \rangle$ Is the Smallest Subgroup Containing $g$

$\langle g\rangle \le G$ , and any subgroup $H\le G$ with $g\in H$ satisfies $\langle g\rangle \subseteq H$ .

Proof

Subgroup: $\langle g\rangle$ is nonempty ( $e=g^0$ is in it). Closure: $g^m g^n = g^{m+n} \in \langle g\rangle$ . Inverses: $(g^n)^{-1} = g^{-n} \in \langle g\rangle$ . By Theorem 3.1, $\langle g\rangle \le G$ .

Smallest: suppose $H\le G$ and $g\in H$ . By closure, $g^2=gg\in H$ , and inductively $g^n\in H$ for every $n\geq1$ . By H2, $g^{-1}\in H$ , and by the same induction $g^{-n}\in H$ for $n\geq1$ . Also $e=gg^{-1}\in H$ . So every power of $g$ lies in $H$ , i.e. $\langle g\rangle \subseteq H$ .

Definition — Cyclic Group

$G$ is cyclic if $G = \langle g\rangle$ for some $g\in G$ , called a generator of $G$ .

Example — $(\mathbb{Z}_n, +)$ is cyclic with generator 1

In additive notation, " $1^k$ " means $1+1+\cdots+1$ ( $k$ times) $=k$ . So $\langle1\rangle = \{0,1,2,\ldots,n-1\} = \mathbb{Z}_n$ — every element of $\mathbb{Z}_n$ is some multiple of $1$ . $\mathbb{Z}_n$ is cyclic, and $1$ is always a valid generator (though frequently not the only one).

Example — $(\mathbb{Z}, +)$ is cyclic with generator 1

$\langle1\rangle = \{\ldots,-2,-1,0,1,2,\ldots\}=\mathbb{Z}$ . Every integer is a multiple of $1$ . $\mathbb{Z}$ is cyclic — the prototypical infinite cyclic group, in contrast to the finite cyclic groups $\mathbb{Z}_n$ .

Common mistake — Not every group is cyclic

Is $S_3$ cyclic? Its elements have orders $1$ (identity), $2$ (three transpositions), or $3$ (two 3-cycles) — checkable directly from Section 04 of Chapter 02. No element has order $6=|S_3|$ , so no single element's powers can produce all six elements. $S_3$ is not cyclic. Being generated by some collection of elements (which every finite group is, trivially, by all its elements) is a much weaker property than being generated by one.

Theorem 3.4 — Order of $\langle g \rangle$ Equals the Order of $g$

$|\langle g\rangle| = |g|$ (with both sides possibly infinite).

Proof

Case $|g|=\infty$ : suppose for contradiction $g^i = g^j$ for some $i<j$ . Then $g^{j-i}=e$ with $j-i>0$ , contradicting that $g$ has infinite order (no positive power equals $e$ ). So all powers $g^n$ ( $n\in\mathbb{Z}$ ) are distinct, and $\langle g\rangle$ is infinite: $|\langle g\rangle| = \infty = |g|$ .

Case $|g|=k$ finite: first, $\{e,g,g^2,\ldots,g^{k-1}\}$ are $k$ distinct elements — if $g^i=g^j$ for $0\le i<j\le k-1$ , then $g^{j-i}=e$ with $0<j-i<k$ , contradicting that $k$ is the smallest positive power giving $e$ . Second, every power of $g$ equals one of these $k$ : for any integer $n$ , write $n=qk+r$ with $0\le r<k$ (division algorithm), so $g^n = (g^k)^q g^r = e^q g^r = g^r$ , one of the $k$ listed elements. Hence $\langle g\rangle = \{e,g,\ldots,g^{k-1}\}$ has exactly $k$ elements: $|\langle g\rangle|=k=|g|$ .

Step-by-step — Finding $\langle 4 \rangle$ in $(\mathbb{Z}_{12}, +)$

Compute powers (multiples) of 4:

1\cdot4=4

2\cdot4=8

3\cdot4=12\equiv0

Stop at the first return to 0: the smallest positive

k

with

4k\equiv0\pmod{12}

k=3

. So

|4|=3

By Theorem 3.4,

|\langle4\rangle| = |4| = 3

, and the elements are

\{0,4,8\}

(the multiples of

4

found before it cycled back to

0

Conclusion:

\langle4\rangle = \{0,4,8\} \le \mathbb{Z}_{12}

, a proper subgroup of order

3

inside a group of order

12

05 · Subgroups of Cyclic Groups

Cyclic groups have an unusually clean subgroup structure — clean enough to fully classify without case-by-case checking.

Theorem 3.5 — Every Subgroup of a Cyclic Group Is Cyclic

If $G=\langle g\rangle$ and $H\le G$ , then $H$ is cyclic.

Proof

If $H=\{e\}$ , then $H=\langle e\rangle$ , cyclic. Otherwise $H$ contains some $g^m\neq e$ ; replacing $m$ with $-m$ if necessary (since $H$ is closed under inverses), assume $H$ contains $g^m$ for some positive $m$ . Let $d$ be the smallest positive integer with $g^d\in H$ — this exists because the set of such positive integers is nonempty (it contains $m$ ) and every nonempty set of positive integers has a least element (well-ordering).

Claim: $H=\langle g^d\rangle$ . Certainly $\langle g^d\rangle \subseteq H$ , since $H$ is closed under the operation and inverses, so every power of $g^d$ lies in $H$ .

For the reverse inclusion, take any $g^n\in H$ . By the division algorithm, $n=qd+r$ with $0\le r<d$ . Then $g^r = g^{n-qd} = g^n(g^d)^{-q}$ . Both $g^n\in H$ and $(g^d)^{-q}\in H$ (a power of $g^d\in H$ ), so their product $g^r\in H$ by closure. But $0\le r<d$ , and $d$ was chosen as the smallest positive integer with $g^d\in H$ — so $r$ must be $0$ (otherwise $r$ would be a smaller positive witness, contradicting minimality of $d$ ). Hence $n=qd$ , and $g^n = (g^d)^q \in \langle g^d\rangle$ .

So $H\subseteq\langle g^d\rangle$ and $\langle g^d\rangle\subseteq H$ , giving $H=\langle g^d\rangle$ — cyclic.

This is the group-theory twin of a familiar fact about Z

Setting $G=\mathbb{Z}=\langle1\rangle$ recovers a classical fact: every subgroup of $\mathbb{Z}$ has the form $n\mathbb{Z}$ for some $n\geq0$ (the subgroups from Section 02). The proof of Theorem 3.5 is literally the proof of that classical fact, written so it works for any cyclic group, not just $\mathbb{Z}$ — the payoff of working abstractly in the first place.

Theorem 3.6 — Subgroups of $\mathbb{Z}_n$ Correspond to Divisors of $n$

For each positive divisor $d$ of $n$ , $\mathbb{Z}_n$ has exactly one subgroup of order $d$ , namely $\langle n/d\rangle$ . These are all the subgroups of $\mathbb{Z}_n$ .

Example — All subgroups of $\mathbb{Z}_{12}$

Divisors of $12$ : $1,2,3,4,6,12$ — six divisors, hence exactly six subgroups.

Divisor $d$ (order of subgroup)	$n/d$	Subgroup $\langle n/d\rangle$
1	12	$\{0\}$
2	6	$\{0,6\}$
3	4	$\{0,4,8\}$
4	3	$\{0,3,6,9\}$
6	2	$\{0,2,4,6,8,10\}$
12	1	$\mathbb{Z}_{12}$

Compare row 3 to Section 04's step-by-step computation of $\langle4\rangle=\{0,4,8\}$ — Theorem 3.6 predicted this exact subgroup from divisibility alone, no repeated computation needed.

Common mistake — Assuming every element of a given order generates a *different* subgroup

$\mathbb{Z}_{12}$ has several elements of order $3$ (not just $4$ ): both $4$ and $8$ have order $3$ (check: $3\cdot8=24\equiv0$ , and no smaller positive multiple works). But $\langle4\rangle = \{0,4,8\} = \langle8\rangle$ — the same subgroup, generated two different ways. Theorem 3.6 counts subgroups, not generators; a single subgroup of order $d$ can have multiple valid generators, exactly as $1$ and $-1$ both generate all of $\mathbb{Z}$ .

06 · Exercises

EXERCISE 3.1

Apply Theorem 3.1: check nonemptiness, closure under matrix multiplication, and closure under matrix inverses.

Let $H = \{A \in GL_2(\mathbb{R}) : \det A = 1\}$ (the special linear group $SL_2(\mathbb{R})$ ).

Nonempty: $I \in H$ since $\det I=1$ .

Closure: if $\det A=1$ and $\det B=1$ , then $\det(AB) = \det A \cdot \det B = 1\cdot1=1$ , so $AB\in H$ .

Inverses: if $\det A=1$ , then $\det(A^{-1}) = 1/\det A = 1$ , so $A^{-1}\in H$ .

Both conditions of Theorem 3.1 hold: $H \le GL_2(\mathbb{R})$ .

Show that $H = \{A \in GL_2(\mathbb{R}) : \det A = 1\}$ is a subgroup of $GL_2(\mathbb{R})$ .

EXERCISE 3.2

Compute $|9|$ in $\mathbb{Z}_{15}$ first (the smallest $k$ with $9k \equiv 0 \pmod{15}$ ), then apply Theorem 3.4.

$1\cdot9=9$ , $2\cdot9=18\equiv3$ , $3\cdot9=27\equiv12$ , $4\cdot9=36\equiv6$ , $5\cdot9=45\equiv0$ . Smallest positive $k$ is $5$ , so $|9|=5$ .

By Theorem 3.4, $|\langle9\rangle| = |9| = 5$ , and $\langle9\rangle = \{0,9,3,12,6\} = \{0,3,6,9,12\}$ .

Find $|\langle 9\rangle|$ in $(\mathbb{Z}_{15},+)$ , and list its elements.

EXERCISE 3.3

List the orders of all six elements of $S_3$ (identity, three transpositions, two 3-cycles) and compare to $|S_3|=6$ .

Orders in $S_3$ : $e$ has order $1$ . Each transposition (e.g. $(1\,2)$ ) has order $2$ ( $((1\,2))^2=e$ ). Each 3-cycle (e.g. $(1\,2\,3)$ ) has order $3$ .

No element has order $6$ . By Theorem 3.4, $|\langle g\rangle| = |g|$ for every $g$ , so the largest cyclic subgroup any single element generates has order $3$ — never all of $S_3$ .

$S_3$ is not cyclic.

Using Theorem 3.4, explain why $S_3$ cannot be cyclic without checking every possible generator by brute force.

EXERCISE 3.4

Apply Theorem 3.6: the divisors of 20 determine the subgroup orders directly.

Divisors of $20$ : $1,2,4,5,10,20$ — six divisors.

By Theorem 3.6, $\mathbb{Z}_{20}$ has exactly six subgroups, of orders $1,2,4,5,10,20$ respectively (namely $\langle20\rangle=\{0\}$ , $\langle10\rangle$ , $\langle5\rangle$ , $\langle4\rangle$ , $\langle2\rangle$ , $\langle1\rangle=\mathbb{Z}_{20}$ ).

How many subgroups does $\mathbb{Z}_{20}$ have? List their orders using Theorem 3.6, without computing each subgroup's elements directly.

EXERCISE 3.5

Mimic the proof of Theorem 3.2 exactly: what does closure under $\ast$ guarantee about the powers of a single element of $H$ , and why must they eventually repeat?

Let $a\in H$ . Closure gives $a,a^2,a^3,\ldots \in H$ . Since $H\subseteq G$ and $G$ is finite, $H$ is finite, so this list of powers cannot all be distinct — some $a^i=a^j$ with $i<j$ . Cancelling (Theorem 2.1, valid since $G$ is a group) gives $a^{j-i}=e$ with $j-i>0$ , so $a$ has some finite order $k=|a|\le j-i$ , and $a^k=e\in H$ .

If $k=1$ : $a=e$ , and $a^{-1}=e\in H$ . If $k>1$ : $a^{-1}=a^{k-1}\in H$ by closure (a power of $a$ ). Either way $a^{-1}\in H$ — exactly the argument of Theorem 3.2, since nothing in it depended on $H$ actually being all of a candidate subgroup versus a general nonempty closed subset.

Restate the proof of Theorem 3.2 in your own words: why does closure alone force inverses to exist, in a finite group?

07 · Chapter Summary

Concept	Statement
Subgroup	$H\subseteq G$ that is itself a group under $G$ 's operation
Subgroup test	Nonempty + closed under $\ast$ + closed under inverses (Thm 3.1)
Finite subgroup test	Nonempty + closed under $\ast$ suffices, if $G$ finite (Thm 3.2)
$\langle g\rangle$	$\{g^n : n\in\mathbb{Z}\}$ ; the smallest subgroup containing $g$ (Thm 3.3)
Cyclic group	$G=\langle g\rangle$ for some generator $g$
Order equality	$
Subgroups of cyclic groups	Always cyclic (Thm 3.5)
Subgroups of $\mathbb{Z}_n$	One per positive divisor of $n$ , namely $\langle n/d\rangle$ (Thm 3.6)

Next: Chapter 04 — Cosets & Lagrange's Theorem uses subgroups to partition an entire group into equal-sized pieces, producing the single most-used counting tool in finite group theory: the order of any subgroup must divide the order of the group.