Subgroups cyclic structure and coset counting

Recognizing a small group inside a larger one is the first step in understanding its structure. The subgroup generated by one element is controlled by integer arithmetic. Cosets then convert a subgroup into a partition of the ambient group. This chapter develops those arguments fully rather than inferring structure from a multiplication table or a few examples.

1 The subgroup test and generated subgroups

Assume the group axioms and rightmost-first function composition from the earlier symmetry chapter. A nonempty subset $H$ is a subgroup exactly when $ab^{-1}\in H$ for every $a,b\in H$.

Proof. Pick $h\in H$, so $e=hh^{-1}\in H$. Then $b^{-1}=eb^{-1}\in H$ for every $b\in H$. Applying the condition to $a$ and $b^{-1}$ gives $ab\in H$. Associativity is inherited from the ambient group, so the remaining axioms hold. Necessity follows from closure and inverses in any subgroup. $\square$

Intersections of subgroups are subgroups: every factor lies in every set in the intersection. Therefore the intersection of all subgroups containing a subset $S$ is the smallest subgroup containing $S$, written $\langle S\rangle$. Equivalently, it consists of finite products of elements of $S$ and their inverses, including the empty product $e$. The product description is closed under multiplication and inverse; every subgroup containing $S$ must contain all such products, proving equivalence.

A union of subgroups usually fails this test. In the additive integers, $2\mathbb Z\cup3\mathbb Z$ contains 2 and 3 but not 5. In fact $H\cup K$ is a subgroup only when one of $H,K$ contains the other. If neither contains the other, choose $h\in H\setminus K$ and $k\in K\setminus H$. Were $hk$ in $H$, cancellation would give $k\in H$; were it in $K$, cancellation would give $h\in K$. Both contradict the choices.

2 Cyclic groups and their complete subgroup structure

For one element $g$, $\langle g\rangle=\{g^k:k\in\mathbb Z\}$. If no positive power is $e$, the map $k\mapsto g^k$ is an isomorphism from the additive integers. Otherwise let $n$ be the smallest positive exponent giving $e$. Division of $k$ by $n$ shows $g^k$ depends only on its residue modulo $n$, and minimality shows the $n$ residues give distinct elements. Thus $\langle g\rangle\cong\mathbb Z/n\mathbb Z$.

Theorem. Every subgroup of a cyclic group is cyclic. In a cyclic group of finite order $n$, for each divisor $d$ of $n$ there is exactly one subgroup of order $d$, namely $\langle g^{n/d}\rangle$.

Proof. The trivial subgroup is cyclic, generated by $e$, including when the ambient cyclic group is infinite. For a nontrivial subgroup $H$, choose the smallest positive integer $m$ with $g^m\in H$. For $g^k\in H$, write $k=qm+r$, $0\le r<m$. Then $g^r=g^k(g^m)^{-q}\in H$, forcing $r=0$. Thus $H=\langle g^m\rangle$. In the finite case, $g^n=e\in H$ implies $m\mid n$ by the same division argument. The distinct powers of $g^m$ number $n/m$, so the order determines $m=n/d$ uniquely. The trivial subgroup corresponds to $m=n$. $\square$

The order formula $|\langle g^k\rangle|=n/\gcd(n,k)$ follows because $(g^k)^r=e$ exactly when $n\mid kr$. Writing $n=dn_0$, $k=dk_0$ with coprime $n_0,k_0$ makes the smallest such positive $r$ equal to $n_0$.

Worked example. In $\mathbb Z/18\mathbb Z$, $[6]$ has order 3 and generates $\{[0],[6],[12]\}$. The element $[12]$ generates the same subgroup because $\gcd(18,12)=6$. The unique subgroup of order 6 is generated by $[3]$. Two different generators need not define two different subgroups.

3 Cosets and Lagrange's theorem

For $H\le G$, a left coset is $aH=\{ah:h\in H\}$. Define $a\sim b$ when $a^{-1}b\in H$. The subgroup laws prove reflexivity, symmetry and transitivity; the equivalence class of $a$ is exactly $aH$. Cosets therefore partition the group. The map $h\mapsto ah$ is a bijection from $H$ to $aH$ by left cancellation and its evident inverse.

If $G$ is finite, counting the partition yields Lagrange's theorem:

$$|G|=[G:H]|H|.$$

In particular, element orders divide $|G|$. If $|G|$ is prime, any nonidentity element generates the whole group. The theorem does not say that every divisor of $|G|$ must occur as a subgroup order.

Worked nonabelian example. In $S_3$, let $H=\{e,(1\ 2)\}$. Its left cosets are

$$H,\quad(1\ 3)H=\{(1\ 3),(1\ 2\ 3)\},\quad (2\ 3)H=\{(2\ 3),(1\ 3\ 2)\}.$$

The right coset $H(1\ 3)=\{(1\ 3),(1\ 3\ 2)\}$ is different. Left and right partitions exist without being identical; a group law on the coset set requires normality, not merely Lagrange's theorem.

4 Graded practice

E1 Foundation. List the subgroups of $\mathbb Z/20\mathbb Z$, including generators and orders. Which elements generate its order-5 subgroup?

E2 Proof. Show $H\cap K$ is a subgroup. Give a specific counterexample showing $H\cup K$ need not be one.

E3 Counting. If a group has 35 elements, what are the possible element orders allowed by Lagrange? Explain what has and has not been proved about existence.

E4 Cyclic arithmetic. In a cyclic group $G=\langle g\rangle$ of order 24, find the orders of $g^6$ and $g^{10}$, and identify $\langle g^6\rangle\cap\langle g^{10}\rangle$.

E5 Coset reasoning. Prove $aH=bH$ if and only if $a^{-1}b\in H$. Explain why equality of cosets need not imply $a=b$.

5 Full solutions

S1. Divisors 1,2,4,5,10,20 give subgroups generated by $[20],[10],[5],[4],[2],[1]$, of the respective orders. The order-5 subgroup is $\{[0],[4],[8],[12],[16]\}$, generated by any of its four nonzero elements because each has gcd 4 with 20. The cyclic subgroup theorem excludes any further subgroup.

S2. The intersection contains identity. If $a,b$ lie in both groups, so does $ab^{-1}$, proving the subgroup test. In $\mathbb Z$, take $H=2\mathbb Z$ and $K=3\mathbb Z$; their union misses $2+3=5$ and is not closed.

S3. The allowed orders are $1,5,7,35$. This is only a necessary restriction. Lagrange alone neither constructs elements of order 5 or 7 nor establishes a generator of order 35; those require further arguments.

S4. Orders are $24/6=4$ and $24/2=12$. The first subgroup consists of powers whose exponent is a multiple of 6, and the second equals $\langle g^2\rangle$. Every multiple of 6 is even, so the first is contained in the second. Their intersection is $\langle g^6\rangle$, order 4. Treating the displayed exponent 10 as the subgroup's unique step size would miss its smaller generator $g^2$.

S5. If $aH=bH$, then $b\in aH$, so $b=ah$ for some $h\in H$. Conversely, if $b=ah$, then $bH=ahH=aH$ because $hH=H$. Distinct representatives in the same coset, such as 0 and 4 modulo $4\mathbb Z$ in the additive integers, give the same class.

Summary

The subgroup test converts axioms into a usable criterion. Cyclic subgroup classification is controlled by division and gcd. Cosets partition a group into equally sized blocks and give Lagrange's theorem. Counting restrictions do not automatically provide existence or a quotient multiplication.