Subgroups cyclic groups and cosets

A subgroup is a smaller group using the same operation. Cosets organize the whole group into equal-sized copies of that subgroup. This partition links structure to counting and proves Lagrange's theorem without requiring that a quotient group already exist.

1 Recognizing a subgroup

A subset $H\subseteq G$ is a subgroup, written $H\le G$, if it is a group under the inherited operation. A convenient test is: $H$ is nonempty and $ab^{-1}\in H$ whenever $a,b\in H$.

Proof of the test. Pick $h\in H$; then $hh^{-1}=e\in H$. For $b\in H$, $eb^{-1}=b^{-1}\in H$. Applying the condition to $a$ and $b^{-1}$ gives $ab\in H$. Associativity is inherited. Conversely any subgroup has these closure properties. $\square$

The even integers form $2\mathbb Z\le\mathbb Z$ under addition. The positive integers do not form a subgroup because they omit zero and negatives. The intersection of subgroups is a subgroup, but their union need not be: $2\mathbb Z\cup3\mathbb Z$ contains 2 and 3 but not their sum 5.

The subgroup generated by $g$ is $\langle g\rangle=\{g^k:k\in\mathbb Z\}$. It is cyclic. If $g$ has finite order $n$, then $g^i=g^j$ exactly when $n$ divides $i-j$, so the powers $e,g,\ldots,g^{n-1}$ are distinct and exhaust it.

Cyclic subgroup theorem. Every subgroup of a cyclic group is cyclic.

Proof. For $H\le\langle g\rangle$, if $H=\{e\}$ the claim holds. Otherwise choose the smallest positive integer $d$ with $g^d\in H$; such a positive exponent exists by taking inverses if necessary. For any $g^k\in H$, divide $k=qd+r$, $0\le r<d$. Then $g^r=g^k(g^d)^{-q}\in H$, so minimality gives $r=0$. Thus $H=\langle g^d\rangle$. The same proof covers finite and infinite cyclic groups. $\square$

In a cyclic group of order $n$, the order of $g^k$ is $n/\gcd(n,k)$: the smallest positive $m$ with $n\mid km$ is that quotient.

2 Cosets partition a group

A left coset is $aH=\{ah:h\in H\}$. A right coset is $Ha$. These can differ. Define $a\sim b$ when $a^{-1}b\in H$. Identity, inverse and product closure show this is an equivalence relation; its class at $a$ is precisely $aH$. Hence left cosets partition $G$.

More directly, if $aH$ and $bH$ overlap, write $ah_1=bh_2$. Then $a=bh_2h_1^{-1}$, giving $aH\subseteq bH$, and the reverse inclusion follows similarly. Equal cosets do not mean $a=b$; they mean $a^{-1}b\in H$.

The map $H\to aH$, $h\mapsto ah$, is a bijection, with inverse left multiplication by $a^{-1}$. Thus every coset has $|H|$ elements.

Lagrange's theorem. If $G$ is finite and $H\le G$, then

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

where $[G:H]$ is the number of left cosets. The proof counts the equal-sized blocks of the partition. In particular, the order of any element divides $|G|$ by applying the theorem to its generated subgroup.

The converse is not automatic: a divisor of $|G|$ need not be the order of a subgroup in an arbitrary finite group. Lagrange also does not apply to a subset that fails the subgroup test.

3 Worked example in a permutation group

In $S_3$, let $H=\{e,(1\ 2)\}$. It has order 2, so index 3. The left cosets can be represented by $e,(1\ 3),(2\ 3)$:

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

They contain all six permutations exactly once. The coset $(1\ 3)H$ differs from $H(1\ 3)$, whose nontrivial product is $(1\ 3\ 2)$. This is a warning: a coset set is always available, but a well-defined coset multiplication needs an extra condition explored next.

Completing the power and cyclic-subgroup arguments

If $g$ has finite order $n$, division with remainder works for every integer exponent $m$: write $m=qn+r$, $0\le r<n$. If $g^m=e$, then $g^r=e$, so minimality of $n$ forces $r=0$. Thus $g^m=e$ if and only if $n\mid m$, including negative exponents. This proves the equality criterion for powers used above.

Write $d=\gcd(n,k)$, $n=dn_0$, $k=dk_0$ with $\gcd(n_0,k_0)=1$. The condition $(g^k)^m=e$ is equivalent to $n_0\mid k_0m$. An integer Bézout relation $un_0+vk_0=1$, multiplied by $m$, shows that this is equivalent to $n_0\mid m$. Therefore the least positive such $m$ is $n_0=n/d$.

In the cyclic-subgroup proof, when $\langle g\rangle$ has finite order $n$, the least positive exponent $d$ for $H$ must divide $n$: divide $n=qd+r$ and use $g^n=e\in H$ to obtain $g^r\in H$, hence $r=0$. This explains why the complete subgroup list in Exercise 1 is indexed by divisors of 12, rather than by arbitrary exponents.

The equivalence-relation proof of the coset partition can also be checked explicitly. Reflexivity uses $a^{-1}a=e\in H$. Symmetry uses $(a^{-1}b)^{-1}=b^{-1}a\in H$. Transitivity uses $(a^{-1}b)(b^{-1}c)=a^{-1}c\in H$. Finally $a^{-1}b\in H$ is equivalent to $b=ah$ for an $h\in H$, so the equivalence class really is $aH$. None of these steps needs normality or finiteness; finiteness enters only when the partition is counted numerically.

4 Graded practice

E1 Foundation. Find all subgroups of the additive group $\mathbb Z/12\mathbb Z$ and their orders.

E2 Cosets. List the cosets of $4\mathbb Z$ in $\mathbb Z$.

E3 Counting. Prove a group of prime order is cyclic.

E4 Reasoning. Suppose $G$ has 15 elements. What can Lagrange's theorem alone say about element orders? Does it prove an element of order 15 exists?

5 Full solutions

S1. Subgroups correspond to positive divisors $d$ of 12 through $\langle[d]\rangle$. They are generated by $1,2,3,4,6,12$ and have orders $12,6,4,3,2,1$ respectively. The last is the zero subgroup, since $[12]=[0]$. The cyclic subgroup theorem proves there are no others.

S2. The four distinct cosets are $4\mathbb Z$, $1+4\mathbb Z$, $2+4\mathbb Z$ and $3+4\mathbb Z$. Division with remainder proves every integer belongs to exactly one.

S3. Choose $g\ne e$. Its order divides the prime $p$ and exceeds 1, so it is $p$. Its generated subgroup therefore has all $p$ elements.

S4. Every element order is one of $1,3,5,15$. Lagrange alone does not supply an element of order 15; it is a divisibility restriction, not an existence theorem. A separate group-theoretic argument is needed for stronger conclusions.

Summary

Subgroups are recognized by closure with inverses. Cosets partition a group into equally sized blocks, yielding Lagrange's divisibility result. Cyclic groups have a particularly complete subgroup description, but coset counting alone must not be mistaken for quotient multiplication or existence results.