Cosets, Lagrange, and Normal Subgroups

Subgroups do more than sit inside a group; each one slices the group into translates of itself. Those translates, the cosets, are all the same size, and that single fact is the most-used counting theorem in finite group theory. When the cosets can be multiplied among themselves, they form a new, smaller group, the quotient.

Cosets and the partition of a group

Left cosets carve $G$ into disjoint blocks that cover it.

Every $g$ lies in its own coset $gN$ (since $1\in N$), so the cosets cover $G$. If two cosets share an element $x=un=vm$, then for any $ut\in uN$ one writes $ut=v(m{n}^{-1}t)\in vN$, and symmetrically, so the two cosets coincide. Distinct cosets are therefore disjoint.¹ The equivalence $uN=vN\leftrightarrow v^{-1}u\in N$ is the practical membership test.

Each coset $gN$ has exactly $|N|$ elements, because $n\mapsto gn$ is a bijection from $N$ onto $gN$ (left cancellation makes it injective). Equal-size blocks partitioning $G$ is all Lagrange's theorem needs.

Everything above holds verbatim for right cosets, which partition $G$ as well. The two partitions differ in general — for a non-normal subgroup some left coset is not a right coset for any representative — but they always have the same number of blocks: $x\mapsto x^{-1}$ maps each left coset $gN$ onto the right coset $N{g}^{-1}$, a bijection between the two collections. So index is unambiguous. These notes work with left cosets throughout.²

Lagrange's theorem

If there are $k$ cosets, each of size $|H|=n$, and they partition $G$, then $|G|=kn$, so $|H|\mid |G|$ and $k=|G|/|H|$.³ The number $k$ is named.

Three consequences are immediate and constant tools.

• Order divides. For $x\in G$, $|x|=|⟨x⟩|$ divides $|G|$, so $x^{|G|}=1$ for every $x$.
• Prime order forces cyclic. If $|G|=p$ is prime, any nonidentity $x$ generates a subgroup of order dividing $p$ and larger than $1$, so $G=⟨x⟩\cong \mathbb{Z}/p\mathbb{Z}$.
• A product-size formula. For finite subgroups $H,K$, $|HK|=\frac{|H||K|}{|H\cap K|}$, where $HK=\{hk\}$ need not be a subgroup.

The partition is concrete for $S_3$ and the order-$2$ subgroup $H=⟨(12)⟩=\{1,(12)\}$. The three left cosets each hold two of the six elements.

The converse of Lagrange is false: a divisor of $|G|$ need not be the order of a subgroup. The alternating group $A_4$ has order $12$ but no subgroup of order $6$: such a subgroup would have index $2$, forcing $g^2\in H$ for every $g\in A_4$, but $A_4$ has eight elements of order $3$ and each is the square of its own square, more than a subgroup of order $6$ can hold.⁴ Two partial converses hold and are proved later: Cauchy's theorem, that a prime $p\mid |G|$ forces an element (hence a subgroup) of order $p$, and Sylow's theorem, giving subgroups of every prime-power order dividing $|G|$.

Products of subgroups

The set $HK=\{hk\mid h\in H,k\in K\}$ collects all products across two subgroups. Counting its elements is a coset argument: $HK$ is a union of cosets $hK$, and $h_{1}K=h_{2}K$ exactly when $h_2^{-1}{h}_1\in H\cap K$, so the number of distinct cosets is $[H:H\cap K]$ and

The formula holds whether or not $HK$ is a subgroup, and often proves it is not: in $S_3$ with $H=⟨(12)⟩$ and $K=⟨(23)⟩$, $|HK|=\frac{2\cdot 2}{1}=4$, which does not divide $6$, so $HK$ is not a subgroup (and, as a byproduct, $⟨(12),(23)⟩=S_3$). The clean criterion: $HK$ is a subgroup if and only if $HK=KH$, which holds in particular whenever $H\le N_G(K)$ — so a product with a normal subgroup is always a subgroup.⁵

Normal subgroups and well-defined coset multiplication

The goal is to multiply cosets: define $(uN)(vN)=(uv)N$. This depends on choosing representatives, and in general the choice matters. The operation is consistent exactly for a distinguished class of subgroups.

If the rule is well defined, comparing the products of the reps $1,n$ of $N$ and $g^{-1},g^{-1}$ of $g^{-1}N$ forces $gn{g}^{-1}\in N$. Conversely, that condition lets one push a representative change from one factor through the other and land in the same coset.⁶ The condition names the subgroup.

Several conditions on $N$ are the same statement, tying normality back to the normalizer.

For the non-normal $H=⟨(12)⟩$ in $S_3$, the cosets $1H$ and $(13)H$ have no consistent product: the representatives $1$ and $(12)$ of $1H$ give $1\cdot (13)=(13)$ and $(12)(13)=(132)$, which lie in different cosets. So the quotient of $S_3$ by $H$ is not a group.⁷

Normal subgroups are precisely the kernels of homomorphisms.

A kernel is normal because $\phi (gn{g}^{-1})=\phi (g)\phi (g{)}^{-1}=1$; conversely, the projection $\pi$ realizes any normal $N$ as a kernel.⁸ Two shortcuts make normality cheap to check: it suffices to test conjugates of a generating set of $N$ by a generating set of $G$, and any subgroup of index 2 is automatically normal, since the two left cosets ($H$ and its complement) must coincide with the two right cosets.

The quotient group

The quotient collapses $N$ to a point and translates that collapse across all of $G$. A small case: in $\mathbb{Z}/6\mathbb{Z}$ take $N=⟨3⟩=\{0,3\}$. Its three cosets $\{0,3\}$, $\{1,4\}$, $\{2,5\}$ multiply (here, add) as representatives, and the quotient is a group of order $3$.

A structural example: $D_8/Z(D_8)$ with $Z(D_8)=⟨r^2⟩$. Each coset $gZ=\{g,g{r}^2\}$ has two elements, so there are four cosets. Every nonidentity coset squares to the identity coset, so the quotient has no element of order $4$ and is the Klein four-group: $D_8/Z(D_8)\cong V_4$.⁹

Application: Fermat and Euler

Lagrange, read inside the group of units $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$, gives two number-theoretic theorems at once. The order of that group is $\phi (n)$, so $x^{\phi (n)}=1$ for every unit.

This is $x^{|G|}=1$ applied to $G=(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$ and $x=\bar{a}$.¹⁰ Specializing to a prime $n=p$, where $\phi (p)=p-1$, gives the classical case.

No number theory beyond the definition of $\phi$ enters: both are the statement that an element's order divides the group's order.

A $48$-digit number is resolved by one group order and a handful of squarings: the repeated-squares ladder, with Euler shrinking the exponent before the ladder even starts.