Cosets

A subgroup $H$ cuts its group into equal pieces. You have seen this already: the residue classes mod $4$ are the pieces of $\mathbb{Z}$ cut out by $4\mathbb{Z}$. The pieces are called cosets, and they are the key to the next three topics.

Definition. For $H \le G$ and $a \in G$, the left coset of $H$ containing $a$ is $$aH = \{ah : h \in H\}.$$ The right coset is $Ha = \{ha : h \in H\}$. In additive notation we write $a + H$. For example, in $\mathbb{Z}_{12}$ with $H = \langle 4 \rangle = \{0, 4, 8\}$, the coset $1 + H = \{1, 5, 9\}$.

Why it works

The cosets of $H$ split $G$ into disjoint pieces of size $|H|$.
(1) They cover $G$: $a = ae \in aH$.
(2) Two cosets are equal or disjoint: define $a \sim b$ when $a^{-1}b \in H$. This is an equivalence relation: $a^{-1}a = e \in H$; if $a^{-1}b \in H$ then its inverse $b^{-1}a \in H$; if $a^{-1}b, b^{-1}c \in H$ then their product $a^{-1}c \in H$. The class of $a$ is $\{b : b = ah \text{ for some } h \in H\} = aH$, and equivalence classes are equal or disjoint. In particular $$aH = bH \iff a^{-1}b \in H.$$
(3) Same size: the map $h \mapsto ah$ from $H$ to $aH$ is onto by definition and one-to-one by cancellation ($ah = ah' \Rightarrow h = h'$). $\square$

$\mathbb{Z}_{12}$ colored by the cosets of $H = \langle 4 \rangle$. Tap any number to light its coset: four cosets, three elements each.

Worked example: Cosets in $\mathbb{Z}_{12}$

List the cosets of $H = \langle 4 \rangle$ in $\mathbb{Z}_{12}$. Is $2 + H = 10 + H$?

$0 + H = \{0, 4, 8\}$, $1 + H = \{1, 5, 9\}$, $2 + H = \{2, 6, 10\}$, $3 + H = \{3, 7, 11\}$.
$10 - 2 = 8 \in H$, so $2 + H = 10 + H$; indeed both are $\{2, 6, 10\}$.

Answer: Four cosets; $2 + H = 10 + H$.

Worked example: Left and right cosets can differ

In $S_{3}$ let $H = \{e, (1\;2)\}$. Compare $(1\;3)H$ and $H(1\;3)$.

$(1\;3)H = \{(1\;3),\ (1\;3)(1\;2)\}$. Apply $(1\;2)$ first: $1 \to 2 \to 2$, $2 \to 1 \to 3$, $3 \to 3 \to 1$, so $(1\;3)(1\;2) = (1\;2\;3)$.
$H(1\;3) = \{(1\;3),\ (1\;2)(1\;3)\}$, and $(1\;2)(1\;3) = (1\;3\;2)$.
So $(1\;3)H = \{(1\;3), (1\;2\;3)\}$ but $H(1\;3) = \{(1\;3), (1\;3\;2)\}$.

Answer: They are different sets.

Left and right cosets of $H = \{e, (1\;2)\}$ in $S_{3}$, side by side. Each side is a partition into three pairs, but the two partitions differ.

Worked example: A coset in $U(15)$

In $U(15)$, let $H = \langle 4 \rangle = \{1, 4\}$. Find $2H$ and $7H$.

$2H = \{2, 8\}$.
$7H = \{7, 28 \bmod 15\} = \{7, 13\}$.
With $H$, $11H = \{11, 14\}$ and these, the four cosets cover all eight elements of $U(15)$.

Answer: $2H = \{2, 8\}$ and $7H = \{7, 13\}$.

Common mistake

Treating a coset as a subgroup. Only $H = eH$ itself contains the identity; every other coset misses $e$, so it cannot be a subgroup. Another trap: assuming $aH = Ha$. That holds for every $a$ exactly when $H$ is a normal subgroup, the subject of a later lesson; in $S_{3}$ above it fails. (In an abelian group it always holds.)

Key idea

$aH = \{ah : h \in H\}$. The left cosets of $H$ partition $G$ into pieces of size $|H|$, and $aH = bH$ exactly when $a^{-1}b \in H$.

Sources and further reading

Thomas W. Judson, Abstract Algebra: Theory and Applications (open textbook), chapter 6, Cosets and Lagrange's Theorem (section 6.1, Cosets)

John A. Beachy and William D. Blair, Abstract Algebra, online study guide (Northern Illinois University), section 3.8, Cosets, Normal Subgroups, and Factor Groups

MIT OpenCourseWare 18.703 Modern Algebra (Spring 2013), lecture on cosets and Lagrange

Cosets

Practice

Builds on

Leads to

Sources and further reading