Chapter 10 — Cosets and the Theorem of Lagrange

Fraleigh, A First Course in Abstract Algebra, 7th edition, Section 10. Study companion for SNU Abstract Algebra (26-1).

Cosets are translated copies of a subgroup. Lagrange’s theorem converts that geometric picture into a divisibility statement: the size of every subgroup divides the size of the group. This is the first time subgroup structure produces a serious counting theorem, and it is the true precursor to quotient groups (Chapter 14).

§10.1 Left and Right Cosets

Definition 10.1 (Left coset). Let $H \le G$ and $a \in G$. The left coset of $H$ determined by $a$ is

Definition 10.2 (Right coset). The right coset of $H$ determined by $a$ is

In an abelian group $aH = Ha$ for every $a$, because $ah = ha$. In a non-abelian group they can differ.

Worked examples of cosets

Example 10.3 (Cosets in $\mathbb{Z}$). Take $G = \mathbb{Z}$ (additive) and $H = 3\mathbb{Z} = \{0, \pm 3, \pm 6, \dots\}$. Since $\mathbb{Z}$ is abelian, left and right cosets agree. Writing cosets additively:

These are exactly the residue classes modulo $3$. Every integer lies in exactly one of them, so the three cosets partition $\mathbb{Z}$.

Example 10.4 (Cosets in $\mathbb{Z}_6$). Let $G = \mathbb{Z}_6 = \{0,1,2,3,4,5\}$ and $H = \langle 2 \rangle = \{0, 2, 4\}$. Then

Note $2 + H = \{2, 4, 0\} = 0 + H$ and $3 + H = \{3, 5, 1\} = 1 + H$. There are exactly two distinct cosets, and $[G:H] = 6/3 = 2$.

Figure: the coset partition of $\mathbb{Z}_6$ by $\langle \bar{2}\rangle$.

The two rounded boxes are the two cosets; this is the finite model of the general statement that cosets partition the group.

Example 10.5 (Left and right cosets in $S_3$). Let $G = S_3$ and $H = \{e, (1\;2\;3), (1\;3\;2)\} = \langle (1\;2\;3) \rangle$. Since $|H| = 3$ and $|S_3| = 6$, there are $[S_3 : H] = 2$ cosets.

Left cosets:

Take the transposition $(1\;2) \notin H$:

Indeed,

so the left coset is exactly the set of transpositions.

Right cosets:

Indeed,

so the right coset is also the set of transpositions.

In this case the left and right cosets happen to be the same as sets (both give $\{H, \{(1\;2),(1\;3),(2\;3)\}\}$). This is because $H$ has index $2$ in $S_3$, and index-$2$ subgroups are always normal (see §10.8 and Chapter 14).

§10.1a Reading Coset Structure from a Cayley Table

The Cayley table (or group table) of a finite group $G$ is the $|G| \times |G|$ multiplication table: the entry in row $a$, column $b$ is the product $ab$. Every finite group is completely determined by its Cayley table, and the table encodes subgroup and coset information in a visually striking way. This section develops the technique of coset shading that Fraleigh uses in Tables 10.5—10.9.

Quick review: what a Cayley table tells you

For a group $G = \{g_1, g_2, \dots, g_n\}$, the Cayley table is:

Two fundamental properties:

1. Latin square property. Every element of $G$ appears exactly once in each row and each column. (Proof: left multiplication $g \mapsto ag$ is a bijection on $G$.)
2. Identity row/column. If $g_1 = e$, then the first row is just $g_1, g_2, \dots, g_n$ in order, and so is the first column.

Coset shading: the main idea

Suppose $H \le G$ with index $[G:H] = k$. The left cosets $a_1H, a_2H, \dots, a_kH$ partition $G$ into $k$ blocks of equal size. Now rearrange the rows and columns of the Cayley table so that elements within the same coset are adjacent (grouped together). Then assign each coset a shade (say light, medium, dark, etc.) and color every entry $ab$ according to the coset that $ab$ belongs to.

The resulting shaded table reveals whether the cosets themselves form a group.

Example: $\mathbb{Z}_6$ with $H = \{0, 3\}$

This is Fraleigh’s Example 10.4. Take $G = \mathbb{Z}_6$ and $H = \{0, 3\}$, so $|H| = 2$ and $[G:H] = 3$. The three cosets are:

Rearrange the Cayley table of $(\mathbb{Z}_6, +_6)$ so elements are grouped by coset: $0, 3 \mid 1, 4 \mid 2, 5$.

Now shade by coset membership. Write LT (light) for $C_0 = \{0,3\}$, MD (medium) for $C_1 = \{1,4\}$, DK (dark) for $C_2 = \{2,5\}$:

The critical observation. Look at the $2 \times 2$ blocks formed by grouping rows and columns by coset. Each block is a single, solid shade. This means that whenever $a$ and $a'$ are in the same left coset and $b$ and $b'$ are in the same left coset, the products $ab$ and $a'b'$ land in the same coset. In other words, the coset to which $ab$ belongs depends only on the cosets of $a$ and $b$, not on the particular representatives.

This is exactly the condition needed for the set of cosets to form a group under the induced operation. Reading off the shading gives a $3 \times 3$ “coset multiplication table”:

Replacing LT $\to 0$, MD $\to 1$, DK $\to 2$, this is exactly the Cayley table of $\mathbb{Z}_3$. So the three cosets of $H$ in $\mathbb{Z}_6$, with the operation “add representatives and take the coset of the result,” form a group isomorphic to $\mathbb{Z}_3$. This is the factor group $\mathbb{Z}_6 / H \cong \mathbb{Z}_3$, studied properly in Chapter 14.

Why solid blocks mean “well-defined coset multiplication”

The factor group construction requires a well-defined binary operation on cosets:

The danger is that $aH = a'H$ and $bH = b'H$ (i.e., $a, a'$ are in the same coset, and $b, b'$ are in the same coset) but $(ab)H \ne (a'b')H$. If that happened, the “operation” would depend on which representative we pick, and would not actually be a function on cosets.

In the shaded Cayley table, a solid block means: for every pair of entries in that block, the product lands in the same coset. So solid blocks $\Leftrightarrow$ well-defined coset multiplication $\Leftrightarrow$ the cosets form a group.

This is precisely the condition that $H$ is a normal subgroup ($aH = Ha$ for all $a \in G$). The full proof is in Chapter 14, but the shaded table gives you a concrete, visual diagnostic.

Counterexample: $S_3$ with a non-normal subgroup

Fraleigh’s Tables 10.8—10.9 illustrate the failure. Take $G = S_3$ with the notation from Chapter 8:

Consider the subgroup $H = \{\rho_0, \mu_1\}$ of order $2$, with index $[S_3 : H] = 3$.

Left cosets:

(Verify: $\rho_1 \mu_1 = (1\;2\;3)(2\;3) = (1\;2) = \mu_3$, and $\rho_2 \mu_1 = (1\;3\;2)(2\;3) = (1\;3) = \mu_2$.)

Now write the Cayley table of $S_3$ with elements grouped by left coset: $\rho_0, \mu_1 \mid \rho_1, \mu_3 \mid \rho_2, \mu_2$, and shade by coset membership.

(Entries computed from the full $S_3$ multiplication table in Chapter 8.)

Look at the $2 \times 2$ blocks. They are not solid. For instance, the block in rows $\{\mu_3\}$ and columns $\{\rho_0, \mu_1\}$ has entries MD and DK — two different shades in the same block. This means coset multiplication is not well-defined: different representatives of the same coset can give products in different cosets.

Concretely: $\rho_1$ and $\mu_3$ are both in coset $C_1 = \{\rho_1, \mu_3\}$, and $\rho_0$ is in $C_0$. But $\rho_1 \cdot \rho_0 = \rho_1 \in C_1$ while $\mu_3 \cdot \rho_0 = \mu_3 \in C_1$ — that one is fine. However, looking at other blocks: $\mu_3 \cdot \mu_1 = (1\;2)(2\;3) = (1\;2\;3) = \rho_1 \in C_1$, but $\mu_3 \cdot \rho_1 = (1\;2)(1\;2\;3) = (2\;3) = \mu_1 \in C_0$. Same row-coset, same column-coset, different result-cosets. The block is not solid, and coset multiplication breaks down.

Diagnosis. $H = \{\rho_0, \mu_1\}$ is not normal in $S_3$: the left coset $\rho_1 H = \{\rho_1, \mu_3\}$ but the right coset $H\rho_1 = \{\rho_1, \mu_2\}$, so $\rho_1 H \ne H\rho_1$.

Contrast: $S_3$ with the normal subgroup $\langle \rho_1 \rangle$

Now take $H = \{\rho_0, \rho_1, \rho_2\} = \langle (1\;2\;3) \rangle$, which has index $2$. The two cosets are:

Group the Cayley table by coset and shade:

Every $3 \times 3$ block is a solid shade. Reading off the coset table:

This is $\mathbb{Z}_2$. So $S_3 / \langle (1\;2\;3) \rangle \cong \mathbb{Z}_2$, as expected for an index-$2$ subgroup.

Summary: the shading diagnostic

The technique works for any finite group. In practice: write the Cayley table, group elements by coset, shade, and check whether the blocks are uniform. Uniform blocks mean you have a factor group; mixed blocks mean the subgroup is not normal and no factor group exists via those cosets.

This visual approach is revisited in full generality in Chapter 14, where we prove that the cosets of $H$ form a group under $(aH)(bH) = abH$ if and only if $H \trianglelefteq G$.

§10.2 Cosets Partition the Group

The key idea: define a relation on $G$ by

Theorem 10.6. The relation $\sim$ is an equivalence relation on $G$, and the equivalence class of $a$ is the left coset $aH$.

Proof that $\sim$ is an equivalence relation

We verify the three axioms.

Reflexive. For any $a \in G$, $a^{-1}a = e \in H$ (since $H$ is a subgroup). So $a \sim a$.

Symmetric. Suppose $a \sim b$, i.e.\ $a^{-1}b \in H$. Since $H$ is a subgroup, $(a^{-1}b)^{-1} = b^{-1}a \in H$. Hence $b \sim a$.

Transitive. Suppose $a \sim b$ and $b \sim c$, i.e.\ $a^{-1}b \in H$ and $b^{-1}c \in H$. Since $H$ is closed under multiplication,

$$a^{-1}c = (a^{-1}b)(b^{-1}c) \in H.$$

Hence $a \sim c$. $\blacksquare$

Proof that the equivalence class of $a$ is $aH$

The equivalence class of $a$ is $[a] = \{b \in G : a \sim b\} = \{b \in G : a^{-1}b \in H\}$.

($\subseteq$) If $b \in [a]$, then $a^{-1}b = h$ for some $h \in H$, so $b = ah \in aH$.

($\supseteq$) If $b \in aH$, then $b = ah$ for some $h \in H$, so $a^{-1}b = h \in H$, meaning $b \in [a]$.

Therefore $[a] = aH$. $\blacksquare$

Corollary 10.7. The distinct left cosets of $H$ in $G$ form a partition of $G$. That is, every element of $G$ belongs to exactly one left coset, and two left cosets are either identical or disjoint.

Proof

This is a direct consequence of the fundamental theorem on equivalence relations: the equivalence classes of any equivalence relation on a set $S$ partition $S$. By Theorem 10.6, the equivalence classes of $\sim$ are exactly the left cosets of $H$. $\blacksquare$

§10.3 All Cosets Have the Same Size

Theorem 10.8. For any $a \in G$, $|aH| = |H|$.

Proof

Define $\varphi : H \to aH$ by $\varphi(h) = ah$.

Well-defined: Clear, since $ah \in aH$ for every $h \in H$.

Injective: Suppose $\varphi(h_1) = \varphi(h_2)$, i.e.\ $ah_1 = ah_2$. Left-multiplying by $a^{-1}$ gives $h_1 = h_2$.

Surjective: Every element of $aH$ has the form $ah$ for some $h \in H$, and $\varphi(h) = ah$.

Since $\varphi$ is a bijection, $|aH| = |H|$. $\blacksquare$

Remark. The same argument shows $|Ha| = |H|$ via the bijection $h \mapsto ha$.

§10.4 Lagrange’s Theorem

Theorem 10.9 (Lagrange’s Theorem). If $G$ is a finite group and $H \le G$, then $|H|$ divides $|G|$.

Proof

By Corollary 10.7, the distinct left cosets $a_1H, a_2H, \dots, a_rH$ partition $G$:

$$G = a_1H \;\dot\cup\; a_2H \;\dot\cup\; \cdots \;\dot\cup\; a_rH.$$

Taking cardinalities,

$$|G| = |a_1H| + |a_2H| + \cdots + |a_rH|.$$

By Theorem 10.8, $|a_iH| = |H|$ for each $i$, so

$$|G| = r \cdot |H|.$$

In particular, $|H|$ divides $|G|$, and $r = |G|/|H|$. $\blacksquare$

Definition 10.10 (Index). The index of $H$ in $G$, written $[G:H]$, is the number of distinct left cosets of $H$ in $G$. By the proof above, if $G$ is finite then

Example 10.11. $[\mathbb{Z}_6 : \langle 2 \rangle] = 6/3 = 2$. $[S_3 : \langle (1\;2\;3) \rangle] = 6/3 = 2$.

§10.5 Corollaries of Lagrange’s Theorem

Corollary 10.12 (Order of an element divides $|G|$). If $G$ is a finite group and $a \in G$, then the order $|a|$ divides $|G|$.

Proof

The cyclic subgroup $\langle a \rangle$ has order $|a|$ and is a subgroup of $G$. By Lagrange, $|a| = |\langle a \rangle|$ divides $|G|$. $\blacksquare$

Corollary 10.13. If $G$ is a finite group with $|G| = n$, then $a^n = e$ for all $a \in G$.

Proof

By Corollary 10.12, $|a|$ divides $n$, so $n = |a| \cdot k$ for some positive integer $k$. Then

$$a^n = a^{|a| \cdot k} = (a^{|a|})^k = e^k = e. \qquad \blacksquare$$

Corollary 10.14 (Groups of prime order are cyclic). If $|G| = p$ where $p$ is prime, then $G$ is cyclic and every non-identity element is a generator.

Proof

Let $a \in G$ with $a \ne e$. Then $|a| > 1$ and $|a|$ divides $p$ (Corollary 10.12). Since $p$ is prime, the only divisors of $p$ are $1$ and $p$. Thus $|a| = p$, so $\langle a \rangle = G$, and $G$ is cyclic with generator $a$. $\blacksquare$

Corollary 10.15 (Fermat’s Little Theorem). If $p$ is prime and $\gcd(a, p) = 1$, then

Proof

Consider the multiplicative group $\mathbb{Z}_p^{*} = \{1, 2, \dots, p-1\}$ under multiplication mod $p$. This is a group of order $p - 1$. Since $\gcd(a, p) = 1$, the residue class $\bar{a} \in \mathbb{Z}_p^{*}$. By Corollary 10.13,

$$\bar{a}^{\,p-1} = \bar{1},$$

which means $a^{p-1} \equiv 1 \pmod{p}$. $\blacksquare$

Example. Take $p = 7$, $a = 3$. Then $3^6 = 729$. Now $729 = 104 \cdot 7 + 1$, so $3^6 \equiv 1 \pmod{7}$, confirming Fermat.

Corollary 10.16 (Euler’s Theorem). If $\gcd(a, n) = 1$, then

where $\phi(n) = |\mathbb{Z}_n^{*}|$ is Euler’s totient function.

Proof

The group $\mathbb{Z}_n^{*}$ of units modulo $n$ has order $\phi(n)$. Since $\gcd(a,n) = 1$, we have $\bar{a} \in \mathbb{Z}_n^{*}$. By Corollary 10.13,

$$\bar{a}^{\,\phi(n)} = \bar{1},$$

i.e., $a^{\phi(n)} \equiv 1 \pmod{n}$. $\blacksquare$

Remark. Fermat’s little theorem is the special case $n = p$ prime, since $\phi(p) = p - 1$.

§10.6 The Converse of Lagrange is FALSE

A natural question: if $d$ divides $|G|$, must $G$ have a subgroup of order $d$? The answer is no.

Theorem 10.17. The alternating group $A_4$ has order $12$ but no subgroup of order $6$.

Proof

$A_4$ consists of the $12$ even permutations of $\{1,2,3,4\}$:

• the identity $e$,
• eight $3$-cycles: $(1\;2\;3)$, $(1\;3\;2)$, $(1\;2\;4)$, $(1\;4\;2)$, $(1\;3\;4)$, $(1\;4\;3)$, $(2\;3\;4)$, $(2\;4\;3)$,
• three products of disjoint transpositions: $(1\;2)(3\;4)$, $(1\;3)(2\;4)$, $(1\;4)(2\;3)$.

Suppose for contradiction that $H \le A_4$ with $|H| = 6$. Then $[A_4 : H] = 12/6 = 2$.

Claim: A subgroup of index $2$ is always normal. Indeed, for any $g \in G$, the two left cosets are $H$ and $gH$, and the two right cosets are $H$ and $Hg$. Since $G = H \;\dot\cup\; gH = H \;\dot\cup\; Hg$ and these partitions have only two blocks, we must have $gH = Hg$ for every $g$, so $H \trianglelefteq G$.

So $H$ would be a normal subgroup of $A_4$ of order $6$.

Claim: $H$ must contain all eight $3$-cycles. If $\sigma$ is a $3$-cycle, then $|\sigma| = 3$. The element $\sigma$ generates $\langle \sigma \rangle$ of order $3$. By Lagrange applied to $H$, the order of $\sigma$ in $H$ must divide $|H| = 6$, which is fine. But we can argue more precisely: since $H$ is normal in $A_4$ and $H$ has index $2$, for any $a \in A_4$, $a^2 \in H$ (because in $A_4/H \cong \mathbb{Z}_2$, the square of every element is the identity). Every $3$-cycle $\sigma$ satisfies $\sigma^2 = \sigma^{-1}$, which is also a $3$-cycle. But $\sigma^2 \in H$ implies $\sigma^{-1} \in H$, which implies $\sigma = (\sigma^{-1})^{-1} \in H$.

So all eight $3$-cycles lie in $H$, plus $e \in H$, giving at least $9$ elements. But $|H| = 6 < 9$, a contradiction. Therefore no subgroup of order $6$ exists. $\blacksquare$

Remark. The converse of Lagrange does hold for some classes of groups — for instance, finite abelian groups, and more generally $p$-groups by Sylow theory (Chapter 36 in Fraleigh). The full Sylow theorems guarantee subgroups of order $p^k$ for prime powers dividing $|G|$.

§10.7 Worked Coset Computations

Cosets of $\langle 3 \rangle$ in $\mathbb{Z}_{12}$

Let $G = \mathbb{Z}_{12}$ and $H = \langle 3 \rangle = \{0, 3, 6, 9\}$. Then $|H| = 4$, so $[G:H] = 12/4 = 3$. The three cosets:

Check: $3 + H = \{3, 6, 9, 0\} = 0 + H$, $4 + H = \{4, 7, 10, 1\} = 1 + H$, etc. Every element of $\mathbb{Z}_{12}$ appears in exactly one coset.

Cosets of $\langle 2 \rangle$ in $\mathbb{Z}_8$

Let $G = \mathbb{Z}_8$ and $H = \langle 2 \rangle = \{0, 2, 4, 6\}$. Then $|H| = 4$, $[G : H] = 8/4 = 2$. The cosets:

These are the even and odd elements of $\mathbb{Z}_8$.

Left vs.\ right cosets: an example where they differ

Let $G = S_3$ and $K = \{e, (1\;2)\}$. Then $|K| = 2$ and $[S_3 : K] = 3$.

Left cosets:

(Verify: $(1\;2\;3)(1\;2)$: $1\to 2\to 2\to 1$… applying $(1\;2)$ first then $(1\;2\;3)$: $1\to 2\to 3$, $2\to 1\to 2$, $3\to 3\to 1$. That is $(1\;3)$. And $(1\;3\;2)(1\;2)$: $1\to 2\to 1$, $2\to 1\to 3$, $3\to 3\to 2$, giving $(2\;3)$.)

Right cosets:

(Verify: $(1\;2)(1\;2\;3)$: apply $(1\;2\;3)$ first, then $(1\;2)$: $1\to 2\to 1$, $2\to 3\to 3$, $3\to 1\to 2$. That is $(2\;3)$. And $(1\;2)(1\;3\;2)$: apply $(1\;3\;2)$ first, then $(1\;2)$: $1\to 3\to 3$, $2\to 1\to 2$, $3\to 2\to 1$. That is $(1\;3)$.)

Compare:

The left and right cosets are different. For instance, $(1\;3)$ sits in the coset $(1\;2\;3)K$ on the left, but in $K(1\;3\;2)$ on the right. This happens because $K = \{e, (1\;2)\}$ is not a normal subgroup of $S_3$.

§10.8 When Left Cosets Equal Right Cosets — Preview of Normal Subgroups

Definition 10.18 (Normal subgroup, preview). A subgroup $H \le G$ is normal, written $H \trianglelefteq G$, if $aH = Ha$ for every $a \in G$.

When $H$ is normal, the set of cosets $G/H = \{aH : a \in G\}$ itself becomes a group under the operation $(aH)(bH) = abH$. This is the factor group or quotient group (Chapter 14).

Observation 10.19. Every subgroup of index $2$ is normal.

Proof

Let $[G:H] = 2$. There are exactly two left cosets: $H$ and $G \setminus H$. Likewise, there are exactly two right cosets: $H$ and $G \setminus H$. For any $a \in G$:

• If $a \in H$, then $aH = H = Ha$.
• If $a \notin H$, then $aH = G \setminus H = Ha$.

In either case $aH = Ha$, so $H \trianglelefteq G$. $\blacksquare$