<!-- NOTATION PROPOSAL: The notation for cosets — writing $gN$ for the left coset of $N$ containing $g$, and $G/N$ for the set of left cosets — is standard in algebra but not yet explicitly recorded in the Androma Notation Standards (§18.1 only mentions $G/N$ for the quotient group, not the coset notation). Propose adding: coset $gN = \{gn : n \in N\}$, set of cosets $G/N = \{gN : g \in G\}$, and the rule that $G/N$ is a group when $N \trianglelefteq G$. --> Suppose you want to study a group by ignoring some of its structure — collapsing a subgroup to a point and asking what algebraic structure remains. This is exactly what the quotient group construction achieves. The idea is deceptively simple: given a group $G$ and a subgroup $N \le G$, partition $G$ into cosets of $N$ and try to make the set of these cosets into a group in its own right. The surprise is that this works cleanly only when $N$ satisfies a symmetry condition — normality — and when it does, the resulting group $G/N$ captures precisely the "information in $G$ beyond $N$." Quotient groups are the engine behind the isomorphism theorems, the classification of abelian groups, and the construction of virtually every important group in mathematics: cyclic groups, dihedral groups, simple groups, and Galois groups all arise naturally as quotients. Before the formalism, it is worth seeing where the naive approach fails. Suppose $G = S_3$ (the symmetric group on three elements) and $H = \{e, (12)\} \le S_3$. The left cosets of $H$ are $H = \{e, (12)\}$, $(13)H = \{(13), (132)\}$, and $(23)H = \{(23), (123)\}$. We might hope to multiply cosets by setting $(aH)(bH) = abH$. But take $(13)H$ and $(12)H = H$: we get $(13)(12)H = (132)H = (13)H$. Now reverse the order: $(12)H \cdot (13)H = (12)(13)H = (123)H = (23)H$. The product depends on the order, which is fine — $S_3$ is nonabelian. But now try $(13)H \cdot (23)H$: using representatives $(13)$ and $(23)$, we get $(13)(23)H = (123)H = (23)H$. Using different representatives $(13)$ and $(123)$ for the same two cosets: $(13)(123)H = (23)H$. The same answer — good. But try yet another representative: $(132)(23)H = (132)(23)H$. Compute $(132)(23) = (13)$, so this is $(13)H$. The product of the same two cosets gives different results depending on which representatives we pick. The coset multiplication is not well-defined. [example: Failure of Coset Multiplication for Non-Normal Subgroups] Let $G = S_3$ with multiplication given by composition of permutations, and let $H = \{e, (12)\}$. The left cosets are: \begin{align*} H &= \{e, (12)\} \\ (13)H &= \{(13), (132)\} \\ (23)H &= \{(23), (123)\}. \end{align*} Attempt to define coset multiplication by $(aH)(bH) = (ab)H$. To check whether this is well-defined, we must verify that if $aH = a'H$ and $bH = b'H$, then $abH = a'b'H$. Take the two cosets $(13)H$ and $(23)H$. Representative $(13)$ gives: \begin{align*} (13)(23) = (132), \end{align*} so $(13)(23)H = (132)H$. But $(132) \in (13)H$, so $(132)H = (13)H$. Now use the representative $(132)$ for the coset $(13)H$ and $(123)$ for $(23)H$. This is legitimate because a coset is determined by its elements as a set, not by any particular label: any element of the coset is an equally valid representative. \begin{align*} (132)(123) = e, \end{align*} so $(132)(123)H = eH = H$. We get $(13)H$ from one pair of representatives and $H$ from another pair — both pairs representing the same two cosets. The multiplication rule is not well-defined, and the coset space $S_3 / H$ cannot be made into a group this way. [/example] The failure above has a precise diagnosis. The coset product $aH \cdot bH = abH$ is well-defined if and only if for every $h \in H$, we have $bhb^{-1} \in H$ — that is, if $H$ is closed under conjugation by every element of $G$. This is exactly the normality condition. ## Definition The coset construction needs to be set up carefully. A left coset of $N$ in $G$ is a set of the form $gN = \{gn : n \in N\}$ for some fixed $g \in G$. Different elements of $G$ can give the same coset: $g_1 N = g_2 N$ if and only if $g_1^{-1} g_2 \in N$. This partitions $G$ into disjoint cosets. [citedefinition:Normal Subgroup] The three conditions — conjugation stability, left-right coset equality, and $gNg^{-1} = N$ — are equivalent, and in practice one uses whichever is most convenient for the problem at hand. Now the main construction. When $N \trianglelefteq G$, the set of left cosets $G/N = \{gN : g \in G\}$ inherits a group structure from $G$. [definition: Quotient Group] Let $G$ be a group and $N \trianglelefteq G$ a normal subgroup. The **quotient group** (or **factor group**) of $G$ by $N$ is the set \begin{align*} G/N = \{gN : g \in G\} \end{align*} of left cosets of $N$ in $G$, equipped with the binary operation \begin{align*} (aN)(bN) &:= (ab)N. \end{align*} The identity element is $eN = N$, and the inverse of $gN$ is $g^{-1}N$. [/definition] [remark: Well-Definedness Requires Normality] The definition of the group operation on $G/N$ requires checking that $(ab)N$ depends only on the cosets $aN$ and $bN$, not on the choice of representatives $a$ and $b$. If $a' \in aN$ and $b' \in bN$, then $a' = an_1$ and $b' = bn_2$ for some $n_1, n_2 \in N$. Then: \begin{align*} a'b' &= an_1 bn_2 = a(n_1 b)n_2 = ab(b^{-1}n_1 b)n_2. \end{align*} Since $N \trianglelefteq G$, the element $b^{-1}n_1 b \in N$, so $(b^{-1}n_1 b)n_2 \in N$, and $a'b'N = abN$. Without normality, this argument fails, as the example with $S_3$ shows. [/remark] [example: Integers Modulo n] The most fundamental example: let $G = (\mathbb{Z}, +)$ and $N = n\mathbb{Z} = \{nk : k \in \mathbb{Z}\}$ for some $n \ge 1$. Since $\mathbb{Z}$ is abelian, every subgroup is normal. The cosets are: \begin{align*} r + n\mathbb{Z} &= \{r + nk : k \in \mathbb{Z}\} \end{align*} for $r \in \{0, 1, \dots, n-1\}$. There are exactly $n$ cosets, and the quotient group is: \begin{align*} \mathbb{Z}/n\mathbb{Z} = \{0 + n\mathbb{Z},\ 1 + n\mathbb{Z},\ \dots,\ (n-1) + n\mathbb{Z}\}, \end{align*} with addition $(r + n\mathbb{Z}) + (s + n\mathbb{Z}) = (r + s) + n\mathbb{Z}$. This is the cyclic group of order $n$, and the group operation is addition modulo $n$. For instance, with $n = 4$: \begin{align*} (3 + 4\mathbb{Z}) + (2 + 4\mathbb{Z}) &= 5 + 4\mathbb{Z} = 1 + 4\mathbb{Z}, \end{align*} since $5 = 4 \cdot 1 + 1$. The element $3 + 4\mathbb{Z}$ has order $4$ in $\mathbb{Z}/4\mathbb{Z}$: adding it to itself four times gives $12 + 4\mathbb{Z} = 0 + 4\mathbb{Z}$. [/example] The example $\mathbb{Z}/n\mathbb{Z}$ shows that quotient groups are not merely a formal device — they produce genuinely new and important groups. The entire theory of cyclic groups is the theory of quotients of $\mathbb{Z}$. [example: Quotient of S3 by the Alternating Subgroup] Let $G = S_3$ and $N = A_3 = \{e, (123), (132)\}$, the alternating group of even permutations. Since $[S_3 : A_3] = 2$, a subgroup of index two is always normal (the only left coset other than $A_3$ itself is $S_3 \setminus A_3$, which must also be the only right coset). The two cosets are: \begin{align*} A_3 &= \{e, (123), (132)\} \\ (12)A_3 &= \{(12), (12)(123), (12)(132)\} = \{(12), (13), (23)\}. \end{align*} The quotient $S_3 / A_3$ has exactly two elements: $A_3$ and $(12)A_3$. The multiplication table is: \begin{align*} (A_3)(A_3) &= A_3, \\ (A_3)((12)A_3) &= (12)A_3, \\ ((12)A_3)(A_3) &= (12)A_3, \\ ((12)A_3)((12)A_3) &= (12)^2 A_3 = eA_3 = A_3. \end{align*} This is the cyclic group $\mathbb{Z}/2\mathbb{Z}$. The quotient map sends even permutations to $0$ and odd permutations to $1$, which is exactly the sign homomorphism $\operatorname{sgn}: S_3 \to \{+1, -1\}$ in group-theoretic language. [/example] ## The Canonical Projection and the First Isomorphism Theorem The quotient group $G/N$ is not just a set — it comes equipped with a natural surjective homomorphism from $G$, and this homomorphism is the key tool for understanding quotients in practice. The map sending each element of $G$ to its coset is called the **canonical projection** or **quotient map**. (As a preview: it is a group homomorphism, its kernel is exactly $N$, and it is surjective — these properties will be recorded formally in the theorem below.) [definition: Canonical Projection] Let $G$ be a group and $N \trianglelefteq G$. The **canonical projection** is the surjective group homomorphism \begin{align*} \pi: G &\to G/N \\ g &\mapsto gN. \end{align*} It satisfies $\ker \pi = N$ and $\pi$ is surjective. [/definition] [quotetheorem:2700] The canonical projection encodes the universal property of quotients: any homomorphism from $G$ that kills $N$ factors through $G/N$. [quotetheorem:2701] The universal property says that $G/N$ is the "most efficient" way to kill $N$: any homomorphism killing $N$ factors through $G/N$, and the factored map is as injective as possible. This is precisely the content of the First Isomorphism Theorem. [quotetheorem:842] The First Isomorphism Theorem is one of the most useful tools in algebra. It converts the problem of understanding the image of a homomorphism into the problem of understanding a quotient group. In practice: whenever you want to prove two groups are isomorphic, a common strategy is to find a surjective homomorphism from one to the other and identify its kernel. [example: Cyclic Groups as Quotients] Every cyclic group of order $n$ is isomorphic to $\mathbb{Z}/n\mathbb{Z}$. Define the homomorphism \begin{align*} \varphi: \mathbb{Z} &\to C_n = \langle a \rangle \\ k &\mapsto a^k. \end{align*} This is surjective (every element of $C_n$ is a power of $a$), and $\ker \varphi = \{k \in \mathbb{Z} : a^k = e\} = n\mathbb{Z}$ (since $a$ has order $n$). The First Isomorphism Theorem gives: \begin{align*} C_n \cong \mathbb{Z} / n\mathbb{Z}. \end{align*} Its elements can be labeled by residues $0, 1, \dots, n-1$ with addition modulo $n$. The isomorphism $C_n \cong \mathbb{Z}/n\mathbb{Z}$ will be put to further use in the Correspondence Theorem section below, where it is shown that $\mathbb{Z}/n\mathbb{Z}$ has exactly one subgroup of each order dividing $n$. [/example] [example: Identifying a Quotient via a Homomorphism] Let $G = \mathbb{R}$ (under addition) and define \begin{align*} \varphi: \mathbb{R} &\to S^1 \\ t &\mapsto e^{2\pi i t}, \end{align*} where $S^1 = \{z \in \mathbb{C} : |z| = 1\}$ is the circle group under multiplication. Then $\varphi$ is a group homomorphism: $\varphi(t + s) = e^{2\pi i(t+s)} = e^{2\pi it} e^{2\pi is} = \varphi(t)\varphi(s)$. The kernel is $\ker \varphi = \{t \in \mathbb{R} : e^{2\pi it} = 1\} = \mathbb{Z}$. Since $\varphi$ is surjective (every $e^{i\theta}$ is $\varphi(\theta/2\pi)$), the First Isomorphism Theorem gives: \begin{align*} \mathbb{R}/\mathbb{Z} \cong S^1. \end{align*} The coset $t + \mathbb{Z}$ consists of all real numbers with the same fractional part as $t$, and it corresponds to the point $e^{2\pi it}$ on the unit circle. The quotient $\mathbb{R}/\mathbb{Z}$ is the interval $[0, 1)$ with endpoints identified — geometrically, a circle. [/example] ## Lagrange's Theorem and the Order of a Quotient One of the first numerical constraints on a quotient group comes from counting. The order of $G/N$ is the number of cosets of $N$ in $G$, which is the index $[G : N]$. [quotetheorem:841] Lagrange's theorem has an immediate consequence for quotient groups: if $N \trianglelefteq G$ and $G$ is finite, then \begin{align*} |G/N| = [G : N] = \frac{|G|}{|N|}. \end{align*} This formula is the first tool for computing the order of a quotient group. It also constrains which subgroups can be normal: if $|G|/|N|$ does not divide $|G|$, then no such quotient exists. More usefully, it tells us that $G/N$ is "smaller" than $G$ by a factor of $|N|$. [explanation: Index Two Subgroups are Always Normal] A particularly clean consequence: if $[G : H] = 2$, then $H \trianglelefteq G$. The proof is direct. There are exactly two left cosets of $H$: $H$ itself and the one complementary coset $G \setminus H$. Similarly, there are two right cosets: $H$ and $G \setminus H$. Since the cosets partition $G$, the left coset $gH$ either equals $H$ (if $g \in H$) or equals $G \setminus H$ (if $g \notin H$). The same holds for right cosets $Hg$. Thus $gH = Hg$ for all $g \in G$, proving $H \trianglelefteq G$. This applies immediately: $A_n \trianglelefteq S_n$ for all $n$, since $[S_n : A_n] = 2$. The quotient $S_n / A_n \cong \mathbb{Z}/2\mathbb{Z}$ is the cyclic group of order two, with the quotient map being the sign homomorphism. [/explanation] ## Normal Subgroups and Conjugation A recurring theme in group theory is that normality is not just a technical condition for making quotients work — it reflects deep symmetry. The conjugation action of $G$ on its subgroups is the natural context. The group $G$ acts on the set of its subgroups by conjugation: $g$ sends $H$ to $gHg^{-1}$. A subgroup $N$ is normal precisely when this action fixes $N$ — that is, when the orbit of $N$ under the conjugation action is just $\{N\}$. [definition: Conjugate Subgroup] Let $G$ be a group and $H \le G$. For $g \in G$, the **conjugate of $H$ by $g$** is the subgroup \begin{align*} gHg^{-1} = \{ghg^{-1} : h \in H\}. \end{align*} The subgroup $H$ is normal in $G$ if and only if $gHg^{-1} = H$ for all $g \in G$. [/definition] Normality can be detected in several equivalent ways, and having all of them available matters in practice: condition (i) is the definition itself and is easiest to verify directly from elements; condition (ii) — that left and right cosets agree — is what makes the coset product well-defined; condition (iii) is the natural language of conjugation as a group action; and condition (iv) is often the fastest check in concrete finite groups, since one need only list conjugacy classes. The equivalence of these four formulations is what gives us so much flexibility when working with normal subgroups. [quotetheorem:787] Condition (iv) is often the fastest way to check normality in concrete groups: list the conjugacy classes of $G$ and check whether $N$ is a union of them. This works because conjugacy classes partition $G$, and normality requires $N$ to be preserved under all conjugations. [example: Normal Subgroups of S4 via Conjugacy Classes] The conjugacy classes of $S_4$ are determined by cycle type: \begin{align*} C_1 &= \{e\}, \\ C_2 &= \{(12),(13),(14),(23),(24),(34)\} \quad \text{(transpositions)}, \\ C_3 &= \{(123),(132),(124),(142),(134),(143),(234),(243)\} \quad \text{(3-cycles)}, \\ C_4 &= \{(1234),(1243),(1324),(1342),(1423),(1432)\} \quad \text{(4-cycles)}, \\ C_5 &= \{(12)(34),(13)(24),(14)(23)\} \quad \text{(double transpositions)}. \end{align*} The sizes are $|C_1| = 1$, $|C_2| = 6$, $|C_3| = 8$, $|C_4| = 6$, $|C_5| = 3$, with total $1 + 6 + 8 + 6 + 3 = 24 = |S_4|$. A normal subgroup must be a union of conjugacy classes containing $e$. The possible sizes of such unions must divide $|S_4| = 24$. The divisors of $24$ are $1, 2, 3, 4, 6, 8, 12, 24$. Unions containing $C_1$: - $C_1$: size $1$. This is the trivial subgroup $\{e\} \trianglelefteq S_4$. - $C_1 \cup C_5$: size $4$. Check: $V_4 = \{e, (12)(34), (13)(24), (14)(23)\}$ is the Klein four-group, and indeed $V_4 \trianglelefteq S_4$. - $C_1 \cup C_3 \cup C_5$: size $12$. This is $A_4 \trianglelefteq S_4$. - $C_1 \cup C_2 \cup C_3 \cup C_4 \cup C_5$: size $24$. This is $S_4$ itself. No other unions give subgroups. Thus the normal subgroups of $S_4$ are exactly $\{e\}$, $V_4$, $A_4$, and $S_4$. [/example] The example reveals the lattice of normal subgroups of $S_4$: [illustration:s4-normal-subgroup-chain] ## The Correspondence Theorem A fundamental structural result is that subgroups of $G/N$ correspond bijectively to subgroups of $G$ that contain $N$. This is the correspondence theorem (sometimes called the fourth isomorphism theorem or the lattice isomorphism theorem), and it is essential for understanding the subgroup structure of quotient groups. [quotetheorem:854] [explanation: Why the Correspondence Theorem is Useful] The correspondence theorem reduces questions about subgroups of $G/N$ to questions about subgroups of $G$ — a smaller problem. Suppose $G = \mathbb{Z}$ and $N = 12\mathbb{Z}$. The subgroups of $\mathbb{Z}$ containing $12\mathbb{Z}$ are $d\mathbb{Z}$ for each divisor $d$ of $12$: these are $\mathbb{Z}, 2\mathbb{Z}, 3\mathbb{Z}, 4\mathbb{Z}, 6\mathbb{Z}, 12\mathbb{Z}$. The correspondence theorem says the subgroups of $\mathbb{Z}/12\mathbb{Z}$ are exactly $\{d\mathbb{Z}/12\mathbb{Z} : d \mid 12\}$, which are cyclic groups of orders $12, 6, 4, 3, 2, 1$ respectively. So $\mathbb{Z}/12\mathbb{Z}$ has exactly six subgroups, one for each divisor of $12$. More generally, the subgroups of $\mathbb{Z}/n\mathbb{Z}$ are in bijection with the divisors of $n$, and the subgroup corresponding to divisor $d$ is $\langle d + n\mathbb{Z} \rangle \cong \mathbb{Z}/(n/d)\mathbb{Z}$. [/explanation] ## The Second and Third Isomorphism Theorems The First Isomorphism Theorem has two important companions. The second relates a subgroup and a normal subgroup; the third relates two nested normal subgroups. [quotetheorem:843] The second isomorphism theorem tells us that a subgroup $H$ of $G$ sees the quotient $G/N$ in a manageable way: the image of $H$ under the quotient map $\pi: G \to G/N$ is isomorphic to a quotient of $H$ itself, namely $H/(H \cap N)$. [quotetheorem:844] [example: Third Isomorphism Theorem in Z] Let $G = \mathbb{Z}$, $M = 4\mathbb{Z}$, $N = 12\mathbb{Z}$. Since $12\mathbb{Z} \subset 4\mathbb{Z} \subset \mathbb{Z}$, the third isomorphism theorem gives: \begin{align*} (\mathbb{Z}/12\mathbb{Z})/(4\mathbb{Z}/12\mathbb{Z}) \cong \mathbb{Z}/4\mathbb{Z}. \end{align*} Let us verify this directly. The group $\mathbb{Z}/12\mathbb{Z}$ has order $12$. The subgroup $4\mathbb{Z}/12\mathbb{Z} = \{0, 4, 8\}$ (in residue notation) has order $3$. The quotient $(\mathbb{Z}/12\mathbb{Z})/(4\mathbb{Z}/12\mathbb{Z})$ has order $12/3 = 4$. The group $\mathbb{Z}/4\mathbb{Z}$ also has order $4$. The isomorphism sends the coset $(r + 12\mathbb{Z}) + 4\mathbb{Z}/12\mathbb{Z}$ to $r + 4\mathbb{Z}$; this is well-defined because any two elements of $(r + 12\mathbb{Z}) + 4\mathbb{Z}/12\mathbb{Z}$ differ by an element of $4\mathbb{Z}$. [/example] ## Simple Groups and the Jordan–Hölder Theorem A group $G$ is called simple if it has no normal subgroups other than $\{e\}$ and $G$ itself. Simple groups are the "atoms" of group theory: they cannot be further decomposed by quotients. Understanding them is the first step toward understanding all finite groups. [definition: Simple Group] A group $G$ is **simple** if $|G| > 1$ and the only normal subgroups of $G$ are $\{e\}$ and $G$ itself. [/definition] The simplest examples of simple groups are the cyclic groups $\mathbb{Z}/p\mathbb{Z}$ for prime $p$: since $|\mathbb{Z}/p\mathbb{Z}| = p$, any proper nontrivial subgroup would have order dividing $p$, but $p$ is prime, so no such subgroup exists. The alternating groups $A_n$ for $n \ge 5$ are also simple, though proving this requires more work. [remark: Simplicity and Quotients] A group $G$ is simple if and only if the only quotient groups of $G$ are $\{e\}$ and $G$ itself. Equivalently, every surjective homomorphism $\varphi: G \to H$ is either trivial ($H = \{e\}$) or an isomorphism. Simple groups are those which cannot be "further reduced" by taking quotients. [/remark] The relationship between a group and its quotients is organized by the concept of a composition series — a chain of subgroups reaching from $\{e\}$ to $G$ where each successive quotient is simple. [definition: Composition Series] A **composition series** of a group $G$ is a finite chain of subgroups \begin{align*} \{e\} = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_k = G \end{align*} where each quotient $G_{i+1}/G_i$ is a simple group. The groups $G_{i+1}/G_i$ are called the **composition factors** of the series. [/definition] [quotetheorem:2625] The Jordan–Hölder theorem says that although the composition series itself is not unique, its factors are — they are an invariant of $G$. This is the group-theoretic analogue of the fundamental theorem of arithmetic: just as every integer factors uniquely into primes, every finite group has a unique multiset of composition factors. The composition factors are the simple groups from which $G$ is "built," and the study of how groups are assembled from simple groups (extension theory) is one of the central problems of group theory. [explanation: The Relationship Between Quotient Groups and Group Extensions] Knowing the composition factors of $G$ does not determine $G$ up to isomorphism — the factors can be assembled in different ways. This is the **extension problem**: given simple groups $S_1, \dots, S_k$, how many non-isomorphic groups $G$ have a composition series with factors $S_1, \dots, S_k$? Even the simplest case is nontrivial. Both $\mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ have composition series with factors $\mathbb{Z}/2\mathbb{Z}, \mathbb{Z}/2\mathbb{Z}$, but they are not isomorphic ($\mathbb{Z}/4\mathbb{Z}$ has an element of order $4$, while $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ does not). The quotient group structure controls the "type" of extension: whether the group splits as a direct product, a semidirect product, or a more exotic non-split extension. [/explanation] ## References Dummit, D. S. and Foote, R. M., *Abstract Algebra*, Third Edition (2004). Hungerford, T. W., *Algebra* (1974). Lang, S., *Algebra*, Revised Third Edition (2002). Rotman, J. J., *An Introduction to the Theory of Groups*, Fourth Edition (1995).