A subgroup $H \le G$ is often too small to describe the whole group, but its copies spread through $G$ in a highly controlled way. The first surprise is that a subgroup does not merely sit inside a group as a smaller object: it partitions the entire group into equal-sized pieces. Those pieces are cosets. They are the mechanism behind congruence classes, quotient groups, the [orbit-stabiliser theorem](/theorems/796), and the counting arguments that make finite group theory work. The motivating failure is simple. Suppose we want to divide a group $G$ by a subgroup $H \le G$ in the same way that we divide $\mathbb{Z}$ by $n\mathbb{Z}$. For integers, two numbers represent the same remainder modulo $n$ when their difference lies in $n\mathbb{Z}$. In a non-abelian group, there are two possible directions: $g^{-1}x \in H$ and $xg^{-1} \in H$. These need not agree. Cosets are the language that records both directions without pretending that they are the same. [example: Remainders as Cosets] Let $G = \mathbb{Z}$ under addition and let $H = 3\mathbb{Z} = \{3k : k \in \mathbb{Z}\}$. The three cosets represented by $0,1,2$ are \begin{align*} H &= 0+H = \{3k : k \in \mathbb{Z}\},\\ 1+H &= \{1+3k : k \in \mathbb{Z}\},\\ 2+H &= \{2+3k : k \in \mathbb{Z}\}. \end{align*} Every integer belongs to one of these three sets: by integer division, for each $m \in \mathbb{Z}$ there are $q \in \mathbb{Z}$ and $r \in \{0,1,2\}$ such that \begin{align*} m &= 3q+r, \end{align*} so $m \in r+3\mathbb{Z}$. The three sets are disjoint. For example, if $3r = 1+3s$, then \begin{align*} 1 &= 3r-3s\\ &= 3(r-s), \end{align*} which is impossible because $1$ is not divisible by $3$. Similarly, $3r=2+3s$ would imply $2=3(r-s)$, and $1+3r=2+3s$ would imply $1=3(r-s)$. Thus \begin{align*} \mathbb{Z} &= 3\mathbb{Z} \;\sqcup\; (1+3\mathbb{Z}) \;\sqcup\; (2+3\mathbb{Z}). \end{align*} Finally, ordinary congruence modulo $3$ is exactly equality of these cosets. If $a \equiv b \pmod{3}$, then $b-a=3t$ for some $t \in \mathbb{Z}$, and \begin{align*} b+3k &= a+3t+3k\\ &= a+3(t+k), \end{align*} so $b+3\mathbb{Z} \subseteq a+3\mathbb{Z}$; replacing $t$ by $-t$ gives the reverse inclusion. Conversely, if $a+3\mathbb{Z}=b+3\mathbb{Z}$, then $b \in a+3\mathbb{Z}$, so $b=a+3k$ for some $k \in \mathbb{Z}$, hence $a \equiv b \pmod{3}$. [/example] The integer example hides two issues that become essential in a general group. First, cosets are sets, not elements. Second, multiplication of cosets is not automatically meaningful unless the subgroup satisfies an additional [symmetry condition](/theorems/1360). The chapter develops these two points from the ground up. ## Definition The basic operation is to translate a subgroup by a group element. Translation preserves the internal shape of the subgroup, but moves its identity element to another point of the group. Since multiplication may depend on order, the first translation to isolate is multiplication on the left. [definition: Left Coset] Let $G$ be a group and let $H \le G$. For $g \in G$, the left coset of $H$ with representative $g$ is the subset \begin{align*} gH &= \{gh : h \in H\} \subset G. \end{align*} [/definition] ## Right Cosets and Coset Spaces The representative $g$ is not part of the coset data in a unique way. Many different elements of $G$ may name the same subset $gH$. To compare this with the other possible order of multiplication in a non-abelian group, we need the corresponding right-handed construction. [definition: Right Coset] Let $G$ be a group and let $H \le G$. For $g \in G$, the right coset of $H$ with representative $g$ is the subset \begin{align*} Hg &= \{hg : h \in H\} \subset G. \end{align*} [/definition] In abelian groups, left and right cosets coincide. In non-abelian groups, their difference measures how far $H$ is from being stable under conjugation. The next computation makes the distinction concrete before we package all cosets into a coset space. [example: Left and Right Cosets in $S_3$] Let $G = S_3$, let $H = \{e,(12)\} \le S_3$, and take $g=(123)$. By the definition of a left coset, \begin{align*} gH &= (123)H\\ &= \{(123)e,\,(123)(12)\}\\ &= \{(123),\,(123)(12)\}. \end{align*} To compute $(123)(12)$, compose right to left: \begin{align*} 1 &\mapsto 2 \mapsto 3,\\ 3 &\mapsto 3 \mapsto 1,\\ 2 &\mapsto 1 \mapsto 2. \end{align*} Thus $(123)(12)$ swaps $1$ and $3$ and fixes $2$, so \begin{align*} (123)(12) &= (13), \end{align*} and therefore \begin{align*} (123)H &= \{(123),(13)\}. \end{align*} By the definition of a right coset, \begin{align*} Hg &= H(123)\\ &= \{e(123),\,(12)(123)\}\\ &= \{(123),\,(12)(123)\}. \end{align*} Again composing right to left, \begin{align*} 1 &\mapsto 2 \mapsto 1,\\ 2 &\mapsto 3 \mapsto 3,\\ 3 &\mapsto 1 \mapsto 2. \end{align*} Thus $(12)(123)$ swaps $2$ and $3$ and fixes $1$, so \begin{align*} (12)(123) &= (23), \end{align*} and therefore \begin{align*} H(123) &= \{(123),(23)\}. \end{align*} Since $(13) \neq (23)$ in $S_3$, the subsets $\{(123),(13)\}$ and $\{(123),(23)\}$ are not equal. Thus the left and right cosets of the same subgroup by the same element can differ in a non-abelian group. [/example] A single coset records one translated copy of $H$, but many arguments need the whole family of such copies at once. To discuss partitions, indices, and group actions, we need a name for the set of all left cosets. [definition: Left Coset Space] Let $G$ be a group and let $H \le G$. The left coset space of $H$ in $G$ is the set \begin{align*} G/H &= \{gH : g \in G\}. \end{align*} [/definition] The notation $G/H$ is suggestive, but at this stage it denotes only a set. Since right cosets can differ from left cosets, the right-handed family needs its own notation rather than being silently identified with $G/H$. [definition: Right Coset Space] Let $G$ be a group and let $H \le G$. The right coset space of $H$ in $G$ is the set \begin{align*} H \backslash G &= \{Hg : g \in G\}. \end{align*} [/definition] The next invariant asks how large the coset partition is. Counting translated copies of $H$ is what turns cosets into divisibility statements, so the number of cosets deserves its own name. [definition: Index of a Subgroup] Let $G$ be a group and let $H \le G$. The index of $H$ in $G$ is the cardinality of the left coset space: \begin{align*} [G : H] &= |G/H|. \end{align*} [/definition] For groups, the number of left cosets and the number of right cosets agree, so the definition of index is independent of this choice. This fact is part of the structural theory below, not a property that should be assumed from the notation. ## Cosets as Equivalence Classes Cosets are best understood as equivalence classes. The relation is not equality inside the group; it is equality up to multiplication by an element of the subgroup. This point is essential because it explains why [cosets partition the group](/theorems/781) and why representatives are non-unique. ### Left Congruence When two elements should count as the same relative to $H$, we ask whether moving from one to the other requires only an element of $H$. For left cosets, the subgroup element appears after cancelling the representative on the left. [definition: Left Congruence Modulo a Subgroup] Let $G$ be a group and let $H \le G$. Define a relation $\sim_L$ on $G$ by \begin{align*} x \sim_L y \quad \text{if and only if} \quad x^{-1}y \in H. \end{align*} [/definition] The relation $\sim_L$ is useful only if it really behaves like equality after ignoring the subgroup $H$. Without reflexivity, symmetry, and transitivity, the phrase "the class of $x$" would not define stable blocks of $G$; and without identifying that class with $xH$, the relation would not recover the cosets already defined. This congruence relation has exactly the expected classes: the class of $x$ is $xH$. Thus the relational language of $\sim_L$ recovers the same left cosets used in computations, without needing a separate theorem card duplicating the right-coset analogue. ### Equality of Representatives The equivalence-class viewpoint gives a partition, but computations still need a practical test for when two representatives name the same block. The next criterion gives that test in several interchangeable forms. [quotetheorem:786] The equality criterion is often the fastest way to change representatives. It says that a coset is not tied to the element used to write it; any element of the coset may serve as a representative. The following elementary computation fixes this habit in the familiar setting of modular arithmetic. [example: Non-Unique Representatives] In $\mathbb{Z}$ under addition, let $H=5\mathbb{Z}=\{5m:m\in\mathbb{Z}\}$. The coset represented by $2$ is \begin{align*} 2+H &= 2+5\mathbb{Z}\\ &= \{2+5m:m\in\mathbb{Z}\}. \end{align*} Thus $2$, $7$, and $-3$ all lie in this same coset, since \begin{align*} 2 &= 2+5\cdot 0,\\ 7 &= 2+5\cdot 1,\\ -3 &= 2+5(-1). \end{align*} More generally, every integer of the form $2+5k$ is a representative of the same coset. Indeed, for any $k\in\mathbb{Z}$, \begin{align*} (2+5k)+5\mathbb{Z} &= \{2+5k+5m:m\in\mathbb{Z}\}\\ &= \{2+5(k+m):m\in\mathbb{Z}\}\\ &\subseteq 2+5\mathbb{Z}. \end{align*} Conversely, if $2+5m\in 2+5\mathbb{Z}$, then \begin{align*} 2+5m &= 2+5k+5(m-k), \end{align*} and $m-k\in\mathbb{Z}$, so $2+5m\in (2+5k)+5\mathbb{Z}$. Hence \begin{align*} (2+5k)+5\mathbb{Z} &= 2+5\mathbb{Z}. \end{align*} Taking $k=1$ gives the advertised equality \begin{align*} 7+5\mathbb{Z} &= 2+5\mathbb{Z}. \end{align*} Equivalently, this happens because \begin{align*} 7-2 &= 5 \in 5\mathbb{Z}. \end{align*} In additive notation, two representatives $a$ and $b$ determine the same coset modulo $5\mathbb{Z}$ exactly when $b-a\in 5\mathbb{Z}$; in multiplicative notation, the corresponding test is $gH=kH$ exactly when $g^{-1}k\in H$, as stated by the *Coset Equality Criterion*. [/example] ### Right Congruence The left relation explains left cosets, but non-abelian groups also require a right-sided version. Without a separate right congruence relation, it is too easy to cancel in the wrong order and identify sets that need not coincide. [definition: Right Congruence Modulo a Subgroup] Let $G$ be a group and let $H \le G$. Define a relation $\sim_R$ on $G$ by \begin{align*} x \sim_R y \quad \text{if and only if} \quad yx^{-1} \in H. \end{align*} [/definition] The right-handed relation is designed so that the equivalence class of $x$ becomes $Hx$. We need the formal statement to keep the right-coset partition parallel to the left-coset partition. [quotetheorem:4965] The equivalence-class viewpoint also clarifies the quotient notation. The slash in $G/H$ first means a set of equivalence classes. It becomes a group operation only after a separate condition is imposed. ## Counting and Lagrange's Theorem The first major use of cosets is counting. If $G$ is finite, every coset of $H$ has exactly as many elements as $H$, because left multiplication by $g$ is a bijection from $H$ to $gH$. Since distinct cosets are disjoint and cover $G$, the order of $G$ is split into equal blocks. The bijection behind this argument deserves to be named, because it is the technical reason that subgroups divide finite groups. [quotetheorem:4966] Once cosets have equal size, the natural counting question is how those equal blocks fill a finite group. The answer is the divisibility theorem that underlies many first arguments in group theory. [quotetheorem:841] [Lagrange's theorem](/theorems/782) is often remembered as a divisibility statement, but its real content is the coset partition. To see the partition rather than just the formula, it helps to list the cosets in a small non-abelian group. [example: Counting Cosets in $S_3$] Let $G=S_3$ and let $H=\{e,(12)\}$. Since $|S_3|=6$ and $|H|=2$, *[Lagrange's Theorem](/theorems/841)* gives \begin{align*} [S_3:H] &= \frac{|S_3|}{|H|}\\ &= \frac{6}{2}\\ &= 3. \end{align*} Thus there should be exactly three distinct left cosets. The subgroup itself is \begin{align*} H &= \{e,(12)\}. \end{align*} For the representative $(13)$, \begin{align*} (13)H &= \{(13)e,\,(13)(12)\}\\ &= \{(13),\,(13)(12)\}. \end{align*} Composing right to left, \begin{align*} 1 &\mapsto 2 \mapsto 2,\\ 2 &\mapsto 1 \mapsto 3,\\ 3 &\mapsto 3 \mapsto 1, \end{align*} so $(13)(12)$ sends $1\mapsto 2$, $2\mapsto 3$, and $3\mapsto 1$. Hence \begin{align*} (13)(12) &= (123), \end{align*} and therefore \begin{align*} (13)H &= \{(13),(123)\}. \end{align*} For the representative $(23)$, \begin{align*} (23)H &= \{(23)e,\,(23)(12)\}\\ &= \{(23),\,(23)(12)\}. \end{align*} Again composing right to left, \begin{align*} 1 &\mapsto 2 \mapsto 3,\\ 2 &\mapsto 1 \mapsto 1,\\ 3 &\mapsto 3 \mapsto 2, \end{align*} so $(23)(12)$ sends $1\mapsto 3$, $3\mapsto 2$, and $2\mapsto 1$. Hence \begin{align*} (23)(12) &= (132), \end{align*} and therefore \begin{align*} (23)H &= \{(23),(132)\}. \end{align*} We have listed \begin{align*} H &= \{e,(12)\},\\ (13)H &= \{(13),(123)\},\\ (23)H &= \{(23),(132)\}. \end{align*} The six displayed elements \begin{align*} e,\ (12),\ (13),\ (123),\ (23),\ (132) \end{align*} are exactly the six elements of $S_3$, with no repetition between the three displayed sets. Hence \begin{align*} S_3 &= H \;\sqcup\; (13)H \;\sqcup\; (23)H. \end{align*} This explicit list shows both the coset partition and the need to avoid duplicate representatives when enumerating cosets. [/example] The next question is what this counting says about a single element. Since a cyclic subgroup is generated by one element, the coset count constrains possible periods of powers in a finite group. [quotetheorem:783] For a single element, Lagrange's theorem becomes an order constraint: the cyclic subgroup generated by $g$ has size $\operatorname{ord}(g)$, so element periods in a finite group can only be divisors of $|G|$. In $S_3$, for instance, no element can have order $4$ or $5$, because neither number divides $6$. The result is only a necessary condition on possible element orders; it does not by itself classify which divisors actually occur as element orders. ## Normality and Quotient Groups The next question is whether cosets can be multiplied. The tempting rule is \begin{align*} (gH)(kH) &= (gk)H. \end{align*} The difficulty is representative-dependence: replacing $g$ by another element of $gH$ and $k$ by another element of $kH$ may change the product unless $H$ is stable under conjugation. The condition that removes this ambiguity is normality. It says that the subgroup is positioned symmetrically inside the ambient group, exactly so that changing representatives does not change a coset product. [definition: Normal Subgroup] Let $G$ be a group and let $N \le G$. The subgroup $N$ is normal in $G$, written $N \trianglelefteq G$, if \begin{align*} gNg^{-1} &= N \end{align*} for every $g \in G$. [/definition] The definition is phrased by conjugation, while cosets are written by multiplying $N$ on the left or on the right. To use normality in quotient constructions, these two languages must match: changing from $gN$ to $Ng$ should not change the subset being named. The following equivalences isolate exactly when that left-right ambiguity disappears. [quotetheorem:787] The equivalences now remove the ambiguity in the tentative product of cosets. The next construction asks us to use this stability to turn the coset space itself into a group. [definition: Quotient Group] Let $G$ be a group and let $N \trianglelefteq G$. The [quotient group](/page/Quotient%20Group) $G/N$ is the set of left cosets of $N$ in $G$ equipped with the multiplication map \begin{align*} (G/N) \times (G/N) &\to G/N \\ (gN,kN) &\mapsto (gk)N \end{align*} for all $g,k \in G$. [/definition] The multiplication formula mentions representatives $g$ and $k$, not only the cosets $gN$ and $kN$. Thus the construction would fail if replacing $g$ by $gn_1$ or $k$ by $kn_2$ could change the resulting coset. Normality is precisely the hypothesis that forces all such representative choices to give the same product. Before using $G/N$ as a group, this representative issue must be settled as an actual theorem rather than left as notation. The result below supplies the group axioms for the coset product and identifies the normality condition that makes the construction legitimate. [quotetheorem:790] The theorem is where cosets become more than a partition. They become algebraic objects that support computation. The standard example is modular arithmetic, where quotient notation is already familiar. [example: The Quotient $\mathbb{Z}/n\mathbb{Z}$] Let $n \in \mathbb{N}$ with $n \ge 1$, and let $N=n\mathbb{Z} \le \mathbb{Z}$. Since $\mathbb{Z}$ is abelian, for every $a \in \mathbb{Z}$ we have \begin{align*} a+N-a &= \{a+x-a:x\in N\}\\ &= \{x:x\in N\}\\ &= N, \end{align*} so $N$ is normal. We compute the [quotient group](/theorems/790) $\mathbb{Z}/n\mathbb{Z}$ by identifying its cosets. For any $a\in\mathbb{Z}$, the [division algorithm](/theorems/725) gives integers $q$ and $r$ with $0\le r<n$ such that \begin{align*} a &= nq+r. \end{align*} Then \begin{align*} a+n\mathbb{Z} &= \{a+nk:k\in\mathbb{Z}\}\\ &= \{nq+r+nk:k\in\mathbb{Z}\}\\ &= \{r+n(q+k):k\in\mathbb{Z}\}\\ &\subseteq r+n\mathbb{Z}. \end{align*} Conversely, if $r+nm\in r+n\mathbb{Z}$, then \begin{align*} r+nm &= nq+r+n(m-q)\\ &= a+n(m-q), \end{align*} and $m-q\in\mathbb{Z}$, so $r+nm\in a+n\mathbb{Z}$. Hence \begin{align*} a+n\mathbb{Z} &= r+n\mathbb{Z}. \end{align*} Thus every coset is one of \begin{align*} 0+n\mathbb{Z},\;1+n\mathbb{Z},\;\ldots,\;(n-1)+n\mathbb{Z}. \end{align*} These displayed cosets are distinct. If $r,s\in\{0,1,\ldots,n-1\}$ and \begin{align*} r+n\mathbb{Z} &= s+n\mathbb{Z}, \end{align*} then $s\in r+n\mathbb{Z}$, so $s=r+nt$ for some $t\in\mathbb{Z}$. Therefore \begin{align*} s-r &= nt. \end{align*} Since $0\le r,s<n$, we have $-(n-1)\le s-r\le n-1$. The only multiple of $n$ in this interval is $0$, so $s-r=0$ and hence $r=s$. Addition in the quotient is determined by representatives: \begin{align*} (a+n\mathbb{Z})+(b+n\mathbb{Z}) &= (a+b)+n\mathbb{Z}. \end{align*} This rule is independent of the representatives. Indeed, if \begin{align*} a'&=a+nu,\\ b'&=b+nv \end{align*} for some $u,v\in\mathbb{Z}$, then \begin{align*} a'+b' &= a+nu+b+nv\\ &= a+b+n(u+v), \end{align*} so \begin{align*} (a'+b')+n\mathbb{Z} &= (a+b)+n\mathbb{Z}. \end{align*} Thus $\mathbb{Z}/n\mathbb{Z}$ is exactly the usual arithmetic of residue classes modulo $n$, written as addition of cosets. [/example] The non-normal case shows why the hypothesis cannot be dropped. A set of cosets may exist without supporting a group law, and the following example exhibits the representative-dependence directly. [example: Failed Coset Multiplication Without Normality] Let $G=S_3$ and let $H=\{e,(12)\}$. We first verify that $(123)$ and $(13)$ represent the same left coset of $H$. By the definition of a left coset, \begin{align*} (123)H &= \{(123)e,\,(123)(12)\}\\ &= \{(123),\,(123)(12)\}. \end{align*} Composing right to left, \begin{align*} 1 &\mapsto 2 \mapsto 3,\\ 3 &\mapsto 3 \mapsto 1,\\ 2 &\mapsto 1 \mapsto 2, \end{align*} so $(123)(12)$ swaps $1$ and $3$ and fixes $2$. Hence \begin{align*} (123)(12)&=(13), \end{align*} and therefore \begin{align*} (123)H&=\{(123),(13)\}. \end{align*} Similarly, \begin{align*} (13)H &= \{(13)e,\,(13)(12)\}\\ &= \{(13),\,(13)(12)\}. \end{align*} Composing right to left, \begin{align*} 1 &\mapsto 2 \mapsto 2,\\ 2 &\mapsto 1 \mapsto 3,\\ 3 &\mapsto 3 \mapsto 1, \end{align*} so $(13)(12)$ sends $1\mapsto 2$, $2\mapsto 3$, and $3\mapsto 1$. Thus \begin{align*} (13)(12)&=(123), \end{align*} and hence \begin{align*} (13)H&=\{(13),(123)\}\\ &=(123)H. \end{align*} Now try to define multiplication of left cosets by the rule \begin{align*} (gH)(kH)&=(gk)H. \end{align*} Using the representatives $(123)$ and $(123)$ for the same input coset gives \begin{align*} ((123)(123))H&=(132)H. \end{align*} Indeed, composing $(123)$ with itself, \begin{align*} 1&\mapsto 2 \mapsto 3,\\ 3&\mapsto 1 \mapsto 2,\\ 2&\mapsto 3 \mapsto 1, \end{align*} so $(123)(123)=(132)$. The resulting coset is \begin{align*} (132)H &= \{(132)e,\,(132)(12)\}\\ &= \{(132),\,(132)(12)\}. \end{align*} Composing right to left, \begin{align*} 1&\mapsto 2 \mapsto 1,\\ 2&\mapsto 1 \mapsto 3,\\ 3&\mapsto 3 \mapsto 2, \end{align*} so $(132)(12)=(23)$. Therefore \begin{align*} ((123)(123))H &=(132)H\\ &=\{(132),(23)\}. \end{align*} Using instead the representatives $(13)$ and $(13)$ for that same input coset gives \begin{align*} ((13)(13))H&=eH. \end{align*} Since the transposition $(13)$ swaps $1$ and $3$, applying it twice fixes every element: \begin{align*} 1&\mapsto 3 \mapsto 1,\\ 2&\mapsto 2 \mapsto 2,\\ 3&\mapsto 1 \mapsto 3. \end{align*} Thus $(13)(13)=e$, and so \begin{align*} ((13)(13))H &=eH\\ &=\{ee,e(12)\}\\ &=\{e,(12)\}\\ &=H. \end{align*} The two outputs are not equal, because \begin{align*} (132)H&=\{(132),(23)\},\\ H&=\{e,(12)\}, \end{align*} and the elements $(132)$, $(23)$, $e$, and $(12)$ are distinct in $S_3$. Thus the same input cosets can produce different output cosets when different representatives are chosen, so $S_3/H$ is a coset space here, not a quotient group. [/example] This distinction is one of the main lessons of the topic: every subgroup gives cosets, but only normal subgroups give quotient groups. ## Cosets and Homomorphisms Cosets also arise as fibres of homomorphisms. This gives a conceptual reason for normal subgroups: kernels are exactly the subgroups whose cosets are fibres of a structure-preserving map. When a homomorphism forgets information, two elements become indistinguishable precisely when their quotient lies in the kernel. To describe the fibre through a given element, we name the corresponding coset of the kernel. [definition: Kernel Coset] Let $\varphi: G \to K$ be a group homomorphism and let $a \in G$. The kernel coset of $a$ with respect to $\varphi$ is the left coset \begin{align*} a\ker\varphi &= \{ah : h \in \ker\varphi\}. \end{align*} [/definition] A fibre of $\varphi$ is defined by an equation in the target group, while a coset of $\ker\varphi$ is defined by multiplication inside the source group. The useful point is that these are not two separate partitions: two elements have the same image exactly when their difference, measured multiplicatively, lies in the kernel. This converts fibre questions into coset questions. The next goal is to compare a quotient by a kernel with the image of the homomorphism. This comparison should preserve multiplication, not just count fibres, and it is the main reason kernel cosets are central to quotient theory. [quotetheorem:842] The [first isomorphism theorem](/theorems/791) is often the cleanest way to compute a quotient: instead of listing cosets by hand, build a homomorphism whose kernel is the subgroup in question. The sign map gives a compact example. [example: Sign Homomorphism and Alternating Group] Let $\operatorname{sgn}:S_n\to\{1,-1\}$ be the [sign homomorphism](/theorems/778) for $n\ge 2$. By definition, $A_n$ is the subgroup of even permutations, so \begin{align*} \ker(\operatorname{sgn}) &= \{\tau\in S_n:\operatorname{sgn}(\tau)=1\}\\ &= A_n. \end{align*} The image is all of $\{1,-1\}$: the identity permutation satisfies $\operatorname{sgn}(e)=1$, and the transposition $(12)$ satisfies $\operatorname{sgn}((12))=-1$. The fibre over $1$ is therefore \begin{align*} \operatorname{sgn}^{-1}(\{1\}) &= \{\tau\in S_n:\operatorname{sgn}(\tau)=1\}\\ &= A_n. \end{align*} Now fix any odd permutation $\sigma\in S_n$. We show that the fibre over $-1$ is the coset $\sigma A_n$. If $\tau\in\sigma A_n$, then $\tau=\sigma\alpha$ for some $\alpha\in A_n$, and hence \begin{align*} \operatorname{sgn}(\tau) &= \operatorname{sgn}(\sigma\alpha)\\ &= \operatorname{sgn}(\sigma)\operatorname{sgn}(\alpha)\\ &= (-1)(1)\\ &= -1. \end{align*} Thus $\sigma A_n\subseteq \operatorname{sgn}^{-1}(\{-1\})$. Conversely, if $\tau\in S_n$ is odd, then \begin{align*} \operatorname{sgn}(\sigma^{-1}\tau) &= \operatorname{sgn}(\sigma^{-1})\operatorname{sgn}(\tau)\\ &= \operatorname{sgn}(\sigma)^{-1}\operatorname{sgn}(\tau)\\ &= (-1)^{-1}(-1)\\ &= 1, \end{align*} so $\sigma^{-1}\tau\in A_n$. Therefore \begin{align*} \tau &= \sigma(\sigma^{-1}\tau) \end{align*} lies in $\sigma A_n$, giving \begin{align*} \operatorname{sgn}^{-1}(\{-1\}) &= \sigma A_n. \end{align*} Since $\ker(\operatorname{sgn})=A_n$ and $\operatorname{im}(\operatorname{sgn})=\{1,-1\}$, the *First Isomorphism Theorem* gives \begin{align*} S_n/A_n &= S_n/\ker(\operatorname{sgn})\\ &\cong \operatorname{im}(\operatorname{sgn})\\ &= \{1,-1\}. \end{align*} Thus the quotient $S_n/A_n$ records exactly the two-valued invariant “even or odd.” [/example] Cosets therefore connect two viewpoints: internal subgroup geometry and external maps out of the group. The same partition can be read as translated copies of a subgroup or as the fibres of a homomorphism. ## Actions on Cosets A coset space is not always a group, but it still carries a natural [group action](/page/Group%20Action). This is often the right replacement when $H$ is not normal. The group $G$ acts on $G/H$ by moving every coset by left multiplication. Group actions on cosets convert subgroup questions into permutation representations. This idea is central in the study of finite groups because it lets the group act on a finite set whose size is the index of a subgroup. [definition: Left Action on a Coset Space] Let $G$ be a group and let $H \le G$. The left action of $G$ on $G/H$ is the map \begin{align*} G \times G/H &\to G/H \\ (g, xH) &\mapsto gxH. \end{align*} [/definition] The action would lose much of its value if the original subgroup could not be recovered from it. The base coset $H$ is the natural test point: an element fixes this point exactly when its left multiplication does not move $H$ to a different coset. Computing that stabiliser shows how the subgroup is encoded in the action. This recovery question is the first structural check on the action. Showing that the [stabiliser of the base coset](/theorems/4967) is exactly $H$ confirms that the action remembers the subgroup it came from. [quotetheorem:4967] The base coset remembers the original subgroup. The same computation at a translated coset gives the conjugate stabiliser: the stabiliser of $xH$ is $xHx^{-1}$. This is best read as the transported version of the base-coset result rather than as a separate theorem card. The stabiliser computation suggests a converse question: if a group action has a single orbit, must it come from cosets of a stabiliser? The answer is yes, and it is the structural reason coset spaces appear throughout group actions. [quotetheorem:3240] This theorem explains why cosets appear throughout group actions. Whenever a group moves a space with only one orbit, the space is a disguised coset space. A familiar geometric action makes the identification concrete. [example: Rotations Acting on Vertices] Let the vertices of the regular $n$-gon be $v_0,v_1,\ldots,v_{n-1}$, with indices read modulo $n$. Write the rotation group as \begin{align*} C_n &= \langle r : r^n=e\rangle, \end{align*} where \begin{align*} r^i\cdot v_j &= v_{i+j}. \end{align*} The action is transitive because, for every vertex $v_j$, \begin{align*} r^j\cdot v_0 &= v_j. \end{align*} The stabiliser of $v_0$ is trivial: if $r^i\cdot v_0=v_0$ with $0\le i<n$, then \begin{align*} v_i &= v_0, \end{align*} so $i=0$ and $r^i=e$. Therefore \begin{align*} (C_n)_{v_0} &= \{e\}. \end{align*} By *Transitive Actions Are Coset Actions*, the map \begin{align*} C_n/\{e\} &\to \{v_0,\ldots,v_{n-1}\}\\ r^i\{e\} &\mapsto r^i\cdot v_0=v_i \end{align*} is an isomorphism of $C_n$-sets. Now let the full dihedral group act on the same vertices. Write \begin{align*} D_{2n} &= \langle r,s : r^n=e,\ s^2=e,\ srs=r^{-1}\rangle, \end{align*} where $r$ is the rotation above and choose the reflection $s$ so that \begin{align*} s\cdot v_j &= v_{-j}. \end{align*} Every element of $D_{2n}$ is either $r^i$ or $r^i s$ with $0\le i<n$. For the base vertex $v_0$, \begin{align*} r^i\cdot v_0 &= v_i,\\ r^i s\cdot v_0 &= r^i\cdot (s\cdot v_0)\\ &= r^i\cdot v_0\\ &= v_i. \end{align*} Thus $r^i$ fixes $v_0$ exactly when $i=0$, and $r^i s$ fixes $v_0$ exactly when $i=0$. Hence \begin{align*} (D_{2n})_{v_0} &= \{e,s\}. \end{align*} This subgroup has order $2$ and is the reflection subgroup \begin{align*} H &= \{e,s\}. \end{align*} Since the rotations already move $v_0$ to every vertex, the $D_{2n}$-action is transitive, so *Transitive Actions Are Coset Actions* gives an isomorphism of $D_{2n}$-sets \begin{align*} D_{2n}/H &\to \{v_0,\ldots,v_{n-1}\}\\ gH &\mapsto g\cdot v_0. \end{align*} Thus the same vertex set can be read as a coset space for two different transitive actions: $C_n/\{e\}$ for rotations only, and $D_{2n}/H$ when reflections are included. [/example] This perspective separates two structures that are often conflated. A coset space may fail to be a quotient group, yet it can still be the correct space on which the original group acts. ## Double Cosets and Decomposition Problems Left and right cosets compare a subgroup with one side of multiplication. Many classification problems require both sides at once. For example, matrices may be simplified by multiplying on the left and on the right by special subgroups, and group elements may be studied up to two different symmetries. The resulting pieces are double cosets. They are larger and less rigid than ordinary cosets, but they organise many decomposition theorems in algebra and geometry. [definition: Double Coset] Let $G$ be a group and let $H,K \le G$. For $g \in G$, the double coset of $g$ with respect to $H$ and $K$ is the subset \begin{align*} HgK &= \{hgk : h \in H,\; k \in K\} \subset G. \end{align*} [/definition] A double coset allows a representative to be modified by one subgroup on the left and another subgroup on the right. To discuss the full classification rather than one class at a time, we need the set of all double cosets. [definition: Double Coset Space] Let $G$ be a group and let $H,K \le G$. The double coset space of $G$ by $H$ and $K$ is the set \begin{align*} H \backslash G / K &= \{HgK : g \in G\}. \end{align*} [/definition] Double cosets usually do not inherit a group structure, so their meaning has to come from a classification problem rather than from multiplication. The relevant equivalence allows one symmetry from $H$ on the left and another from $K$ on the right. To make this precise, we need an action whose orbits are exactly the subsets $HgK$. [quotetheorem:4968] The inverse on the right in the action formula does not change the orbit set, since $K$ is closed under inverses. It makes the formula a genuine left action of the product group. Matrix reduction gives a concrete place where double cosets encode canonical forms. [example: Row and Column Operations as Double Cosets] Let $k$ be a field, let $G=GL(n,k)$, and let $B\le G$ be the subgroup of invertible upper triangular matrices. Two matrices $A,A'\in GL(n,k)$ lie in the same double coset in $B\backslash G/B$ exactly when there are $b_L,b_R\in B$ such that \begin{align*} A' &= b_L A b_R. \end{align*} Thus left multiplication by $B$ performs invertible upper-triangular row operations, and right multiplication by $B$ performs invertible upper-triangular column operations. For $1\le i,j\le n$, write $A_{i..n,\,1..j}$ for the submatrix of $A$ using rows $i,i+1,\ldots,n$ and columns $1,2,\ldots,j$. If $A'=b_L A b_R$, then \begin{align*} A'_{i..n,\,1..j} &= (b_L)_{i..n,\,i..n}\, A_{i..n,\,1..j}\, (b_R)_{1..j,\,1..j}. \end{align*} Here $(b_L)_{i..n,\,i..n}$ and $(b_R)_{1..j,\,1..j}$ are invertible upper triangular matrices, so multiplication by them does not change rank. Hence \begin{align*} \operatorname{rank}(A'_{i..n,\,1..j}) &= \operatorname{rank}(A_{i..n,\,1..j}). \end{align*} These ranks are invariants of the double coset. For a permutation $w\in S_n$, let $P_w$ be the permutation matrix with a $1$ in position $(w(j),j)$ and $0$ elsewhere. Then \begin{align*} \operatorname{rank}\bigl((P_w)_{i..n,\,1..j}\bigr) &= \#\{q\in\{1,\ldots,j\}: w(q)\ge i\}, \end{align*} because the selected submatrix contains exactly those standard basis columns whose nonzero entry lies in one of the rows $i,\ldots,n$. The *Bruhat decomposition* states that every $A\in GL(n,k)$ can be written in the form \begin{align*} A &= b_L P_w b_R \end{align*} for some $b_L,b_R\in B$ and a unique permutation $w\in S_n$. Therefore \begin{align*} GL(n,k) &= \bigsqcup_{w\in S_n} B P_w B, \end{align*} and the double coset space is \begin{align*} B\backslash GL(n,k)/B &= \{BP_wB:w\in S_n\}. \end{align*} Thus $B\backslash G/B$ records exactly the pivot pattern that remains after all allowed upper-triangular row and column operations, and the indexing set is the Weyl group $S_n$. [/example] Double cosets reveal the broader role of the ordinary coset construction. A coset is not only a counting device; it is a way of quotienting a group by allowed transformations. ## Beyond and Connected Topics Cosets are the entry point to quotient structures across algebra. In group theory, the next step is [normal subgroup](/page/Normal%20Subgroup) and quotient group theory: once a subgroup is normal, its cosets become elements of a new group. This is the setting for the isomorphism theorems, composition series, and simple groups. In ring and module theory, cosets reappear as additive cosets of ideals and submodules. If $I \trianglelefteq R$ is an ideal, the quotient ring $R/I$ consists of additive cosets $a+I$. If $N$ is a submodule of an $R$-module $M$, the quotient module $M/N$ consists of cosets $m+N$. The same partition idea drives much of [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). In linear algebra, quotient vector spaces are coset spaces of subspaces. A [vector space](/page/Vector%20Space) quotient $V/W$ consists of affine translates $v+W$, and the dimension formula $\dim(V/W)=\dim V-\dim W$ is the vector-space analogue of Lagrange's theorem. This connects naturally with [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). In homological algebra, quotients by images are everywhere. The homology group of a chain complex is a quotient $\ker d_n / \operatorname{im} d_{n+1}$, so its elements are cosets of boundaries inside cycles. This makes cosets part of the language of [Homological Algebra I: Complexes and Resolutions](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions). In commutative algebra, localisations and quotient rings use cosets to impose equations algebraically. Passing from $R$ to $R/I$ forces every element of $I$ to become zero, and the coset $a+I$ is the residue of $a$ after imposing those equations. This viewpoint is developed further in [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). ## References Androma, [Normal Subgroup](/page/Normal%20Subgroup). Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Androma, [Homological Algebra I: Complexes and Resolutions](/page/Homological%20Algebra%20I%3A%20Complexes%20and%20Resolutions). Dummit and Foote, *Abstract Algebra* (2004). Lang, *Algebra* (2002). Artin, *Algebra* (2011).