We often meet groups first as abstract multiplication tables, but the first groups most people actually use are groups of rearrangements. If three roots of a polynomial are unnamed points on a page, there are six ways to relabel them. If four vertices of a tetrahedron are numbered, every symmetry of the labelled set is a permutation. The symmetric group is the universal home for these rearrangements: it records every possible bijective relabelling of a finite set, and it turns the idea of a group into something visible and computable. The basic question is this: how complicated can the group of all rearrangements of $n$ objects be, and how can we calculate inside it without listing all $n!$ permutations? The answer begins with cycles. A permutation may look like a scrambled table of values, but its repeated action decomposes the underlying set into disjoint directed loops. This single observation explains orders of elements, conjugacy classes, signs, generators, and many homomorphisms out of $S_n$. [example: Three Letters Already Give Non-Commutativity] Let $X=\{1,2,3\}$. Define $\sigma,\tau:X\to X$ by \begin{align*} \sigma(1)=2,\quad \sigma(2)=1,\quad \sigma(3)=3,\quad \tau(1)=1,\quad \tau(2)=3,\quad \tau(3)=2. \end{align*} We compare the two possible orders of composition on the single input $1$. Since $\tau(1)=1$ and $\sigma(1)=2$, \begin{align*} (\sigma\circ\tau)(1)=\sigma(\tau(1))=\sigma(1)=2. \end{align*} In the other order, since $\sigma(1)=2$ and $\tau(2)=3$, \begin{align*} (\tau\circ\sigma)(1)=\tau(\sigma(1))=\tau(2)=3. \end{align*} The two composite functions have different values at $1$, so $\sigma\circ\tau\ne \tau\circ\sigma$. Thus even in $S_3$, composition of bijections depends on the order in which the relabellings are performed. [/example] This failure of commutativity is not a defect; it is the point. A permutation remembers order of operations, and different orders of relabelling can produce different outcomes. The symmetric group is therefore the natural testing ground for group theory: every abstract group can be represented inside some symmetric group, but symmetric groups themselves already contain rich internal structure. ## Definition The parent concept is a [group](/page/Group): a set with an associative binary operation, identity, and inverses. For a finite set, the most natural group operation on bijections is composition. The symmetric group isolates this construction for the standard $n$-element set. [definition: Symmetric Group] Let $n \in \mathbb{N}$. The symmetric group on $n$ letters is \begin{align*} S_n := \{\sigma: \{1,\dots,n\} \to \{1,\dots,n\} : \sigma \text{ is a bijection}\}, \end{align*} equipped with the group operation given by composition of functions. [/definition] The identity element is the identity map $\operatorname{id}_{\{1,\dots,n\}}$, usually written simply as $e$ once the ambient group is fixed. The inverse of $\sigma \in S_n$ is its inverse function $\sigma^{-1}: \{1,\dots,n\} \to \{1,\dots,n\}$. The convention in this page is that products are composed right-to-left: $\sigma\tau$ means $\sigma \circ \tau$, so $\tau$ acts first. A definition in terms of maps is precise, but calculations still need a compact way to record the images of all letters. Two-line notation solves the bookkeeping problem by putting the domain in a fixed order and writing the corresponding images underneath in prose-readable form. [definition: Two-Line Notation] Let $\sigma \in S_n$. The two-line notation for $\sigma$ is the array whose first row is $1,2,\dots,n$ and whose second row is $\sigma(1),\sigma(2),\dots,\sigma(n)$. [/definition] Two-line notation is close to the definition, but it hides the repeated motion of a point under the permutation. To see that motion, follow $i$, then $\sigma(i)$, then $\sigma(\sigma(i))$, and continue until returning to $i$. This leads to cycle notation, which records the orbits of this repeated motion. [definition: Cycle] Let $n \in \mathbb{N}$, and let $a_1,\dots,a_k \in \{1,\dots,n\}$ be distinct. The cycle $(a_1\ a_2\ \cdots\ a_k)$ is the permutation $\sigma: \{1,\dots,n\} \to \{1,\dots,n\}$ defined by sending $a_i$ to $a_{i+1}$ for $1 \le i < k$, sending $a_k$ to $a_1$, and fixing every element of $\{1,\dots,n\} \setminus \{a_1,\dots,a_k\}$. [/definition] A cycle of length $1$ fixes its single element, so such cycles are usually omitted from notation. The cycle $(1\ 3\ 4)$ means $1 \mapsto 3$, $3 \mapsto 4$, $4 \mapsto 1$, and every other letter is fixed. Since the same loop can be started at any point, $(1\ 3\ 4)$, $(3\ 4\ 1)$, and $(4\ 1\ 3)$ represent the same permutation. [example: Converting a Table into Cycles] Let $\sigma \in S_5$ be given by $\sigma(1)=4$, $\sigma(2)=5$, $\sigma(3)=3$, $\sigma(4)=1$, and $\sigma(5)=2$. To convert this table into cycles, follow each unused letter under repeated application of $\sigma$. Starting at $1$, the given values give \begin{align*} 1 \mapsto \sigma(1)=4. \end{align*} Then \begin{align*} 4 \mapsto \sigma(4)=1. \end{align*} So the letters $1$ and $4$ form the cycle $(1\ 4)$. The smallest unused letter is now $2$. Since $\sigma(2)=5$ and $\sigma(5)=2$, we have \begin{align*} 2 \mapsto 5 \mapsto 2. \end{align*} This gives the cycle $(2\ 5)$. Finally, \begin{align*} \sigma(3)=3, \end{align*} so $3$ is a fixed point and may be omitted from cycle notation. Therefore \begin{align*} \sigma=(1\ 4)(2\ 5). \end{align*} This cycle expression records exactly the same images as the original table, but it groups the letters by their repeated motion under $\sigma$. [/example] ## Cycles and Element Structure The first structural problem is to decide whether cycle notation is merely convenient or actually canonical. A general product of cycles can be hard to read because cycles may move the same letters. The useful case is when the cycles are disjoint, since then each letter belongs to at most one moving loop. [definition: Disjoint Cycles] Two cycles $(a_1\ \cdots\ a_r)$ and $(b_1\ \cdots\ b_s)$ in $S_n$ are disjoint if \begin{align*} \{a_1,\dots,a_r\} \cap \{b_1,\dots,b_s\}=\varnothing. \end{align*} [/definition] Disjoint cycles act on separate letters, so their product can be read independently on each part of the set. The next theorem answers the normal-form question: every permutation can be broken into these noninterfering loops, and there is no hidden choice except the order in which the loops are written. [quotetheorem:775] The decomposition theorem converts questions about a permutation into questions about several independent loops. To compare two permutations, though, we need a way to remember the sizes of those loops while ignoring the accidental names of the letters. That invariant is the cycle type. [definition: Cycle Type] Let $\sigma \in S_n$. The cycle type of $\sigma$ is the partition of $n$ whose parts are the lengths of the disjoint cycles in the [disjoint cycle decomposition](/theorems/775) of $\sigma$, including cycles of length $1$. [/definition] For example, $(1\ 4)(2\ 5) \in S_5$ has cycle type $(2,2,1)$. Cycle type records how the permutation decomposes the set into orbits under repeated application. Once the loop lengths are known, the return time of the whole permutation becomes a least common multiple problem. [quotetheorem:8539] The order formula says that a permutation returns every letter to its starting point exactly when every cycle has completed an integer number of turns. Thus a $3$-cycle and a disjoint $4$-cycle combine to form an element of order $12$. [example: Computing an Element Order] In $S_8$, let \begin{align*} \sigma=(1\ 5\ 3)(2\ 8\ 4\ 6). \end{align*} The two displayed cycles are disjoint: the first moves $\{1,5,3\}$ and the second moves $\{2,8,4,6\}$. The only letter of $\{1,\dots,8\}$ not appearing is $7$, so $7$ is fixed and contributes one cycle of length $1$. By *Order from Cycle Type*, the order of $\sigma$ is the least common multiple of the disjoint cycle lengths: \begin{align*} \operatorname{ord}(\sigma)=\operatorname{lcm}(3,4,1). \end{align*} Since $3$ is prime and $4=2^2$, the least common multiple must contain the prime factors $3$ and $2^2$, so \begin{align*} \operatorname{lcm}(3,4,1)=2^2\cdot 3=12. \end{align*} Equivalently, the $3$-cycle returns to its starting positions exactly at powers divisible by $3$, the $4$-cycle returns exactly at powers divisible by $4$, and the fixed point returns at every power. The first positive integer divisible by both $3$ and $4$ is $12$, so \begin{align*} \operatorname{ord}(\sigma)=12. \end{align*} Thus the two independent loops force the whole permutation to return only when the $3$-step and $4$-step motions have both completed whole numbers of turns. [/example] ## Transpositions, Generators, and Sign Cycles describe what a permutation is, but in calculations we often want elementary moves. Swapping two letters is the smallest nonidentity rearrangement, and a long cycle can be built by repeatedly swapping one chosen letter with the others. [definition: Transposition] A transposition in $S_n$ is a cycle of length $2$. [/definition] Transpositions are simple enough to track by hand, but they are still powerful enough to express every possible relabelling. This gives a generating set for $S_n$ that is far smaller than the full group. [quotetheorem:8540] The theorem gives many expressions for the same permutation, so it raises a well-definedness problem. To turn swap-counting into an invariant, we need a distinction that survives all possible transposition expressions. The invariant that survives is parity. [definition: Even and Odd Permutations] A permutation $\sigma \in S_n$ is even if it can be written as a product of an even number of transpositions. It is odd if it can be written as a product of an odd number of transpositions. [/definition] ### Well-Defined Parity This definition only makes sense if the choice of transposition factorization cannot change the parity. Otherwise the same permutation could be both even and odd, and the words would describe a particular calculation rather than an intrinsic feature of the permutation. The obstruction to remove is therefore the possibility of two factorizations with opposite parity. [quotetheorem:7876] ### The Sign Homomorphism Once parity is well-defined, we need notation that records it without carrying around a chosen transposition expression. Encoding even permutations by $1$ and odd permutations by $-1$ turns the parity distinction into a value that can later be multiplied and used to identify kernels. [definition: Sign Homomorphism] The [sign homomorphism](/theorems/778) is the map $\operatorname{sgn}: S_n \to \{1,-1\}$ that sends even permutations to $1$ and odd permutations to $-1$. [/definition] The codomain $\{1,-1\}$ is viewed as a group under multiplication. A parity label would be much less useful if composing two permutations destroyed the relationship between their signs. The point to check is that sign respects the group operation, so that parity can be used in the same way as other homomorphism invariants. This is the step where parity must stop being just a classification and become a structural invariant. To use sign for subgroup structure, quotient arguments, and later kernel computations, we need more than a definition: we need compatibility with composition in $S_n$. The next result supplies that compatibility by showing that the sign of a product is the product of the signs, so the even permutations form the kernel of a genuine [group homomorphism](/page/Group%20Homomorphism) rather than an accidental collection. [quotetheorem:778] ### The Alternating Group The sign homomorphism creates a new object: its kernel. We need a name for the permutations that become invisible to sign, because kernels are the source of normal subgroups and quotient groups. That named kernel is the alternating group. [definition: Alternating Group] Let $n \in \mathbb{N}$. The alternating group $A_n$ is \begin{align*} A_n := \{\sigma \in S_n : \operatorname{sgn}(\sigma)=1\}. \end{align*} [/definition] The alternating group is the first major subgroup sitting inside $S_n$. It is normal because it is the kernel of a homomorphism, and it has half the elements of $S_n$ when $n \ge 2$. [example: Signs from Cycle Lengths] A $k$-cycle factors into $k-1$ transpositions as \begin{align*} (a_1\ a_2\ \cdots\ a_k)=(a_1\ a_k)(a_1\ a_{k-1})\cdots(a_1\ a_2). \end{align*} With products read right-to-left, the right-hand side sends $a_1$ to $a_2$, because $(a_1\ a_2)$ sends $a_1$ to $a_2$ and every later transposition in the product fixes $a_2$. If $2 \le j < k$, then the factors to the right of $(a_1\ a_j)$ fix $a_j$, the transposition $(a_1\ a_j)$ sends $a_j$ to $a_1$, and the next factor $(a_1\ a_{j+1})$ sends $a_1$ to $a_{j+1}$; all remaining factors fix $a_{j+1}$. Finally, $a_k$ is fixed by every factor except the leftmost one, and $(a_1\ a_k)$ sends $a_k$ to $a_1$. Every letter outside $\{a_1,\dots,a_k\}$ is fixed by each transposition, so the product has exactly the same action as $(a_1\ a_2\ \cdots\ a_k)$. Each transposition is odd, so it has sign $-1$. Using *Sign is a Homomorphism*, the sign of the displayed product is the product of the signs of its $k-1$ transposition factors: \begin{align*} \operatorname{sgn}(a_1\ a_2\ \cdots\ a_k)=(-1)^{k-1}. \end{align*} For \begin{align*} \sigma=(1\ 4\ 6)(2\ 5) \in S_6, \end{align*} the $3$-cycle contributes \begin{align*} \operatorname{sgn}(1\ 4\ 6)=(-1)^{3-1}=(-1)^2=1, \end{align*} and the transposition contributes \begin{align*} \operatorname{sgn}(2\ 5)=(-1)^{2-1}=(-1)^1=-1. \end{align*} Applying *Sign is a Homomorphism* to the product gives \begin{align*} \operatorname{sgn}(\sigma)=\operatorname{sgn}(1\ 4\ 6)\operatorname{sgn}(2\ 5)=1\cdot(-1)=-1. \end{align*} Thus this permutation is odd: the even $3$-cycle and the odd transposition combine to give an odd permutation. [/example] ## Conjugacy and Relabelling A central idea in group theory is that conjugation means changing coordinates inside the group. In $S_n$, this interpretation is literal: conjugating a permutation by another permutation just renames the letters. To study this relabelling process, we first name the relation it produces. [definition: Conjugacy in $S_n$] Let $\sigma,\tau \in S_n$. The conjugate of $\sigma$ by $\tau$ is \begin{align*} \tau\sigma\tau^{-1}. \end{align*} Two permutations $\sigma,\rho \in S_n$ are conjugate in $S_n$ if there exists $\tau \in S_n$ such that \begin{align*} \rho=\tau\sigma\tau^{-1}. \end{align*} [/definition] Conjugation preserves the shape of the cycle decomposition but changes the names appearing in it. The next formula makes that slogan exact on a single cycle, and then disjoint decompositions extend the statement to all permutations. [quotetheorem:8541] The relabelling formula suggests a complete classification question: when do two permutations differ only by a change of names? The answer is that the names do not matter, but the cycle lengths do. [quotetheorem:798] This theorem turns the classification of conjugacy classes into the combinatorics of partitions of $n$. It also shows why symmetric groups are a meeting point between algebra and enumeration. [example: Conjugacy Classes in $S_4$] The partitions of $4$ are \begin{align*} 4, \quad 3+1, \quad 2+2, \quad 2+1+1, \quad 1+1+1+1. \end{align*} By *Conjugacy Classes in $S_n$*, conjugacy classes in $S_4$ are determined exactly by cycle type. The five partitions therefore give five classes: \begin{align*} 4 \leftrightarrow \text{$4$-cycles}. \end{align*} \begin{align*} 3+1 \leftrightarrow \text{$3$-cycles with one fixed point}. \end{align*} \begin{align*} 2+2 \leftrightarrow \text{products of two disjoint transpositions}. \end{align*} \begin{align*} 2+1+1 \leftrightarrow \text{single transpositions with two fixed points}. \end{align*} \begin{align*} 1+1+1+1 \leftrightarrow \text{the identity}. \end{align*} For example, $(1\ 2\ 3)$ sends $1 \mapsto 2$, $2 \mapsto 3$, $3 \mapsto 1$, and fixes $4$, so its cycle type is $3+1$. Hence its [conjugacy class](/page/Conjugacy%20Class) is precisely the class attached to the partition $3+1$, namely all $3$-cycles in $S_4$. [/example] ## Actions and Cayley's Theorem The symmetric group is not only an example of a group; it is the ambient group for every action on a finite set. Whenever a group acts on $n$ objects, each group element determines a permutation of those objects. This gives a homomorphism into a symmetric group. For an arbitrary set $X$, we write $S_X$ for its symmetric group; the notation $\operatorname{Sym}(X)$ means the same group of all bijections $X\to X$, and will appear in the formulation of [Cayley's theorem](/theorems/795) below. [definition: Symmetric Group on a Set] Let $X$ be a set. The symmetric group on $X$ is \begin{align*} S_X := \{\sigma: X \to X : \sigma \text{ is a bijection}\}, \end{align*} equipped with composition of functions. [/definition] For a set that already has meaningful labels, $S_X$ avoids replacing those labels by $1,\dots,n$ too early. The reason for introducing $S_X$ is that any action on $X$ turns group elements into actual members of this symmetric group. We need a map that records this action as a homomorphism into $S_X$. [definition: Permutation Representation] Let $G$ be a group acting on the left on a finite set $X$. A permutation representation associated to the action is the homomorphism $\varphi: G \to S_X$ such that $\varphi(g)(x)=g \cdot x$ for every $g \in G$ and $x \in X$. [/definition] A permutation representation can lose information if different elements of $G$ act in the same way. The central universality question is whether every finite group has at least one action that loses no information. Acting on itself by left multiplication answers yes. [quotetheorem:795] [Cayley's theorem](/theorems/846) changes the status of symmetric groups. They are not merely a family of examples; they are large enough to contain every finite group, up to isomorphism. Many abstract group questions can therefore be translated into questions about subgroups of symmetric groups. [example: A Cyclic Group Inside $S_4$] Let $G=C_4=\langle g:g^4=e\rangle$, so the elements of $G$ are $e,g,g^2,g^3$. Consider the left multiplication action of $G$ on itself. The element $g$ acts by the function $L_g:G\to G$ defined by $L_g(x)=gx$. We compute its values on all four elements: \begin{align*} L_g(e)=ge=g. \end{align*} \begin{align*} L_g(g)=gg=g^2. \end{align*} \begin{align*} L_g(g^2)=gg^2=g^3. \end{align*} \begin{align*} L_g(g^3)=gg^3=g^4=e. \end{align*} Thus $L_g$ sends \begin{align*} e\mapsto g\mapsto g^2\mapsto g^3\mapsto e. \end{align*} Now label the four elements $e,g,g^2,g^3$ by $1,2,3,4$, respectively. Under this labelling, the same permutation sends \begin{align*} 1\mapsto 2,\quad 2\mapsto 3,\quad 3\mapsto 4,\quad 4\mapsto 1, \end{align*} so it is the cycle $(1\ 2\ 3\ 4)$ in $S_4$. The powers of this cycle are \begin{align*} (1\ 2\ 3\ 4)^0=e. \end{align*} \begin{align*} (1\ 2\ 3\ 4)^1=(1\ 2\ 3\ 4). \end{align*} \begin{align*} (1\ 2\ 3\ 4)^2=(1\ 3)(2\ 4). \end{align*} \begin{align*} (1\ 2\ 3\ 4)^3=(1\ 4\ 3\ 2). \end{align*} \begin{align*} (1\ 2\ 3\ 4)^4=e. \end{align*} Therefore the image of $C_4$ under this action is \begin{align*} \langle (1\ 2\ 3\ 4)\rangle=\{e,(1\ 2\ 3\ 4),(1\ 3)(2\ 4),(1\ 4\ 3\ 2)\}\le S_4. \end{align*} Thus the [cyclic group](/page/Cyclic%20Group) of order $4$ appears concretely as the subgroup generated by one four-step rotation of the labelled set. [/example] ## Small Symmetric Groups The first few symmetric groups show the range of behaviour. At $n=1$ and $n=2$, the groups are abelian. At $n=3$, non-commutativity begins. At $n=4$, normal subgroups and conjugacy structure become richer. Before looking at examples, we need the basic size calculation. [quotetheorem:8542] The factorial growth explains why structural notation is needed. Even $S_5$ has $120$ elements, too many to handle by a multiplication table in ordinary work. [example: The Group $S_3$] Every permutation in $S_3$ is determined by the ordered triple $(\sigma(1),\sigma(2),\sigma(3))$, and this triple must be a rearrangement of $1,2,3$. If $\sigma(1)=1$, then $(\sigma(2),\sigma(3))$ is either $(2,3)$ or $(3,2)$. If $\sigma(1)=2$, then $(\sigma(2),\sigma(3))$ is either $(1,3)$ or $(3,1)$. If $\sigma(1)=3$, then $(\sigma(2),\sigma(3))$ is either $(1,2)$ or $(2,1)$. Thus the six elements are \begin{align*} e, \quad (2\ 3), \quad (1\ 2), \quad (1\ 2\ 3), \quad (1\ 3\ 2), \quad (1\ 3). \end{align*} The identity is even because it is a product of zero transpositions. Each transposition is odd because it is already a product of one transposition: \begin{align*} (1\ 2)=(1\ 2), \quad (1\ 3)=(1\ 3), \quad (2\ 3)=(2\ 3). \end{align*} The two $3$-cycles are even because each is a product of two transpositions: \begin{align*} (1\ 2\ 3)=(1\ 3)(1\ 2). \end{align*} Indeed, the right-hand side sends $1\mapsto 2$, $2\mapsto 3$, and $3\mapsto 1$. Similarly, \begin{align*} (1\ 3\ 2)=(1\ 2)(1\ 3). \end{align*} The right-hand side sends $1\mapsto 3$, $3\mapsto 2$, and $2\mapsto 1$. Therefore \begin{align*} A_3=\{e,(1\ 2\ 3),(1\ 3\ 2)\}. \end{align*} Non-commutativity is already visible in $S_3$. With products read right-to-left, \begin{align*} (1\ 2)(2\ 3)=(1\ 2\ 3), \end{align*} since $1\mapsto 2$, $2\mapsto 3$, and $3\mapsto 1$. In the other order, \begin{align*} (2\ 3)(1\ 2)=(1\ 3\ 2), \end{align*} since $1\mapsto 3$, $3\mapsto 2$, and $2\mapsto 1$. These two products are different, so $S_3$ is the smallest symmetric group where sign, cycle structure, and non-commutative multiplication all appear together. [/example] To understand $S_4$, it is not enough to list elements; we need to find subsets that remain stable under the whole group operation. The most important new example is built from the identity and the products of two disjoint transpositions, a four-element pattern that deserves its own name. [definition: Klein Four Subgroup of $S_4$] The Klein four subgroup of $S_4$ is \begin{align*} V_4 := \{e,(1\ 2)(3\ 4),(1\ 3)(2\ 4),(1\ 4)(2\ 3)\}. \end{align*} [/definition] After naming $V_4$, the structural question is whether it is merely closed under its own multiplication or whether it remains stable when the whole group $S_4$ relabels it by conjugation. A subgroup can fail this stronger test if conjugating one of its elements produces an element outside the subgroup, so normality is the extra obstruction to check for $V_4$. [quotetheorem:8543] The [normal subgroup](/page/Normal%20Subgroup) $V_4$ is a useful warning against guessing subgroup structure from size alone. Symmetric groups have many subgroups, but their normal subgroups are much more constrained, especially when $n$ grows. ## Beyond and Connected Topics The symmetric group is the gateway from elementary group axioms to concrete finite group theory. In [Cambridge IA Groups](/page/Cambridge%20IA%20Groups), group actions, homomorphisms, quotient groups, and conjugacy provide the general language behind the computations on this page. The point is not only that $S_n$ is a useful example, but that actions on finite sets turn many abstract questions into permutation questions. The next algebraic layer is the alternating group $A_n$, where the sign homomorphism becomes a structural dividing line. For $n \ge 5$, the simplicity of $A_n$ is one of the central theorems of finite group theory and is a major reason $S_n$ appears in Galois theory. These ideas sit naturally alongside the group theory and module material in [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules), where normal subgroups, quotient groups, and actions become systematic tools rather than isolated calculations. Linear algebra gives another continuation through permutation matrices. Each $\sigma \in S_n$ acts on $\mathbb{R}^n$ by permuting the standard basis vectors. This embeds $S_n$ into $GL(n,\mathbb{R})$ and turns questions about permutations into questions about matrices, determinants, eigenvalues, and representations; see [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra) for the surrounding linear algebra. Commutative algebra and Galois theory use symmetric groups to measure how roots of polynomials can be permuted while preserving algebraic relations. The full symmetric group $S_n$ often represents the largest possible [Galois group](/page/Galois%20Group) of a degree $n$ polynomial, while smaller subgroups encode special algebraic structure. This is one reason symmetric groups remain visible in [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). ## References Androma, [Cambridge IA Groups](/page/Cambridge%20IA%20Groups). Androma, [Cambridge IB Groups, Rings and Modules](/page/Cambridge%20IB%20Groups%2C%20Rings%20and%20Modules). Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Cambridge III Commutative Algebra](/page/Cambridge%20III%20Commutative%20Algebra). Dummit and Foote, *Abstract Algebra* (2004). Artin, *Algebra* (2011). Rotman, *An Introduction to the Theory of Groups* (1995).