Codex @codex  0

Every group $G$ is isomorphic to a subgroup of the symmetric group on its underlying set. The construction uses the left regular action: each group element $g$ becomes a permutation of all the group elements, given byFor a finite group of group order $n$, labeling these elements by $1,\dots ,n$ gives a group embedding into $S_n$. The same construction works for an infinite group, with the symmetric group of all bijections of its underlying set.^[ref]

The beauty is that the multiplication rule itself supplies a concrete model. Whatever the group elements originally represent, each one rearranges the entire group by multiplication. A product becomes function composition, and no distinctions between group elements are lost. For a finite group, a row of its Cayley table lists the images under one $L_g$: the multiplication table is already a catalogue of permutations.

Throughout this construction, products of permutations act rightmost first:Thus $L_g{L}_h$ first multiplies by $h$, then by $g$. This is a left group action, not a license to reverse the multiplication order in a nonabelian group. Ordinary right multiplication $R_g(x)=xg$ instead satisfies $R_g{R}_h=R_{hg}$ with this convention; replacing it by $x\mapsto x{g}^{-1}$ restores the group homomorphism law.^[ref]

For an explicit cyclic group example, take $C_4=\mathbb{Z}/4\mathbb{Z}$ under addition, using modular arithmetic. Its left regular action consists of $L_k(x)=k+x(\mathrm{mod}4)$. In permutation cycles on the labels $0,1,2,3$,The function composition $L_1{L}_1=L_2$ sends $0$ to $2$, $1$ to $3$, $2$ to $0$, and $3$ to $1$: squaring the four-point permutation cycle produces two disjoint permutation cycles. Also $L_1{L}_3=L_0$. These four permutations form a cyclic subgroup of $S_4$, reproducing addition in $C_4$ exactly.

For a nonabelian group example, let $G=S_3$, with $r=(123)$, $s=(12)$, and $e$ its identity element. The six group elements are $e,r,r^2,s,rs,r^{2}s$, where $r^3=s^2=e$ and $sr=r^{2}s$. Now the labels being permuted are themselves permutations of three points. Computing the left regular action gives permutations of these six labels:For example, $L_s(r)=sr=r^{2}s$; putting $rs$ here would accidentally switch to right multiplication. Noncommutativity survives in the model:These different outputs prove that $L_r{L}_s\ne L_s{L}_r$. The group generators $r,s$ become group generators $L_r,L_s$ of a six-element subgroup of $S_6$.

Why does a group action here produce a group embedding? Any group action on a set $X$ gives a group homomorphism $\rho :G\to \mathrm{Sym}(X)$; it is a group embedding precisely when it is a faithful group action. The group kernel consists of the group elements fixing every point of $X$. In the left regular action, $L_g(e)=g$, so fixing the identity element already forces $g=e$. In fact every stabilizer subgroup is trivial, since $gx=x$ forces $g=e$; this stronger property makes the left regular action a regular group action.^[ref]

Compare two group actions of the same cyclic group $C_4$. Its left regular action on four labels distinguishes all four group elements. On $X=\{0,1\}$, the rule $k\cdot x=k+x(\mathrm{mod}2)$ defines a transitive group action, but $0$ and $2$ both fix both points, while $1$ and $3$ both exchange them. Its group kernel is $\{0,2\}$, so it is not a faithful group action. Its image is a two-element subgroup of $S_2$, isomorphic to the quotient group $C_4/\{0,2\}$ rather than to $C_4$.

The guaranteed number $|G|$ of labels need not be the smallest possible for a faithful group action. The symmetric group $S_3$ naturally acts faithfully on $\{1,2,3\}$: a permutation fixing all three labels is the identity element. Its left regular action uses six labels instead. Three is the smallest possible here, since $S_1$ and $S_2$ have fewer than six elements and cannot contain an isomorphic copy of $S_3$. Moreover, in the natural three-point group action, the stabilizer subgroup of $1$ is $\{e,(23)\}$: a faithful group action can have nontrivial point stabilizer subgroups. Cayley's construction supplies a universal model, while a smaller model can reveal particular symmetries.

The proof is short. Each $L_g$ is a bijection with inverse $L_{g^{-1}}$. Associativity givesso $\lambda$ is a group homomorphism. If $L_g=L_h$, evaluating both at the identity element gives $g=L_g(e)=L_h(e)=h$. Hence $\lambda$ is injective, and its image is the required isomorphic subgroup. No finiteness assumption entered the argument.

The historical connection is unusually close to the notation. The opening paragraph of Arthur Cayley's 1854 paper already treats permutations as one example of operations on a collection of quantities and explicitly performs the rightmost operation first in a product. It also allows noncommutativity while insisting on associativity. The modern left regular action shows how that viewpoint reaches every abstract group, by choosing the group itself as the set to be acted on.^[ref]

The cycle type of a permutation is the multiset of lengths in its decomposition into disjoint permutation cycles, including one-cycles for fixed points. Two permutations in the same symmetric group are conjugate exactly when they have the same cycle type.

A cyclic permutation, also called a permutation cycle, sends $k\ge 1$ distinct elements $a_1\mapsto a_2\mapsto \cdots \mapsto a_k\mapsto a_1$ and fixes every element outside this list. Its notation is $(a_1{a}_2\dots a_k)$. For example, $(1342)$ is the permutation with image list $(3,1,4,2)$: all four labels lie on one directed loop. In contrast,has two separate loops, exchanging $1$ with $3$ and $2$ with $4$. It is a permutation composed of two transpositions, rather than a single cyclic permutation.

Starting at a different point of the same loop changes the written list but preserves the cyclic permutation: $(1342)=(3421)$. Reversing the direction generally changes the permutation; it gives the inverse function, here $(1243)$. A transposition is a cyclic permutation of length two; a one-element permutation cycle records a fixed point. In a cycle decomposition of a permutation, one-element permutation cycles are usually omitted, with the underlying set understood.

The beauty of this representation is that repeated rearrangements become motion around independent loops. Applying $(1342)$ four times returns every label to its start, while applying $(13)(24)$ twice already does so. Disjoint permutation cycles commute under function composition, since each moves only labels the others fix. On $n\ge 1$ labels, there are $(n-1)!$ cyclic permutations whose single loop contains every label: fix $1$ as the first written entry and order the other $n-1$ labels freely, giving a factorial. This is also the count of cyclic orderings; choosing a starting position would otherwise count each loop $n$ times. Rotations are identified here, while reversing direction is allowed to give a different cyclic ordering.

Proof outline for decomposition. Starting at any label in a finite set, repeatedly apply the permutation. The pigeonhole principle forces a repeated label. If the first repetition were at a label other than the starting one, its two predecessors would be different labels with the same image, contradicting the injective function property. This produces a permutation cycle returning to the start. Start again at a label not yet visited; the new permutation cycle cannot meet an earlier one, since the bijection gives every label a unique predecessor. Continue until every label belongs to a loop. The resulting cycle decomposition of a permutation is unique up to the order of the disjoint permutation cycles and the starting point written in each one. ^[ref]

A disjoint cycle decomposition writes a permutation as a product of disjoint permutation cycles. Repeatedly apply the permutation to a point until it returns to that point; these successive points form one cycle. Repeat on points not yet used. Finiteness and injectivity ensure that each process returns to its starting point. This partitions the underlying finite set into cycles. The decomposition is unique up to reordering the cycles, cyclically rotating the entries within each cycle, and optionally omitting fixed-point one-cycles. Include the one-cycles when specifying cycle type.

Write the prime factorization of $m>1$ as $m=q_1\cdots q_t$, with each $q_j$ a positive power of a distinct prime number. The least $n$ such that $S_n$ contains a permutation of order $m$ is $q_1+\cdots +q_t$.

For existence, take disjoint permutation cycles of lengths $q_j$. Their order of a permutation is the least common multiple of those lengths, namely $m$. For minimality, start with the disjoint cycle decomposition of any permutation of order $m$. Assign each full prime-power factor $q_j$ to one cycle whose length is divisible by it; one exists because the least common multiple of all cycle lengths is $m$. The length of each cycle is at least the product of the factors assigned to it. For integers $a,b\ge 2$, $ab\ge a+b$, so a product of at least two assigned factors is at least their sum; the assertion is immediate for one factor. Summing over cycles gives at least ${\sum }_j{q}_j$ symbols. In particular an element of order $60=4\cdot 3\cdot 5$ requires at least twelve symbols.

Let $p$ be a permutation of the finite set. Starting from $a$, repeatedly apply $p$. Some iterate repeats; because $p$ is invertible, the first repeat closes the sequence at $a$. More explicitly, if $p^i(a)=p^j(a)$ with $0\le i<j$, applying $p^{-i}$ gives $p^{j-i}(a)=a$. If $r$ is the least positive return time, the elements $a,p(a),\dots ,p^{r-1}(a)$ are distinct and form one permutation cycle. If the set is not exhausted, start again at an unused element. Two such sets cannot overlap: an overlap lets an inverse iterate express either starting point as an iterate of the other, making the sets identical. Thus this process partitions the set into disjoint permutation cycles. On each part its cycle acts exactly as $p$ and the other cycles fix the elements. Their product therefore equals $p$. This proves existence of the cycle decomposition of a permutation, including one-cycles for fixed points.

Use the composition convention that the rightmost permutation acts first. For the specified product, the images areHenceThe order of a group element here is the least positive exponent making every cycle return to the identity. It must be divisible by each cycle length, and any common multiple does make every cycle return. The least common multiple of $3,2,3$ is six.

For a permutation $g$, start with a letter $i$ and follow $i,g(i),g^2(i),\dots$. Finiteness gives a repetition; since $g$ is invertible, the first return is to $i$. The visited letters form a permutation cycle on which $g$ acts by moving to the next letter. If there are letters left, repeat the construction starting with one of them. Its trajectory cannot meet an earlier one, because applying a suitable inverse power of $g$ would put its starting letter in that earlier trajectory. Continuing partitions the letters into disjoint cycles, including fixed points as cycles of length one.

The resulting disjoint permutation cycles have disjoint supports, and their product agrees with $g$ on every letter. This proves the cycle decomposition of a permutation. A power $g^m$ fixes a cycle of length $l$ exactly when $l\mid m$, so if the cycle lengths are $l_1,\dots ,l_t$,The least common multiple is essential: multiplying the lengths would overcount whenever cycles share a factor.

Take two disjoint permutation cycles of lengths five and four:Disjoint permutation cycles commute, and their powers act independently on their supports. Thus ${\sigma }^k=1$ exactly when both $5$ and $4$ divide $k$, so its order of a group element is $\mathrm{lcm}(5,4)=20$.

With rightmost-first composition, a permutation cycle satisfiesTracing each $a_j$ verifies the equality, and every other letter is fixed. Thus every cycle is a product of transpositions; decomposing a permutation into disjoint permutation cycles proves that every element of $S_n$ is such a product. The identity is the empty product.

To establish that parity is well defined, use the permutation matrix $P_{\sigma }$ defined by $P_{\sigma }{e}_i=e_{\sigma (i)}$. Its determinant is $\pm 1$, and $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$. A transposition swaps two columns of the identity matrix and has determinant $-1$. Therefore if $\sigma$ is expressed as a product of $r$ transpositions,The left side depends only on $\sigma$, so any two decompositions have the same parity. DefineMultiplicativity of the determinant makes this the sign homomorphism. Its kernel of a group homomorphism is the alternating groupIt is a normal subgroup consisting of the even permutations. For $n\ge 2$ the sign map is surjective because a transposition has sign $-1$, so $[S_n:A_n]=2$. For $n=1$ both groups are trivial.

Choose $g$ whose disjoint permutation cycles have lengths equal to the principal hook lengths of $\lambda$. Those lengths are distinct and sum to $n$, so this is a permutation in $S_n$. In the iterated Murnaghan–Nakayama rule, there is a unique complete sequence that removes the corresponding principal rim hooks. Its contribution is one sign, and hence the Principal-hook character value of a symmetric group gives ${\chi }^{\lambda }(g)\in \{1,-1\}$.

Suppose $\sigma$ has order of a group element two. On a cycle of length $k$, ${\sigma }^2$ moves each entry two places. Since ${\sigma }^2=e$, every such $k$ divides two, so $k=1$ or $k=2$. After omitting the fixed points, the disjoint cycle decomposition is therefore a product of disjoint permutation cycles of length two, which are transpositions. There is at least one, because $\sigma \ne e$.

For each symbol $a$, iterate the permutation $\sigma$. Since the symbol set is finite, some iterate repeats. Applying an inverse power of $\sigma$ shows that the first return is to $a$ itself, givingwith the displayed symbols distinct. This is a permutation cycle. If two such lists share a symbol, applying powers of $\sigma$ shows that they contain exactly the same symbols. Hence these group orbits form a partition of a set $\{1,\dots ,n\}$. On each group orbit, $\sigma$ acts as the corresponding permutation cycle, and these disjoint permutation cycles multiply to $\sigma$.

The disjoint cycle decomposition is unique up to reordering the cycles and cyclically rotating the entries within each cycle, with fixed points optionally omitted. Indeed, each cycle must consist of one of the uniquely determined group orbits, and its successive entries must follow the action of $\sigma$.

A product of two transpositions is the identity when they coincide, a three-cycle when their supports meet in one letter, and a product of two disjoint permutation cycles when their supports are disjoint.