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The alternating group is the subgroup of the symmetric group consisting of even permutations:For a permutation cycle of length $l$,using rightmost-first composition. Its sign of a permutation is therefore $(-1{)}^{l-1}$. Multiplicativity of the sign homomorphism givesThusEquivalently, $n-t$ is even when $t$ counts all cycles, including fixed points. Odd-length cycles contribute positive sign and impose no further restriction.

Write $D_n=\det [K(x_i,x_j){]}_{i,j=1}^n$ and expand it by permutations. The assumptions ensure that the indicated single-variable integrals exist; the finite-rank projection kernel constructed above satisfies them with $r=m$. No symmetry of a general kernel is needed for the following argument.

If a permutation fixes $n$, its product contains the separate factor $K(x_n,x_n)$. Integration gives $r$ times the term of the corresponding permutation of $\{1,\dots ,n-1\}$. These terms contribute $r{D}_{n-1}$.

Otherwise $n$ belongs to a longer permutation cycle. Let $i$ precede $n$ and let $j$ follow it. The only factors containing $x_n$ are $K(x_i,x_n)K(x_n,x_j)$, whose integral is $K(x_i,x_j)$ by the reproducing assumption. Delete $n$ from the cycle to obtain a permutation $\tau$ on $n-1$ letters. Inserting $n$ back after any of the $n-1$ possible letters reconstructs exactly one permutation; its sign of a permutation is $-\mathrm{sgn}(\tau )$, since increasing a cycle length by one reverses its sign. Thus the nonfixed terms contribute $-(n-1)D_{n-1}$.

Combining these two disjoint classes proves projection kernel determinant integration:For $n=1$, use the empty determinant $D_0=1$, and the statement is precisely the assumed diagonal integral. For our kernel, orthonormality directly yieldsso every step of the marginal integration is justified.

Take $G=S_3$ and its normal subgroup $N=A_3=⟨r⟩$, where $r=(123)$. Normality follows either from the sign of a permutation or directly from conjugation sending $r$ to $r$ or $r^{-1}$. The quotient group has order two, and $t=(12)$ provides a section $i:C_2\to S_3$ because $t^2=1$ and $tN$ is its nonidentity coset.

However, the map $\Phi (q,n)=i(q)n$ from the direct product of groups is not a group homomorphism. In its domain $(1,r)$ commutes with $(tN,1)$, whereas their images $r$ and $t$ do not commute: $tr{t}^{-1}=r^{-1}\ne r$. This is a split group extension giving a genuine semidirect product instead of a direct product. The pair $S_3,A_3$ supplies the required split but non-direct example.

For the general bijectivity assertion, let $\pi :G\to G/N$ be the quotient projection and assume $\pi \circ i$ is the identity. Given $g\in G$, put $q=\pi (g)$ and $n=i(q{)}^{-1}g$. Then $\pi (n)=q^{-1}q=1$, so $n\in N$ and $g=i(q)n$. This proves surjectivity of $\Phi$. If $i(q)n=i(q^{\prime })n^{\prime }$, applying $\pi$ gives $q=q^{\prime }$, after which cancellation gives $n=n^{\prime }$. ThusThis section normal form for a split group extension does not require finiteness. The obstruction to its being a homomorphism is the nontrivial conjugation action of the section on $N$.

True. The sign of a permutation is a group homomorphism $\mathrm{sgn}:S_3\to \{1,-1\}$, as follows, for example, from the multiplicative determinant of permutation matrices. Its kernel of a group homomorphism is the alternating groupA transposition maps to $-1$, so this homomorphism is onto.

Every permutation is a product of transpositions, since a cycle can be writtenwith rightmost factors acting first. Define the sign of a permutation to be $(-1{)}^r$ for a product of $r$ transpositions. To prove that it is well defined without assuming parity, let $P_{\sigma }$ be the permutation matrix specified by $P_{\sigma }{e}_i=e_{\sigma (i)}$. Then $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$, and a transposition exchanges two columns of the identity, so its determinant is $-1$. Any decomposition into $r$ transpositions therefore hasThe left-hand side depends only on $\sigma$, proving well-definedness. Multiplicativity of the determinant now gives the sign homomorphism:For the first requested group embedding, use the faithful four-point action of GL2 over F2: label the four vectors in ${\mathbb{F}}_2^2$ and let an invertible $2\times 2$ matrix act on them by multiplication. Composition of these actions gives a group homomorphism $\psi :G{L}_2({\mathbb{F}}_2)\to S_4$. If a matrix fixes every vector, it fixes both standard basis vectors and is the identity, so the action is faithful. The matrixfixes $(0,0)$ and $(1,1)$ and exchanges $(1,0)$ with $(0,1)$. Its image is a single transposition, hence $\mathrm{sgn}(\psi (B))=-1$. The composite sign map is nontrivial.

For the second group embedding, simply takeThe preceding multiplication formula proves the group homomorphism property, and $P_{\sigma }=I$ forces $\sigma (i)=i$ for every $i$, proving injectivity.

A permutation representation is a group homomorphism $\rho :G\to \mathrm{Sym}(\Omega )$, equivalently a group action on a set $\Omega$. The left regular action takes $\Omega =G$ and $\rho (g)(h)=gh$. Each such map is a bijection, and $\rho (g_1{g}_2)=\rho (g_1)\rho (g_2)$. If $\rho (g)$ is the identity, evaluating at the identity element gives $g=1$. Thus the representation is faithful. Numbering the elements of $G$ identifies $\mathrm{Sym}(G)$ with the symmetric group $S_n$, proving Cayley theorem:For a nonabelian simple group, compose this faithful representation with the sign of a permutation homomorphism $S_n\to \{1,-1\}$. Its kernel is a normal subgroup of $G$, hence either trivial or all of $G$. A trivial kernel would embed $G$ in a group of order two, contradicting noncommutativity. Therefore every represented element is even, and the same representation embeds $G$ in the alternating group $A_n$.

Finally let $S_3$ act on the two cosets of its normal subgroup $A_3$. An even permutation fixes both cosets, whereas an odd permutation exchanges them. This gives $S_3\to S_2$ with kernel $A_3$; explicitly, send even permutations to the identity and odd permutations to $(12)$.

A finite nonabelian simple group cannot be a p-group: a nontrivial finite p-group has nontrivial centre of a group, since its class equation makes every noncentral conjugacy class size divisible by $p$, and hence makes the centre size a positive multiple of $p$. That centre is a normal subgroup, so must be the whole group if $G$ is a simple group, making it abelian; an abelian simple group has prime order. Therefore every Sylow subgroup here is nontrivial and proper. It cannot be normal, so $n_p>1$.

To obtain a simple group embedding from Sylow conjugation, the conjugation action on Sylow subgroups gives a group homomorphism $\rho :G\to S_{n_p}$. This group action is transitive by the Sylow theorems and is nontrivial since $n_p>1$. Its kernel of a group homomorphism is a normal subgroup, so the defining property of a simple group makes the kernel of a group homomorphism trivial: $\rho$ is an embedding. Now compose with the sign of a permutation $S_{n_p}\to \{\pm 1\}$. A nontrivial composite would again be injective because $G$ is a simple group, embedding $G$ in a group of order two, impossible for a nonabelian group. Thus $\rho (G)$ lies in the alternating group $A_{n_p}$. By Lagrange's theorem,The argument only invokes the factorial formula once $n_p\ge 2$; the impossible case $n_p=2$ itself is also excluded by the embedding.

Star the adjacent regions $o,e$. Choose the Kauffman state of a knot diagram putting the marker at $A$ in region $d$, at $B$ in region $a$, at $C$ in region $b$, and at $D$ in region $c$. Every unstarred region is used exactly once, and every crossing has exactly one marker.

In the full Alexander matrix, circle the entries $(A,d),(B,a),(C,b),(D,c)$, namelyIn the reduced Alexander matrix $B$, these are the diagonal entries in column order $(d,a,b,c)$. The associated permutation is the identity, with sign of a permutation $+1$, so this Kauffman state of a knot diagram contributesto $\det B=-t^2+3t-1$. The other four nonzero permutation terms are $t,t,t,-t^2$. The drawing circles the same four entries literally; the boxes above identify them in the typeset matrix. Different row or generator conventions multiply the Alexander polynomial of a knot and its state weights by the corresponding overall unit.

Interchanging two rows reverses the sign of an alternating multilinear map, soIn the Leibniz formula for determinants, the same result follows by composing every permutation with the corresponding transposition, which reverses its sign.

Expand the determinant of the matrix $A_1$ from part (b) by the Leibniz formula for determinants. A matrix entry is a signed sum of monomials arising from the possible corners at its crossing. Choosing one summand in every row chooses one corner at every crossing, while choosing distinct columns puts exactly one chosen corner in every region $R_i$ with $i>1$ and none in the two deleted regions $R_0,R_1$. The surviving determinant terms are therefore in bijection with the Kauffman states $s\in \mathcal{S}(D)$.

The sign of a permutation in the determinant together with the corner signs gives $(-1{)}^{ϵ(s)}$, and multiplying the corner monomials gives $t^{\delta (s)}$. Since part (b) identifies this determinant with the Alexander polynomial of a knot up to a unit,

Let $s_i=(ii+1)$. Since $\mu -\mathrm{id}=s_i(\lambda -\mathrm{id})$, reindex the alternating sum defining ${\psi }^{\mu }$ by $\pi \mapsto s_i\pi$. This preserves all permutation-character terms and reverses every sign of a permutation, because $\mathrm{sgn}(s_i\pi )=-\mathrm{sgn}(\pi )$. Hence ${\psi }^{\mu }=-{\psi }^{\lambda }$.

For $\sigma \in S_n$, let $P_{\sigma }$ be its permutation matrix and define the sign of a permutation bySince $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$,so this is a homomorphism. A transposition interchanges two columns of the identity matrix and therefore has determinant $-1$. If $\sigma$ is expressed as a product of $r$ transpositions, then $\mathrm{sgn}(\sigma )=(-1{)}^r$; because the determinant depends only on $\sigma$, the parity of $r$ is independent of the chosen expression.

A group homomorphism $h:G\to F$ preserves multiplication: $h(g_1{g}_2)=h(g_1)h(g_2)$. It sends the identity element to the identity and inverse elements to inverse elements. Its group kernel is $K=\{g\in G:h(g)=e_F\}$. This contains $e_G$ and is closed under multiplication and inverse elements, so it is a subgroup. For $k\in K$ and $g\in G$,Thus $gK{g}^{-1}\subseteq K$; applying the same argument with $g^{-1}$ gives equality. Hence the group kernel is a normal subgroup.

Let $S_2=\{e,\tau \}$ with $\tau =(12)$. The parity of a permutation defines the desired map from $S_3$:The sign of a permutation is multiplicative, so parity adds modulo two under composition, exactly matching ${\tau }^2=e$. Thus this is a group homomorphism; it is surjective because a transposition maps to $\tau$. Its group kernel isThis is the alternating group on three objects, a normal subgroup of index two in $S_3$.

A normal subgroup $H$ satisfies $gH{g}^{-1}=H$ for every $g\in G$. A group homomorphism preserves multiplication, $\Phi (ab)=\Phi (a)\Phi (b)$. It follows that $\Phi (e_1)=e_2$ and $\Phi (g^{-1})=\Phi (g{)}^{-1}$.

The kernel of a group homomorphism $K=\{g:\Phi (g)=e_2\}$ contains the identity and satisfies $a{b}^{-1}\in K$ for $a,b\in K$, so is a subgroup. For $k\in K$, $\Phi (gk{g}^{-1})=\Phi (g)e_2\Phi (g{)}^{-1}=e_2$. Thus $gK{g}^{-1}\subseteq K$, and applying the same inclusion to $g^{-1}$ gives equality. Hence the kernel is a normal subgroup.

The product $f={\prod }_{i<j}(x_i-x_j)$ is a Vandermonde determinant up to a fixed overall sign. A transposition exchanging $x_p,x_q$ reverses their mutual factor. Every other factor involving one of these variables is paired with the corresponding factor involving the other; those paired products are unchanged. ThereforeIf a permutation has decompositions into $r$ and $s$ transpositions, they act as $(-1{)}^{r}f$ and $(-1{)}^{s}f$. This polynomial is nonzero, for example at pairwise distinct real inputs, so the signs must agree. Thus all decompositions have the same parity, and an even decomposition forces every decomposition to be even. This defines the sign of a permutation; its kernel is the alternating group, linking the parity argument to the normal-kernel result above.

The kernel is a normal subgroup, so it contains the identity class and is a union of the classes listed in part (c). Its candidate orders are therefore $1$, $1+3=4$, $1+2=3$ and $1+3+2=6$. Lagrange's theorem excludes four, since it does not divide six. The remaining possibilities areAll three are normal: the first and last trivially, while $A_3$ is the kernel of the map giving the sign of a permutation, or the index-two subgroup generated by a 3-cycle.

Because $\Phi$ is onto, the first isomorphism theorem gives $G^{\prime }\cong S_3/K$. Explicitly, the map $gK\mapsto \Phi (g)$ is well defined, multiplicative and surjective; equality of its values means $h^{-1}g\in K$, so it is also injective. ThusEach is realized by the identity map, the sign map and the trivial map, respectively. This completes the classification of normal subgroups and quotients of the symmetric group on three letters.

The two one-dimensional representations are the trivial representation, ${\rho }_{+}(g)=1$, and the sign representation, ${\rho }_{-}(g)=\mathrm{sgn}(g)$. The first respects multiplication immediately. For the second, the sign of a permutation is the determinant of its permutation matrix, and multiplicativity of the determinant proves ${\rho }_{-}(gh)={\rho }_{-}(g){\rho }_{-}(h)$. They are inequivalent since a transposition has images $1$ and $-1$, and conjugation by a nonzero scalar cannot change a one-dimensional matrix.

Their group homomorphism kernels areBoth are subgroups, indeed normal subgroups, because they are kernels. In the second case $A_4$ is the alternating group of even permutations, with twelve elements.

The characters are their scalar values, since the trace of a one-dimensional matrix is its only entry. In the class order from part (b),The character inner product is the group average of ${\chi }_{+}\overline{{\chi }_{-}}$, soEach has squared norm $(1+6+3+8+6)/24=1$, in agreement with character orthogonality for distinct irreducible representations.

For $i<j$, interchange the entries in slots $i,j$ while preserving the order of all intermediate fermionic creation operators. Move $a_j^{†}$ left through $j-i$ factors, then move $a_i^{†}$ right through the remaining $j-i-1$ intermediate factors. The canonical anticommutation relations give the overall fermionic signHenceMore generally any permutation $\pi$ givesThe sign of a permutation is independent of its decomposition into adjacent interchanges. Thus the $n$-particle sector is an exterior power of the one-particle state space, the defining antisymmetry of fermionic Fock space. Repeated complete mode labels make the state zero, so they satisfy the same identity trivially.

The permanent of a matrix adds the products obtained by choosing one entry in each row and one in each column. For an $n\times n$ matrix $A=(a_{ij})$ with entries in the real numbers or complex numbers,Here each permutation $\sigma$ chooses column $\sigma (i)$ in row $i$, and the symmetric group $S_n$ lists every such bijection. The formula resembles the Leibniz formula for determinants, but omits the sign of a permutation. For a two-by-two matrix,The absence of alternating signs makes the permanent of a matrix particularly useful for counting allowed assignments.

For a zero-one matrix, let rows represent people, columns represent jobs, and $a_{ij}=1$ mean that person $i$ can do job $j$. This is the biadjacency matrix of a bipartite graph. A product in the permanent is one exactly when all the selected assignments are allowed; otherwise it is zero. Each surviving permutation assigns every person a different job, which is a perfect matching. Thus the permanent counts perfect matchings. This interpretation and the definition are given in Yufei Zhao, Probabilistic Methods in Combinatorics, §10.2, pp. 178–179.

For a fully enumerated example, takeAmong the six permutations of three columns, exactly three give nonzero products: $(1,2,3)$, $(1,3,2)$ and $(2,1,3)$. These tuples list the chosen column in each successive row. Each product is one, so $\mathrm{perm}A=3$. Their permutation signs are $+1,-1,-1$, giving $\det A=-1$. The diagram displays all three allowed assignments.

A zero determinant need not mean that no assignment exists. The two-by-two all-ones matrix has permanent $2$ and determinant $0$: its two assignments cancel in the determinant. The nonzero determinant criterion for a perfect matching is sufficient for existence, but the converse fails for a fixed zero-one matrix.

For nonnegative entries, the permanent instead sums assignment weights, with each assignment weighted by the product of its selected entries. It is then a total weight rather than necessarily an integer count. With arbitrary entries in the real numbers or complex numbers, even the permanent can have cancellation; the counting interpretation specifically uses zero-one entries. The permanent as probability of distinct independent choices gives another useful nonnegative interpretation.

There are $n!$ terms in the defining sum, linking direct evaluation to the factorial. For the all-ones $n\times n$ matrix, every term is one and the permanent is exactly $n!$; the identity matrix has permanent one. Finding one allowed assignment and counting all allowed assignments are different questions. Their relationship to computational complexity theory was studied in L. G. Valiant, The complexity of computing the permanent, Theoretical Computer Science 8(2), 189–201, 1979. The elementary sum is useful for small examples, while the existing dense permanent approximation describes a more specialized way of estimating large counts.

Conjugation preserves order of a group element, because $(gx{g}^{-1}{)}^k=g{x}^k{g}^{-1}$, and preserves the sign of a permutation, because sign is a group homomorphism into $\{1,-1\}$. These invariants do not determine cycle type. In $S_6$, $(123)$ and $(123)(456)$ both have order three and positive sign, but have three and zero fixed points respectively. Conjugation relabels cycles, so these permutations are not conjugate.

For a group $G$ and a normal subgroup $N$, the quotient group $G/N$ is the set of cosets $gN$, with operationIts identity element is the coset $N$, and the inverse element of $gN$ is $g^{-1}N$. A quotient group treats all elements within one coset as the same input while retaining a consistent operation on the classes. The normal subgroup condition is what makes this operation independent of which representatives are chosen. ^[ref].

Take the additive group $G=\mathbb{Z}/6\mathbb{Z}$ and $N=\{0,3\}$. Since $G$ is an abelian group, $N$ is a normal subgroup. There are three elements in the quotient group:Adding two copies of $2+N$ gives $4+N=1+N$; adding $1+N$ and $2+N$ gives $3+N=N$. Hence the quotient group has the addition table of $\mathbb{Z}/3\mathbb{Z}$. It does not select one representative and discard the others: its elements are the entire cosets.

This is useful whenever distinctions within a normal subgroup should be ignored. For example, the sign of a permutation retains only even or odd parity of a permutation in $S_3$. Its group kernel is the alternating group $A_3$, and the quotient group $S_3/A_3$ has two elements. The first isomorphism theorem explains why a group homomorphism's image is a quotient group of its domain.