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For the symmetric group $S_3$, take the transpositions $x=(12)$ and $y=(23)$. Products of permutations apply the rightmost factor first. ThenThe last permutation sends $1\mapsto 3$, $3\mapsto 2$, and $2\mapsto 1$: two swaps fail to commute by an entire three-cycle.

Every group commutator has sign of a permutation $+1$, so $S_3^{\prime }\subseteq A_3$, the alternating group of even permutations. Conversely, the displayed group commutator generates $A_3=\{e,(123),(132)\}$. Therefore $S_3^{\prime }=A_3$. The two cosets are the even permutations and the odd permutations, and their multiplication is just multiplication of signs:This is the first isomorphism theorem applied to the sign of a permutation. Once commutativity is forced, all three transpositions become indistinguishable, and both three-cycles become the identity element. The surviving information is exactly whether the permutation is even or odd. Compare the very different center of a group $Z(S_3)=\{e\}$.

Let $G=S_3$, the symmetric group on three letters, and take the normal subgroup $N=A_3=\{e,(123),(132)\}$, the alternating group of even permutations. Its two left cosets areProducts of permutations apply the rightmost factor first. Left multiplication by an even permutation preserves both cosets, while left multiplication by an odd permutation exchanges them. For example,Consequently all six group elements give only two distinct permutations of $\{E,O\}$. Both stabilizer subgroups equal $A_3$, so their intersection, the group kernel, is $A_3$. This is a transitive group action because $(12)$ reaches the other coset, but it is not a faithful group action because $(123)$ fixes both.

The orbit-stabilizer theorem still works perfectly: $6=2\cdot 3$. It counts reachable labels, without asserting that the induced group homomorphism is injective. The quotient group $S_3/A_3\cong C_2$ captures exactly the two visible operations, identity and exchange. This is the same group homomorphism as the sign of a permutation, with $+1$ represented by fixing both labels and $-1$ by exchanging them. Contrast the three-label example in faithful coset action of the symmetric group on three letters.

Compose permutations from right to left. The indicated cycle lengths immediately give $X^3=P^2=Q^2=e$. Under conjugation, $X^{-1}(ab)X=(X^{-1}a{X}^{-1}b)$. Since $X^{-1}$ sends $1,2,3$ to $3,1,2$ and fixes $4$,Direct composition also gives $PQ=QP=(14)(23)$. Thus $V=\{e,P,Q,PQ\}$ is a Klein four-group. Conjugation by $X$ cycles its three nonidentity elements, so $X$ normalizes $V$.

The cosets $V,VX,V{X}^2$ have four elements each and are distinct. If two coincided, a nontrivial power of $X$ would belong to $V$, impossible because it has order three whereas every nonidentity element of $V$ has order two. Hence the subgroup generated by $X,P,Q$ contains at least twelve elements. These generators are even permutations, so the subgroup lies in the alternating group $A_4$, whose order is $4!/2=12$; multiplying by one fixed transposition pairs even and odd permutations to justify this count. ThereforeThe calculation also describes $A_4$ as the semidirect product $V⋊⟨X⟩$, with the order-three factor permuting the three involutions.

Compose permutations from right to left. Tracking the four symbols through the two permutation cycles gives $1\mapsto 2\mapsto 1$ and $3\mapsto 4\mapsto 3$. Thus the cycle decomposition of a permutation isThe disjoint transpositions commute, and squaring their product gives the identity element, while the product itself is nonidentity. Its order of a group element is therefore two. Each transposition has negative sign of a permutation, so the product has sign $(-1{)}^2=\boxed{+1}$: it is an even permutation.

For a permutation $\sigma \in S_n$, let $P_{\sigma }$ be its permutation matrix, defined by $P_{\sigma }{e}_j=e_{\sigma (j)}$. Define the signature of a permutation by $ϵ(\sigma )=\det P_{\sigma }$. A permutation matrix is orthogonal, so $(\det P_{\sigma }{)}^2=1$ and its permutation sign lies in $\{1,-1\}$. A transposition exchanges two columns of the identity and therefore has determinant $-1$. Every permutation is a product of transpositions, by decomposing each permutation cycle asHence $ϵ(\sigma )=(-1{)}^m$ for any expression as $m$ transpositions; the determinant definition proves that this parity of a permutation is independent of the expression. Equivalently the permutation sign is positive for an even permutation and negative for an odd permutation.

Composition is taken rightmost first. Since $P_{\sigma \tau }{e}_j=e_{\sigma (\tau (j))}=P_{\sigma }{P}_{\tau }{e}_j$, we have $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$. Multiplicativity of the determinant therefore proves the sign homomorphism identityThe alternating group consists of the even permutations:It contains the identity. If $\sigma ,\tau \in A_n$, their product has permutation sign $1$, and $ϵ({\sigma }^{-1})=ϵ(\sigma {)}^{-1}=1$, so their inverses also belong to $A_n$. This proves it is a subgroup. Finally, for every $g\in S_n$ and $a\in A_n$,Thus $A_n$ is a normal subgroup of $S_n$. This also exhibits directly why a kernel of a group homomorphism is normal.

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.

A permutation cycle $(a_1{a}_2\cdots a_p)$ sends $a_1$ to $a_2$, each successive listed point to the next, and $a_p$ to $a_1$, fixing every unlisted point. Its length is $p$. For any permutation $h$,Thus conjugate group elements in the symmetric group have the same cycle length. Conversely, two cycles of the same length are conjugate: send their entries, in cyclic order, to one another and extend this bijection arbitrarily to the unused points. Length-one cycles represent the identity and cause no exception to this conclusion.

An odd-length permutation cycle is an even permutation, since its sign is $(-1{)}^{p-1}$. Let $c,d$ be such cycles on $p+2$ points and choose $h\in S_{p+2}$ with $hc{h}^{-1}=d$. If $h$ is odd, let $t$ be the transposition of the two points unused by the listed cycle $c$. Then $tc=ct$, and $ht$ is even withHence all the $p$-cycles are conjugate within $A_{p+2}$. The two spare letters supply the parity correction; they are the reason the same argument need not work with only one spare letter.

In $A_4$ the answer is no. For example, $(123)$ and $(132)$ are conjugated by the odd transposition $(23)$ in $S_4$. The centralizer of $(123)$ in $S_4$ is exactly its order-three cyclic subgroup: a commuting permutation fixes the unique fixed point $4$ and acts as a power of the cycle on the other three points. All these centralizing elements are even. Every other conjugating element differs from $(23)$ by a centralizing element, so it is also odd. There is no conjugator in $A_4$. Equivalently, the eight three-cycles split into two conjugacy classes of size $12/3=4$, illustrating the alternating conjugacy class splitting criterion.

No such finite group exists. Cayley theorem embeds every finite group $G$ of order $n$ into $S_n$: the left-regular group action sends $g$ to the permutation $x\mapsto gx$, and this is faithful because the image of the identity element is $g$.

To make every image permutation even, let $t=(n+1n+2)$, with support disjoint from the first $n$ letters. For $\sigma \in S_n$, let $e(\sigma )=0$ for an even permutation and $e(\sigma )=1$ for an odd permutation, and setThe sign homomorphism gives $e(\sigma \tau )\equiv e(\sigma )+e(\tau )(\mathrm{mod}2)$, and $t$ commutes with all permutations of the first $n$ letters. Thus $\iota$ is a group homomorphism. It is injective by restriction to those letters, and its sign is $(-1{)}^{e(\sigma )}(-1{)}^{e(\sigma )}=1$. ConsequentlyThis is the alternating-group embedding of a finite group.

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.

A three-cycle is an even permutation, as is a product of two disjoint three-cycles. Thus every element counted in part (c) lies in $A_7$, and every order-three element of $A_7$ was already counted there. The answer remains $\boxed{350}$.

The scalar triple product is a determinant:A cyclic permutation of its three columns is an even permutation, so it leaves the determinant unchanged. Thus $\det (\mathbf{a},\mathbf{b},\mathbf{c})=\det (\mathbf{b},\mathbf{c},\mathbf{a})=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$, proving the identity.

For every abelian group $A$ and group homomorphism $f:G\to A$, there is a unique group homomorphism $\bar{f}:G^{\mathrm{ab}}\to A$ such thatThis is a universal property. To see why it holds, compute $f([x,y])=[f(x),f(y)]=e_A$, using commutativity in $A$. Hence $G^{\prime }\subseteq \ker f$, and $f$ is constant on each coset of $G^{\prime }$. The displayed formula is therefore well-defined and preserves multiplication. It is unique because $q$ is surjective. This is an instance of factoring a group homomorphism through a quotient group; see Conrad, Universal mapping properties, theorem 3.4.

In particular, every abelian quotient $G/N$ comes from a surjective group homomorphism $G^{\mathrm{ab}}\to G/N$, $g{G}^{\prime }\mapsto gN$. For $S_3$, any group homomorphism to an abelian group is determined by the common image $a$ of its three transpositions, with $a^2=e_A$. Every even permutation must map to $e_A$. Thus sign of a permutation captures all possible information of this kind, including maps with infinite abelian groups as their targets.