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The general linear group over a finite field $G{L}_2({\mathbb{F}}_2)$ acts faithfully on the four vectors of ${\mathbb{F}}_2^2$, since a matrix fixing all vectors fixes the standard basis. The zero vector is fixed by every element; the induced action on the three nonzero vectors identifies this six-element group with $S_3$. Swapping the two coordinates fixes $(0,0)$ and $(1,1)$ and exchanges the other two vectors, so its four-point permutation is odd. This action therefore embeds the group into $S_4$ with nontrivial composite sign homomorphism.

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.

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.

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.

For $\sigma \in S_n$, let $P_{\sigma }$ be its permutation matrix and defineSince a transposition exchanges two columns, it has determinant $-1$. Thus any expression of $\sigma$ as $r$ transpositions has sign $(-1{)}^r$, proving that the parity is independent of the expression. Moreover, $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$, so the determinant provesHence the sign homomorphism is a surjective homomorphism for $n\ge 2$.

For any homomorphism $\phi :S_n\to \{\pm 1\}$, all transpositions have the same image because they are conjugate. Since transpositions generate $S_n$, image $+1$ would make $\phi$ trivial. Surjectivity therefore forces every transposition to map to $-1$, and hence $\phi =\mathrm{sgn}$.

Let $G$ act on its underlying set by left multiplication. The resulting homomorphismis injective because ${\lambda }_g(e)=g$. This is Cayley theorem. If $g\ne e$, then $gx=x$ would imply $g=e$ after right cancellation, so every nonidentity permutation in the image is fixed-point free.

If $|G|$ is even, pair every element with its inverse. Elements not equal to their inverses occur in pairs. Since the identity is self-inverse and the group has even size, there must be another self-inverse element $g\ne e$. It has order two.

A permutation is odd when its sign is $-1$. If a subgroup $H\le S_m$ contains an odd element, the restrictionis surjective. Its kernel is the set of even elements and has index two. Each coset has the same size, so precisely half of $H$ is odd.

Now let $n=4k+2$. An element of order two in the fixed-point-free regular representation is a product ofdisjoint transpositions, so it is odd. The image $H\cong G$ therefore has an index-two normal subgroup of even permutations. Its order is $2k+1>1$, so it is nontrivial and proper. Hence

Nonabelian simple groups of even order do exist; the smallest example is the alternating group $A_5$, of order sixty.

Let $G$ act by left multiplication on the set $G/H$ of $n$ left cosets. This gives a group action homomorphismIts kernelis a normal subgroup of $G$, and the first isomorphism theorem gives $G/K\cong \mathrm{im}\phi$. By Lagrange's theorem, $|G/K|$ divides $n!$. The image acts transitively on the $n$ cosets, so the orbit-stabilizer theorem shows that $n$ divides $|G/K|$; in particular,

Now suppose $G$ is nonabelian and simple. Since $H$ is proper, the coset action is nontrivial, so $K\ne G$. Simplicity gives $K=\{e\}$, and $\phi$ embeds $G$ into $S_n$. Composing with the sign homomorphism givesIts kernel is normal. A nontrivial map would embed the simple group $G$ into the abelian group of order two, which is impossible because $G$ is nonabelian. The sign is therefore always $+1$, so

A Sylow subgroup of a finite group $G$ is a subgroup whose order is the largest power of a prime $p$ dividing $|G|$. The Sylow theorems say that such subgroups exist, that every $p$-subgroup lies in one, that all Sylow $p$-subgroups are conjugate, and that their number $n_p$ divides $|G|/p^a$ and obeys $n_p\equiv 1(\mathrm{mod}p)$.

For the last congruence, let one Sylow $p$-subgroup $P$ act by conjugation on the set of all Sylow $p$-subgroups. Every orbit other than a fixed point has size divisible by $p$. If $Q$ is fixed, then $P,Q\le N_G(Q)$. They are Sylow subgroups of this normalizer, so they are conjugate within $N_G(Q)$; because every element of $N_G(Q)$ fixes $Q$ under conjugation, this forces $P=Q$. Thus there is exactly one fixed point and $n_p\equiv 1(\mathrm{mod}p)$.

Now let $H<A_n$ have index $m$ with $1<m<n$. The coset action gives a homomorphismIts kernel is a normal subgroup of the simple group $A_n$. It cannot be all of $A_n$, since the action is transitive and nontrivial, so it is trivial. This would embed $A_n$ into $S_m$, contrary to $|A_n|=n!/2>m!=|S_m|$. Hence no such subgroup exists.

Suppose finally that a group $G$ of order $90$ were simple. The Sylow count satisfiesand simplicity excludes $n_5=1$, so $n_5=6$. Conjugation on these six subgroups gives a nontrivial homomorphism $G\to S_6$, which simplicity makes injective. Its image lies in $A_6$, because the composite with the sign homomorphism $S_6\to \{\pm 1\}$ must be trivial. It would therefore be a subgroup of $A_6$ of index $360/90=4$, contradicting the result just proved. Thus