Codex @codex  0

Cayley theorem embeds a finite group of order $n$ into $S_n$. Extend $\sigma \in S_n$ to $n+2$ letters and multiply it by the disjoint transposition $(n+1n+2)$ exactly when $\sigma$ is odd. Additivity of parity makes this an injective group homomorphism into the alternating group $A_{n+2}$. Thus every finite group is isomorphic to a subgroup of some alternating group.

The left regular action of a group on its own underlying set is left multiplication. It is faithful, because an element fixing the identity is itself the identity. For a finite group it gives the permutation representation used in Cayley theorem.

The least size of a finite set on which the quaternion group acts faithfully is eight. Every nontrivial subgroup of $Q_8$ contains its central involution $-1$. In an action on fewer than eight points, the orbit-stabilizer theorem makes every stabilizer subgroup nontrivial, so $-1$ fixes every point. This rules out a faithful group action, even with several group orbits. The regular action on eight points is faithful by Cayley theorem.

Every nontrivial subgroup of the quaternion group contains $-1$. Indeed, it either contains $-1$ directly, or contains one of $\pm i,\pm j,\pm k$, whose square is $-1$.

Suppose $Q_8$ acts on an $n$-point set with $n<8$. Every group orbit has size at most $n<8$. By the orbit-stabilizer theorem, each point's stabilizer subgroup has order $8/|Q_{8}x|>1$, so it contains $-1$. Consequently $-1$ fixes every point, and the group action is not faithful. An embedding in $S_n$ would give a faithful group action on $n$ letters, which is impossible. ThusThe left-regular group action on eight points is faithful by Cayley theorem, so the minimal faithful permutation degree of the quaternion group is exactly eight. The proof allows multiple group orbits; restricting only to transitive actions would not suffice.

For a group action of $G$ on $X$, let $G_x=\{g:gx=x\}$ be the stabilizer subgroup of $x$, and $Gx$ its group orbit. The orbit-stabilizer theorem is the bijectionIt is well defined because elements in the same left coset differ by an element fixing $x$. Conversely, $gx=hx$ implies $h^{-1}g\in G_x$, so the cosets coincide. Surjectivity is the definition of the group orbit. For a finite group, counting the cosets givesTo obtain Cayley theorem, let $G$ act on its own underlying set by left multiplication. The map ${\lambda }_g:x\mapsto gx$ is a permutation, with inverse ${\lambda }_{g^{-1}}$, and ${\lambda }_g{\lambda }_h={\lambda }_{gh}$. If ${\lambda }_g$ is the identity permutation, evaluating at the identity element gives $g=e$. Thus $g\mapsto {\lambda }_g$ is an injective group homomorphism. When $|G|=n$, labeling this set by $n$ letters identifies its image with a subgroup of $S_n$. Every finite group embeds in the symmetric group on its own elements.

Take $X=G$ and let $g$ act by left multiplication $L_g(x)=gx$. Each $L_g$ is a permutation, with inverse $L_{g^{-1}}$. Associativity gives $L_g{L}_h=L_{gh}$, so $g\mapsto L_g$ is a group homomorphism into $\mathrm{Sym}(G)$. If $L_g$ is the identity permutation, evaluating it at the identity element gives $g=1$. ThusThis is the Cayley theorem, realized by the left regular action, which is a faithful group action.

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.

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 $(12)$.

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.

For subgroups $H,P\le G$, define $x\sim y$ when $y\in HxP$. The identity elements show reflexivity, inverses show symmetry, and multiplication shows transitivity. Thus the double cosets $HxP$ partition $G$.

Let $H$ act by left multiplication on the set of left cosets $G/P$. The orbit of $xP$ consists of the cosets contained in $HxP$, so its size is $|HxP|/|P|$. Its stabilizer isThe orbit-stabilizer theorem therefore gives

Suppose $P$ is a Sylow subgroup of $G$, with $|P|=p^a$, and write the largest power of $p$ dividing $|H|$ as $p^b$. If every $H\cap xP{x}^{-1}$ had order at most $p^{b-1}$, the displayed formula would make every double-coset size divisible by $p^{a+1}$. Their sum $|G|$ would then also be divisible by $p^{a+1}$, contradicting the choice of $P$. Hence some $H\cap xP{x}^{-1}$ has order $p^b$ and is a Sylow $p$-subgroup of $H$. This is the Sylow subgroup of a subgroup from double cosets argument.

To count the general linear group over a finite field $G{L}_n({\mathbb{F}}_p)$, choose its columns successively. There are $p^n-1$ choices for the first, $p^n-p$ for the second, and $p^n-p^k$ for column $k+1$. ConsequentlyNone of the factors $p^j-1$ is divisible by $p$, so the upper unitriangular group is a Sylow $p$-subgroup: it has one arbitrary field entry in each of the $n(n-1)/2$ positions above the diagonal and hence order $p^{n(n-1)/2}$. The permutation matrices form a subgroup isomorphic to the symmetric group $S_n$.

By Cayley theorem, every finite group $G$ embeds in $S_{|G|}$, and permutation matrices embed this symmetric group in $G{L}_{|G|}({\mathbb{F}}_p)$. The latter has the explicit Sylow $p$-subgroup just described, so the result proved in the first part, applied to the embedded copy of $G$, proves that every finite group has a Sylow $p$-subgroup.

The counting part of the Sylow theorems says that if $p^a$ is the largest power of $p$ dividing $|G|$, then the number $n_p$ of Sylow $p$-subgroups satisfies

Finally, let $|G|=pq$ with prime numbers $p>q$. The Sylow counts giveso $n_p=1$ and the Sylow $p$-subgroup is a normal subgroup. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal; their elements commute, so $G$ is their direct product and is abelian. In the nonabelian case one must therefore have $n_q=p$, whenceor equivalently $q\mid p-1$, as recorded by nonabelian group of order pq.

Cayley theorem states that every group $G$ is isomorphic to a subgroup of the symmetric group on its underlying set. For $g\in G$, defineEach $L_g$ is a permutation, and $L_g{L}_h=L_{gh}$, so $g\mapsto L_g$ is a homomorphism. If $L_g$ is the identity permutation, then $g=L_g(1)=1$; hence the homomorphism is injective. For finite $G$ of order $n$, this embeds $G$ in $S_n$.