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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.

Choose $x_0\in X$ with stabilizer subgroup $H$. Transitivity means that every point is $g{x}_0$ for some $g$. Its stabilizer subgroup issince $a(g{x}_0)=g{x}_0$ exactly when $g^{-1}ag\in H$. The kernel of a group homomorphism arising from the group action is the set of elements fixing every point, henceThis is the subgroup core. A group homomorphism is injective precisely when its kernel is trivial. The desired equivalence is therefore exactly the equivalence between a trivial subgroup core and a faithful group action. No finiteness assumption is required in this part.

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.

Use the standard implicit assumption that $p$ is prime, so ${\mathbb{F}}_p$ is a finite field. The matrices in question form $G{L}_2({\mathbb{F}}_p)$, the general linear group over a finite field. Products stay in the set because determinants multiply, the identity has determinant one, andhas entries in the finite field and nonzero determinant. Together with associativity this verifies the group axioms. The multiplication of matrices with column vectors is a group action because $Iv=v$ and $(gh)v=g(hv)$.

Let $C=⟨g⟩$ be the order-$p$ cyclic subgroup. By the orbit-stabilizer theorem, its group orbits on ${\mathbb{F}}_p^2$ have size either one or $p$. If $f$ is the number of fixed vectors, counting all $p^2$ vectors gives $f\equiv p^2\equiv 0(\mathrm{mod}p)$. Since the zero vector is fixed, $f\ge 1$ and hence $f\ge p$. There is a nonzero fixed vector $v$.

Extend $v$ to a basis $(v,w)$. In this basis $g$ has matrix $\left(\begin{array}{cc}1& b\\ 0& d\end{array}\right)$. From $g^p=I$ we get $d^p=1$; in the prime finite field, $d^p=d$, so $d=1$. Since $g$ has order $p>1$, it is not the identity, and $b\ne 0$. Replace $w$ by $b^{-1}w$. Then $g(b^{-1}w)=v+b^{-1}w$, and in the new basisThus all the order-p matrices in GL2 over the prime field form one conjugacy class. This proof also works at $p=2$. Primality is essential to the first assertion: for modulus four, $\mathrm{diag}(2,1)$ has nonzero determinant but no inverse, so the set defined using merely nonzero determinants would not be a group.

The coordinate ring is $\mathbb{C}[W]=\mathrm{Sym}(W^{\ast })$. The induced action is contragredient on functions:It is a group action by degree-preserving algebra automorphisms, extending the dual action on linear polynomial functions. Its polynomial invariant ring isFor a monic polynomial $f(t)={\prod }_{i=1}^d(t-r_i)$, its polynomial discriminant is $\mathrm{disc}(f)={\prod }_{i<j}(r_i-r_j{)}^2$. This is symmetric in the roots, hence polynomial in the coefficients, and is zero exactly when a root is repeated.

For the alternating group action, let $e_1,\dots ,e_n$ be the elementary symmetric polynomials and put $\Delta ={\prod }_{i<j}(X_i-X_j)$. If $p$ is $A_n$-invariant and $\tau$ is any transposition, decomposeNormality and index two of $A_n$ show that $p_{+}$ is symmetric and $p_{-}$ transforms by the sign character of $S_n$. Any alternating polynomial vanishes when $X_i=X_j$, so every $X_i-X_j$ divides it. These pairwise nonassociate prime factors have product $\Delta$, so $p_{-}=\Delta q$ with $q$ symmetric. The Fundamental theorem of symmetric polynomials yieldsThe sum is direct, since a polynomial that is both symmetric and alternating is zero in characteristic zero. Also ${\Delta }^2=D(e_1,\dots ,e_n)$, where $D$ is the discriminant polynomial ofThus, more precisely than the requested quotient assertion,Surjectivity follows from the direct-sum expression. Divide any putative kernel element by the monic quadratic in $Z$; its remainder is $a(T)+Zb(T)$. The direct sum forces $a(e)=b(e)=0$, and algebraic independence of the $e_i$ gives $a=b=0$. Hence the displayed relation is the entire kernel. The argument also covers $n=2$, where $A_2$ is trivial.

Now let $G$ be finite with no nontrivial linear characters, and write $R=\mathbb{C}[W]$, $S=R^G$. The polynomial ring $R$ is a unique factorization domain. Factor a nonzero invariant $f$ in $R$; invariance permutes the associate classes of its irreducible factors and makes their exponents constant on each orbit. For an orbit $\mathcal{O}$, choose representatives and form its orbit product of polynomial factorsFor every $g$, $g\cdot q_{\mathcal{O}}=\theta (g)q_{\mathcal{O}}$ for a nonzero scalar $\theta (g)$. Applying two group elements proves that $\theta :G\to {\mathbb{C}}^{\times }$ is a homomorphism. The hypothesis forces $\theta =1$, so $q_{\mathcal{O}}\in S$.

This orbit product is prime in $S$. If it divides $ab$ with $a,b\in S$, one factor $p\in \mathcal{O}$ divides $a$ or $b$ in $R$. Invariance of that chosen polynomial makes every factor in the orbit divide it. Their product therefore divides it in $R$, and the quotient is invariant because numerator and denominator are invariant and cancellation is valid in the integral domain $R$. Thus $q_{\mathcal{O}}$ divides $a$ or $b$ in $S$. Every nonzero $f\in S$ is a scalar times a product of these prime orbit products, and the only units of $S$ are nonzero constants, as they are units in $R$. Therefore $\mathbb{C}[W{]}^G$ is a unique factorization domain. This gives a direct proof, without assuming a divisor-class-group theorem.

For a failure in characteristic zero, let the cyclic group of order two act on ${\mathbb{C}}^2$ by $-I$. Its coordinate invariants areEvery invariant monomial has even total degree: its exponents are either both even, or both odd, giving the indicated generators. Reducing powers of $b$ to at most one shows that $ac-b^2$ is the only relation, since $a^i{c}^j$ and $b{a}^i{c}^j$ map to distinct monomials. The three quadratic invariants are irreducible in $S$: each nonconstant invariant has degree at least two, so a product of two nonunits has degree at least four. They are pairwise nonassociate, butgives two different irreducible factorizations. Hence this invariant ring is not a unique factorization domain. The nontrivial sign character is precisely the kind of character excluded in the preceding theorem.

The SU(2) group consists of complex $2\times 2$ matrices $U$ with $U^{†}U=I$ and $\det U=1$. The SO(3) group consists of real $3\times 3$ matrices $R$ with $R^{T}R=I$ and $\det R=1$. Differentiate these equations along a path through the identity. This givesConversely the matrix exponential of each displayed infinitesimal matrix satisfies the corresponding group equations, so these are exactly the Lie algebras, with bracket the matrix commutator.

Using the Pauli matrices, put $t_a=-i{\sigma }_a/2$. They form a real basis of the SU(2) Lie algebra. The Pauli matrix commutator identity gives $[t_a,t_b]={ϵ}_{abc}{t}_c$. Define $J_a$ on ${\mathbb{R}}^3$ by $J_{a}v=e_a\times v$. These are a basis of the SO(3) Lie algebra; the vector triple-product identity gives $[J_a,J_b]={ϵ}_{abc}{J}_c$. Thusis a real linear bijection preserving the bracket. This is the SU(2)-SO(3) Lie algebra isomorphism. It is a Lie-algebra isomorphism, not a group isomorphism: the Adjoint double cover from SU(2) to SO(3) has kernel $\{I,-I\}$.

The SU(3) group is defined similarly by $U^{†}U=I$ and $\det U=1$ on complex $3\times 3$ matrices. One SU(2) subgroup is $\{\mathrm{diag}(V,1):V\in SU(2)\}$. The real orthogonal matrices with determinant one form an SO(3) group subgroup, since real orthogonality is also complex unitarity.

Under the defining group action, the group orbit of $e_1$ is exactly the unit sphere in ${\mathbb{C}}^3$:Unitarity proves containment. Conversely, extend a unit vector $z$ to an orthonormal basis and use it as the first column of a unitary matrix. Multiplying the last column by the inverse of its determinant makes that determinant one without changing the first column. The isotropy group of $e_1$ must also preserve its orthogonal complement, and therefore isThis proves the unit sphere orbit of the defining special unitary action, including $S^5\cong SU(3)/SU(2)$.

For the complex quadric, write $z=x+iy$. Its defining relation separates intoThe Hermitian norm is then $z^{†}z=1+2|y{|}^2$, which is not constant on this set. For example, $e_1$ and $\sqrt{2}{e}_1+i{e}_2$ both satisfy the complex bilinear relation but have Hermitian norms one and three. Since the SU(3) group preserves that norm, the quadric cannot be a single SU(3) group orbit. In fact it is not even invariant under the whole group: $\mathrm{diag}(e^{it},e^{-it},1)e_1$ fails the bilinear relation when $e^{2it}\ne 1$.

The real SO(3) group does preserve the quadric, acting simultaneously on $x$ and $y$. Its group orbits are classified completely by $r=|y|\ge 0$. For $r=0$, $y=0$ and $x$ is a real unit vector, giving the group orbit $S^2$ and stabilizer subgroup $SO(2)$. For $r>0$, the vectors $x/\sqrt{1+r^2}$ and $y/r$ are orthonormal. Adjoining their cross product gives an oriented orthonormal frame. There is a unique rotation taking the standard frame to this one; hence the action is transitive at fixed $r$ and the stabilizer subgroup is trivial. Thus the real rotation orbits on a complex unit quadric areIn particular the real subgroup does not act transitively on the whole quadric. The parametrization $x=\sqrt{1+|y{|}^2}n$ with $n\in S^2$ and $y\perp n$ also identifies the quadric, as a real manifold, with the tangent bundle of $S^2$.

Place the cube at the origin with face normals $\pm e_1,\pm e_2,\pm e_3$, and label opposite faces by $(1,2),(3,4),(5,6)$. Every cube symmetry is a signed permutation matrix: there are $2^{3}3!=48$ choices. Its group action on faces is faithful because fixing all faces fixes all their normal directions. Central inversion $-I$ exchanges every opposite pair and therefore acts as $h$. Since $-I$ commutes with every linear cube symmetry, the face-action image lies in $C_{S_6}(h)$. The preceding count gives $|C_{S_6}(h)|=48$, so the faithful face action is onto this centraliser:For the remaining isomorphism, let $R$ be the rotational symmetry group of a cube. Half the signed permutation matrices have determinant $1$, so $|R|=24$. Central inversion has determinant $-1$, and every orientation-reversing symmetry is $(-I)r$ for a unique $r\in R$. Since $-I$ is central and $R\cap \{I,-I\}=\{I\}$, multiplication gives a direct product of groups:The group action of $R$ on the four body diagonals is faithful. To prove this, choose direction vectors $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$. They span ${\mathbb{R}}^3$, and their only linear relation is that their sum is zero. A rotation fixing all four diagonal lines sends each listed vector to itself or its negative. Applying the relation forces all four signs to agree. The all-negative choice is $-I$, which is not in $R$, so the rotation is the identity. Thus $R$ embeds in $S_4$; both have order $24$, giving $R\cong S_4$. Finally $C_{S_6}(g)\cong C_2\times S_4$ from part (i). ConsequentlyThe direct-product decomposition of the cube symmetry group supplies the second isomorphism, while the first comes from the specified face action.

Let $n=[G:H]<\infty$. The group action by left multiplication on $G/H$ gives a group homomorphism $G\to S_n$. Its kernel is the normal core of a subgroup,The containment follows by looking at the stabilizer of the coset $H$, and finite index follows from the finite image in $S_n$.

The Higman group isIt is visibly a finitely presented group. To prove infinitude, first formIt is the amalgamated free product of $⟨a,b\mid ab{a}^{-1}=b^2⟩$ and $⟨b,c\mid bc{b}^{-1}=c^2⟩$, identifying their infinite cyclic subgroups generated by $b$. Each factor is an HNN extension of an infinite cyclic group, so its base and stable letter both have infinite order. No nonzero power of $a$ belongs to $⟨b⟩$ in the first factor, by the map to $\mathbb{Z}$ sending $a$ to one and $b$ to zero. In the second factor, no nonzero power of $c$ belongs to $⟨b⟩$: the stable-letter map forces a hypothetical equality $c^k=b^l$ to have $l=0$, and the base $c$ has infinite order. The normal form theorem for an amalgamated free product therefore shows that $⟨a,c⟩\le K$ is a rank-two free group.

Similarly,contains $⟨c,a⟩$ as a rank-two free group. Identifying these two free subgroups yieldsThe normal form theorem for an amalgamated free product embeds $K$ in $J$. In particular, $J$ contains a free group of rank two and is infinite.

The finite quotients of cyclic squaring presentations argument now rules out every nontrivial finite quotient of $J$. In a finite image, a relation $xy{x}^{-1}=y^2$ forces the order of $y$ to be odd, since conjugate elements have the same order. If any generator has nontrivial image, let $p$ be the least prime number dividing the order of any of the four generator images, and choose $y$ whose order is divisible by $p$. Its predecessor $x$ conjugates it to its square. If $r$ is the order of $x$, iterating conjugation gives $2^r\equiv 1(\mathrm{mod}p)$. Hence the multiplicative order of $2$ modulo $p$ divides $r$. It is greater than one and divides $p-1$, so it has a prime factor smaller than $p$, which also divides $r$. This contradicts the minimal choice of $p$. Thus all four generator images are trivial. $J$ has no nontrivial finite quotient, and the normal-core argument above implies that $J$ has no proper finite-index subgroup.

For the final argument, Conjugation preserves the order of an element. Thus if one nonidentity element has finite order $n$, every nonidentity element has that same order, and $n\ge 2$. Moreover $n$ is prime: if a prime factor $p$ properly divides $n$, then $g^p$ is nonidentity but has the smaller order $n/p$.

When $n\ge 3$, the element $g^2$ is nonidentity, so choose $x$ with $xg{x}^{-1}=g^2$. The conjugator is not the identity, since $g^2\ne g$, and therefore $x^n=1$. Induction gives $x^{k}g{x}^{-k}=g^{2^k}$, and at $k=n$ this yieldsBut Fermat's little theorem, with the odd prime $n$, gives $2^n\equiv 2(\mathrm{mod}n)$, contradicting that divisibility.

For $n=2$, $g^2=1$ is not in the nonidentity conjugacy class, so the required conjugator cannot be chosen. Instead, a group in which every element has square one is an abelian group: $(ab{)}^{-1}=ab$ also equals $b^{-1}{a}^{-1}=ba$. In an abelian group every conjugacy class is a singleton, so one nonidentity class permits only one nonidentity element, giving a group of order two. This contradicts infinitude. Consequently the infinite group in question is a $\boxed{\text{torsion-free group}}$.

This establishes the torsion-freeness from one nonidentity conjugacy class criterion.

A residually finite group $G$ has the property that every $1\ne g\in G$ survives under a group homomorphism to some finite group. Equivalently, the intersection of its finite-index normal subgroups is trivial. A Hopfian group is a group for which every surjective endomorphism is an automorphism.

Suppose $G$ is generated by $d$ elements. A group homomorphism $G\to S_n$ is determined by the images of these generators, so there are at most $(n!{)}^d$ such maps. Every subgroup of index $n$ gives a transitive coset group action on an $n$-element set, and the subgroup is the stabilizer of a point in that action. There are at most $n$ point stabilizers per action. The finite-index subgroup count for a finitely generated group therefore gives

Now let $\psi :G\to G$ be a surjective endomorphism. For any fixed $n$, inverse image under $\psi$ preserves the index of a normal subgroup. It is also an injective function on the finite set of normal subgroups of index $n$: if ${\psi }^{-1}(M)={\psi }^{-1}(N)$, surjectivity gives $M=N$. It is therefore a permutation of that finite set. Given any finite-index normal subgroup $N$, there is another such subgroup $M$ with $N={\psi }^{-1}(M)$, so $\ker \psi \le N$. If $G$ is a residually finite group, intersecting all these $N$ gives $\ker \psi =1$. Hence $\psi$ is an automorphism. Every finitely generated group that is a residually finite group is a Hopfian group.

A useful residual finiteness of semidirect products theorem is: if $K$ is a finitely generated group, thenMore generally, the forward construction only requires that $K$ have a separating family of finite-index normal subgroups invariant under the $H$ action, and that $H$ be a residually finite group. Necessity follows by restricting finite separating maps to the embedded subgroups $K$ and $H$.

For sufficiency, first consider $(k,h)$ with $h\ne 1$: projection to $H$ and then a suitable finite quotient separates it. If $h=1$ and $k\ne 1$, choose a finite-index normal subgroup $U⊴K$ with $k\notin U$. Because $K$ is finitely generated, it has only finitely many subgroups of index at most $[K:U]$. Their intersection $C$ is a finite-index characteristic subgroup of $K$, is contained in $U$, and is invariant under every automorphism in the $H$ action. The quotient $K/C$ is finite. Let $\rho :H\to \mathrm{Aut}(K/C)$ be the induced action. The mapis a group homomorphism to a finite group and separates $(k,1)$. This proves the theorem and the more general invariant-subgroup criterion. The finite-generation condition is used to produce the characteristic subgroup $C$, not assumed for $H$.

A left group action is a map $(g,x)\mapsto g\cdot x$ with $e\cdot x=x$ and $(gh)\cdot x=g\cdot (h\cdot x)$. The group orbit and stabiliser subgroup areThe identity belongs to $G_x$. If $g,h\in G_x$, then $(gh)\cdot x=g\cdot x=x$; also $g^{-1}\cdot x=g^{-1}\cdot (g\cdot x)=x$. Thus $G_x$ is a subgroup.

For the orbit-stabilizer theorem, the map $g{G}_x\mapsto g\cdot x$ from left cosets to the group orbit is a bijection: $g\cdot x=h\cdot x$ if and only if $h^{-1}g\in G_x$, which is precisely $g{G}_x=h{G}_x$. Each left coset has $|G_x|$ elements, since multiplication by $g$ is a bijection from $G_x$ to $g{G}_x$, and the left cosets partition $G$. For finite $G$ this proves

The dimension of a topological space by irreducible chains is the supremum of the integers $d$ for which there is a chain $Z_0⊊Z_1⊊\cdots ⊊Z_d$ of nonempty irreducible closed subsets of the space. Closedness here is relative to the given locally closed space. For varieties this is their Krull dimension.

An algebraic group is a group whose underlying space is an algebraic variety and whose multiplication $G\times G\to G$ and inversion $G\to G$ are morphisms. An algebraic group action on a variety $V$ is a morphism $G\times V\to V$ satisfying the identity and associativity axioms of a group action.

For the homomorphism $\phi$, the kernel is ${\phi }^{-1}(1_H)$, so the kernel is closed. To prove closedness of the image, use the Chevalley constructibility theorem: the image of a morphism of varieties is constructible. Thus $S=\phi (G)$ is a constructible subset of a variety and an abstract subgroup. Its closure $C$ is also a subgroup: translation by elements of $S$ preserves $C$, and continuity then extends multiplication and inversion to $C$.

A dense constructible subset contains a dense open subset $U$ of its closure. For $c\in C$, both $U$ and $cU$ are dense open subsets of $C$, so their intersection is nonempty. If $u=cv$ with $u,v\in U\subseteq S$, then $c=u{v}^{-1}\in S$. Hence $S=C$, proving that the image is closed. This is the principle that a constructible subgroup is closed.

Every nonempty fiber of $G\to S$ is a translate of $\ker \phi$ and has that same dimension. The fiber dimension theorem therefore givesThis dimension formula for an algebraic group homomorphism is a dimension statement, so it does not require separability of $\phi$.

For a dimension vector of a quiver representation $\mathbf{n}$, setThe base change action on quiver representations isThe entries are regular functions on the product of the general linear groups and the quiver representation space, because inverse entries are cofactors divided by the invertible determinant. Thus this is an algebraic group action.

For nonzero $\mathbf{n}$, let $\Delta k^{\times }$ be the common scalar subgroup $(\lambda I_{n_i}{)}_i$ and define $\mathrm{PGL}(\mathbf{n})=\mathrm{GL}(\mathbf{n})/\Delta k^{\times }$. Common scalars act trivially, so the formula descends to an algebraic group action of this projective base change group of a quiver. This is a quotient by one common scalar, not a product of the individual projective groups. If every $n_i=0$, the representation space is a point and both actions are taken to be trivial.

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.

For a group action of a finite group $G$ on a set and a point $x$, the orbit-stabilizer theorem iswhere $G_x=\{g:gx=x\}$ is the stabilizer subgroup. The map $g{G}_x\mapsto gx$ is well defined because elements of the same coset act identically on $x$. If $gx=hx$, then $h^{-1}g\in G_x$, so their cosets agree; every orbit point has the form $gx$, giving surjectivity. This is a bijection. Lagrange's theorem, $|G|=[G:H]|H|$ for any subgroup $H$, yields the counting formula.

Apply the conjugation action of the finite p-group $G$ to its nontrivial normal subgroup $N$. Normality ensures the action stays inside $N$. The singleton orbits of a group action are exactly the elements of $N\cap Z(G)$. Every larger orbit of a group action has size a positive power of $p$, by the orbit-stabilizer theorem and Lagrange's theorem. ThereforeSince $N$ is nontrivial, its order is divisible by $p$. The intersection contains the identity and has cardinality a positive multiple of $p$, so it contains a nonidentity element. This proves the central intersection property of normal subgroups of finite p-groups.

For the proper subgroup $H$, let $H$ act by left multiplication on the left cosets $G/H$. A coset $gH$ is fixed by every $h\in H$ exactly when $g^{-1}Hg\subseteq H$. Both groups have the same finite size, so this inclusion is equality; the fixed cosets are precisely those with $g\in N_G(H)$. Hence their number is $[N_G(H):H]$. All non-singleton orbits of a group action have size divisible by $p$, soSince $H$ is proper, $[G:H]$ is a positive power of $p$ greater than one. The fixed-coset count is nonzero, because $H$ itself is fixed, and is divisible by $p$. Therefore the normalizer strictly contains $H$:This is the normalizer condition for finite p-groups. The proof also covers $H=\{1\}$, whose action fixes all cosets.

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 $(12)$.

The free product $G_1\ast G_2$ is obtained by taking disjoint copies of the elements of the two groups as generators and imposing precisely their internal multiplication relations. Equivalently, if $G_i=⟨X_i\mid R_i⟩$ with disjoint $X_i$, thenA nontrivial free product has both factors different from the trivial group.

The normal form theorem for a free product says that every element has a unique expressionwhere the empty expression represents the identity. The $g_j$ are called syllables in a free product. Thus nonempty alternating products cannot be the identity, and the natural maps of the factors into the free product are injective.

For existence, multiply adjacent syllables from the same factor and delete identity syllables until the word alternates. Each change uses a defining relation and decreases the number of syllables. To establish uniqueness without assuming that these reductions are confluent, let $\mathcal{W}$ be the set of alternating nonidentity syllable sequences, including the empty sequence. For $h\in G_i$, define a permutation $L_h$ of $\mathcal{W}$ by prepending $h$, multiplying into the first syllable if it belongs to $G_i$, and deleting it if that product is the identity; for $h=1$, do nothing.

The multiplication law inside $G_i$ gives $L_h{L}_k=L_{hk}$ for $h,k\in G_i$, and $L_h^{-1}=L_{h^{-1}}$. This can be checked at the first syllable: if a product deletes it, the next syllable belongs to the other factor and is treated as a new first syllable. Hence the defining relations give a group action of $G_1\ast G_2$ on $\mathcal{W}$. An alternating word $g_1\cdots g_r$ sends the empty sequence to its own syllable sequence. Two such words representing the same group element induce the same permutation, and therefore have identical sequences. This proves the normal form theorem for a free product.

The universal property of a free product states that, for group homomorphisms $f_i:G_i\to H$, there is a unique group homomorphism $f:G_1\ast G_2\to H$ extending both. Explicitly,The internal multiplication relations are respected because each $f_i$ is a group homomorphism, so this assignment descends from words to the presented group. Alternatively, multiplication of two normal forms is concatenation followed by precisely those internal relations, which do not change its value in $H$. Uniqueness holds because the two factors generate the free product.

The HNN extension in the convention used here isThe new generator $t$ is its stable letter, and $A,B$ are its associated subgroups of an HNN extension. We prove the normal form theorem for an HNN extension, which also proves that the natural map $G\to G_{\ast \varphi }$ is injective.

Set $C_{+}=A$, $C_{-}=B$, ${\psi }_{+}=\varphi$, and ${\psi }_{-}={\varphi }^{-1}$. For each sign $ϵ$, choose a right coset transversal $R_{ϵ}$ for $C_{ϵ}\backslash G$, containing $1$ as the representative of $C_{ϵ}$. Thus every $g\in G$ has a unique decomposition $g=cr$, with $c\in C_{ϵ}$ and $r\in R_{ϵ}$. A normal form iswhere $g_0\in G$ is unrestricted and the case $k=0$ is just $g_0$.

Here is a constructive proof. Let $\mathcal{N}$ be the set of these formal sequences. Left multiplication by $h\in G$ changes only $g_0$ to $h{g}_0$. To prepend $t^{ϵ}$, uniquely decompose $g_0=cr$ using $R_{ϵ}$ and useIf the old first stable letter is $t^{-ϵ}$ and $r=1$, cancel this pair and multiply ${\psi }_{ϵ}(c)$ into the now-leading coefficient $r_1$. Otherwise retain the new first stable letter. Both cases give a sequence in $\mathcal{N}$; after a cancellation the leading coefficient is unrestricted, so no further normalization is needed at that end. Prepending the letters of a word from right to left proves existence.

These operations define permutations of $\mathcal{N}$. The base-group operations satisfy $L_h{L}_{h^{\prime }}=L_{h{h}^{\prime }}$. The operations $L_{t^{+}}$ and $L_{t^{-}}$ are inverse: if prepending did not cancel, the reverse operation decomposes the new leading coefficient ${\psi }_{ϵ}(c)$ with representative $1$ and cancels the newly inserted letter; if it did cancel, the reverse operation decomposes ${\psi }_{ϵ}(c)r_1$ with representative $r_1$, restoring the original prefix. The normal-form restriction rules out an unwanted second cancellation in the latter case.

For $a\in A$, decomposing $a{g}_0$ uses the same representative as decomposing $g_0$, and changes its subgroup coefficient from $c$ to $ac$. Both the cancelling and noncancelling cases therefore giveThus all defining relations act identically on $\mathcal{N}$, and we obtain a group action of the presented HNN extension. Each written normal form sends the empty form, whose leading coefficient is $1$, to that normal form itself. Equal group elements must have the same image of the empty form, proving uniqueness and the embedding of $G$.

A reduced sequence in an HNN extension is a word $g_0{t}^{{ϵ}_1}{g}_1\cdots t^{{ϵ}_k}{g}_k$ containing no pinch in an HNN extension: no $ta{t}^{-1}$ with $a\in A$ and no $t^{-1}bt$ with $b\in B$. Equivalently, whenever ${ϵ}_{i+1}=-{ϵ}_i$, one requires $g_i\notin C_{{ϵ}_i}$.

Britton's lemma states thatMore strongly it cannot represent an element of $G$. To prove this from the normal form theorem for an HNN extension, normalize the coefficients from right to left. Splitting $g_i=cr$ moves ${\psi }_{{ϵ}_i}(c)$ to the coefficient immediately on its left. If the preceding stable letter has the opposite sign, this transported factor belongs to the subgroup relevant to that inverse pair; multiplying by it cannot turn a coefficient outside that subgroup into one inside it. If the signs agree, cancellation is impossible anyway. Hence no stable letter disappears during normalization of a reduced sequence. Its unique normal form has $k>0$, whereas every element of $G$ has stable-letter length zero. This proves Britton's lemma. The same proof works with finitely many stable letters, with pinches requiring the same letter and its inverse.

First supply the unlettered preliminaries. Assuming the relevant categorical limits and colimits exist, limits of shape $I$ commute with colimits of shape $J$ when for every functor $D:I\times J\to \mathcal{C}$ the canonical mapis an isomorphism. To construct this map, for each $j$ compose the limiting projections with the $j$th colimit injections. They form an $I$-cone, giving a map into the right-hand categorical limit; these maps form a $J$-cocone, giving the comparison. This specifies the actual comparison required for commutation of limits and colimits.

A group regarded as a one-object category has its group elements as arrows, with composition given by multiplication. A functor from it to the Category of sets is a group action. For a $G$-set $A$, a cone consists of a map into $A$ whose values are fixed by every $g\in G$. Hence its categorical limit is the set $A^G$ of fixed points of a group action. A cocone is a map out of $A$ constant on orbits of a group action; its universal quotient is the orbit set. Thus

A diagram in a category $G\times H\to \mathbf{Set}$ is a set $A$ with commuting group actions of $G$ and $H$. Its limit–colimit comparison isCommutation makes $A^G$ stable under $H$. The comparison is always injective: if two $G$-fixed representatives are in the same $H$-orbit in $A$, they already determine the same orbit in $A^G$.

For surjectivity under the finite coprime hypotheses, let $O$ be an $H$-orbit fixed setwise by $G$. Its size divides $|H|$ by the orbit-stabilizer theorem. Since the group actions commute and $H$ is transitive on $O$, all $G$-orbits inside $O$ have the same size $d$: the bijection $a\mapsto ha$ identifies the $G$-orbits of $a$ and $ha$. Thus $d$ divides $|O|$, while the orbit-stabilizer theorem also gives $d\mid |G|$. The orders are coprime, so $d=1$. Every point in $O$ is therefore $G$-fixed, and $O$ lies in the image of $\theta$. We have proved commutation of fixed points and orbit quotients for coprime groups:The bijection is the canonical comparison and is natural in the set with commuting group actions, so it proves the requested commutation of limits and colimits in the Category of sets.