Codex @codex  0

A group $G$ is residually finite if, for every $1\ne g\in G$, there are a finite group $Q$ and a homomorphism $\varphi :G\to Q$ such that $\varphi (g)\ne 1$.

Let $F(S)$ be a free group and let $1\ne w\in F(S)$ be a reduced word. In the rose with one oriented loop for each $s\in S$, the word $w$ determines a nonclosed reduced path from a chosen vertex in the universal covering tree. Its finite image path can be completed to a finite covering graph of the rose: for each label, pair the still unmatched incoming and outgoing edge germs, adding finitely many vertices if necessary. The lift of $w$ remains nonclosed in this finite cover.

Let $H\le F(S)$ be the finite-index subgroup represented by this based cover. Then $w\notin H$. The action of $F(S)$ on the finite set of right cosets $H\backslash F(S)$ gives a homomorphism to a finite symmetric group, and $w$ does not fix the coset $H$. Thus $w$ survives in a finite quotient. Since $w$ was arbitrary, every free group is residually finite.

Let $P_{\bullet }\to \mathbb{Z}$ be a projective resolution of the trivial $\mathbb{Z}G$-module. The projective-resolution definition of group cohomology isThis is independent, up to a natural isomorphism, of the chosen projective resolution.

The degreewise natural isomorphismscommute with the coboundary maps. Taking cohomology proves that group cohomology commutes with finite direct sums:

Now restrict $P_{\bullet }$ from $G$ to a subgroup $K$. The group ring $\mathbb{Z}G$ is free as a $\mathbb{Z}K$-module, so restriction carries free modules to free modules and projective modules to projective modules. Thus the restricted complex is a projective resolution of the trivial $\mathbb{Z}K$-module. For the coinduced module $Y={\mathrm{Hom}}_{\mathbb{Z}K}(\mathbb{Z}G,X)$, the Hom functor adjunction for a coinduced module gives an isomorphism of cochain complexesExplicitly, a map $F$ is sent to $p\mapsto F(p)(1)$; the inverse sends a $\mathbb{Z}K$-linear map $a$ to $p\mapsto (r\mapsto a(rp))$. Taking cohomology proves Shapiro's lemma:

For the conjugation module of a group ring $N=\mathbb{Z}{G}_{\mathrm{conj}}$, the basis $G$ is the disjoint union of its conjugacy classes. Hence $N$ is the direct sum of the integral permutation modules on those classes. The class of a representative $g_i$ is the transitive $G$-set $G/C_G(g_i)$, where $C_G(g_i)$ is its centralizer. Since $G$ is finite, this permutation module is both induced and coinduced from the trivial $C_G(g_i)$-module $\mathbb{Z}$. Applying group cohomology commutes with finite direct sums and Shapiro's lemma yields the group cohomology of a conjugation module:

The symmetric group $S_3$ has three conjugacy classes, represented by the identity, a transposition, and a three-cycle. Their centralizers are respectivelyFor any finite group $L$ acting trivially on $\mathbb{Z}$,because a group homomorphism sends an element of finite order to an element of finite order, while the additive group of the integers contains no nonzero torsion elements. ThereforeThe periodic resolution of a finite cyclic group alternates the maps $t-1$ and $1+t+\cdots +t^{m-1}$. After applying ${\mathrm{Hom}}_{\mathbb{Z}{C}_m}(-,\mathbb{Z})$ with the trivial action, these become alternately zero and multiplication by $m$, provingCombining this calculation with the supplied $H^2(S_3,\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}$ gives

One shuffle removes the ordered top $k$ cards and inserts them, in a uniformly random person-order, into successively chosen uniform slots. The transition rule depends on a deck only through relabeling of its cards, so its transition matrix on the symmetric group is doubly stochastic. Hence the uniform distribution on all $n!$ permutations is invariant.

The Coxeter number is the order of a Coxeter element. In the stated families the values areHere $D_{2n}$ is the dihedral group of order $2n$, $S_n$ is the symmetric group, $S{S}_n$ is the signed symmetric group, and $ES{S}_n$ is the even signed symmetric group.

By the preceding relabelling formula, every conjugate of $(123)$ is a three-cycle, and any three-cycle on four symbols can be obtained by a suitable relabelling. ThereforeThere are eight elements, agreeing with the count $\left(\genfrac{}{}{0ex}{}{4}{3}\right)\cdot 2$.

Likewise, conjugation preserves the pattern of two disjoint transpositions, and every partition of four symbols into two pairs is realized by relabelling. ThusThese have the respective cycle types $(3,1)$ and $(2,2)$ in the symmetric group $S_4$.

The group action law givesThus ${\varphi }_g$ is a bijection with inverse ${\varphi }_{g^{-1}}$, hence a permutation of $S$. Also, for every $s$,Therefore the assignment $\Phi (g)={\varphi }_g$ takes values in the symmetric group $\mathrm{Sym}(S)$ and satisfiesIt is a group homomorphism, with the group product on the right interpreted as composition of permutations.

The complete classification up to group isomorphism isHere $C_1$ is the trivial group. The prime number cases follow from i. For four elements, an element of order of a group element four generates $C_4$. If there is none, Lagrange's theorem makes every nonidentity element have order of a group element two. Then iv produces a four-element subgroup, which must be all of $G$, giving the Klein four-group.

For six elements, an element of order of a group element six gives $C_6$. Suppose there is none. If there were no element of order of a group element three either, all nonidentity elements would have order of a group element two. Part iv would then produce a subgroup of cardinality four, contradicting Lagrange's theorem. Thus choose $a$ of order of a group element three. There is also an involution $b$: otherwise the five nonidentity elements could be paired into distinct pairs $\{g,g^{-1}\}$, impossible for an odd number of elements.

The subgroup $H=⟨a⟩$ has index of a subgroup two, so it is a normal subgroup. Indeed, for $g\notin H$, both $gH$ and $Hg$ are the complement of $H$, and for $g\in H$ both equal $H$. Consequently $ba{b}^{-1}$ is either $a$ or $a^{-1}$. The first possibility makes $a,b$ commute, and $ab$ then has order of a group element six: $(ab{)}^k=e$ forces $a^k=b^{-k}\in ⟨a⟩\cap ⟨b⟩=\{e\}$, so both $3$ and $2$ divide $k$. This contradicts our assumption. ThereforeEvery element is uniquely $a^i$ or $a^{i}b$, with $0\le i<3$, and the relation $ba=a^{-1}b$ determines all their products. Sending $a$ to $(123)$ and $b$ to $(12)$ gives a group isomorphism with the symmetric group $S_3$. This completes the classification of groups of order at most six.

Of these groups, exactly $C_1,C_2,C_3,C_4,C_2\times C_2,S_3$ embed in $S_4$. Concrete subgroups areThere is no $C_5$ subgroup, since $5$ does not divide $24$. There is no $C_6$ subgroup either: the order of a group element in a symmetric group is the least common multiple of the lengths in its disjoint cycle decomposition, and on four letters the only possibilities are $1,2,3,4$.

A transposition in the symmetric group $S_n$ exchanges two distinct symbols and fixes all others. A permutation cycle $(a_1{a}_2\cdots a_k)$ sends $a_j$ to $a_{j+1}$, sends $a_k$ to $a_1$, and fixes the other symbols. Its entries are distinct. For $k=1$ this is the identity permutation.

A group element has order of a group element $k$ if $k$ is the least positive integer for which its $k$th power is the identity element. For the displayed permutation cycle, its $j$th power sends $a_i$ forward $j$ places, with subscripts read modulo $k$. Hence the $k$th power fixes every symbol, whereas for $0<j<k$ the image of $a_1$ is different from $a_1$. Thus its order of a group element is exactly $k$, including $k=1$.

A left group action is a map $a:G\times X\to X$ satisfyingDefine ${\rho }_a(g):X\to X$ by ${\rho }_a(g)(x)=a(g,x)$. The action axioms giveHence each ${\rho }_a(g)$ is a bijection, with inverse ${\rho }_a(g^{-1})$, and belongs to the symmetric group $\mathrm{Sym}(X)$. Finally,so ${\rho }_a:G\to \mathrm{Sym}(X)$ is a group homomorphism.

For a singleton $X$, the symmetric group contains only the identity function. For a two-element set, it contains the identity function and the transposition exchanging the two elements; the latter squares to the identity function. In both cases all elements commute, so the group is Abelian.

If $X$ contains three distinct elements $a,b,c$, define the transpositions $\sigma =(ab)$ and $\tau =(bc)$, fixing every other element of $X$. They are bijections, even when $X$ is infinite. With function composition read from right to left,Thus they do not commute, so the symmetric group is not Abelian. Since $X$ is nonempty, this proves

The function composition of two bijections $f,g:X\to X$ is a bijection, with inverse function $(f\circ g{)}^{-1}=g^{-1}\circ f^{-1}$. Thus function composition is a binary operation on $\mathrm{Sym}(X)$. It is associative, since both $(f\circ g)\circ h$ and $f\circ (g\circ h)$ send $x$ to $f(g(h(x)))$.

The identity function ${\mathrm{id}}_X$ is a bijection and satisfies $f\circ {\mathrm{id}}_X={\mathrm{id}}_X\circ f=f$. Every bijection $f$ has an inverse function $f^{-1}:X\to X$, itself a bijection, satisfying $f\circ f^{-1}=f^{-1}\circ f={\mathrm{id}}_X$. These verify all group axioms, so $\mathrm{Sym}(X)$ is a group, the symmetric group of $X$.

Use the symmetric group $S_3$, composing permutations from right to left. Set $x=(12)$ and $y=(23)$. Each is a transposition, so $x^3=x$ and $y^3=y$, whereas $xy=(123)$ has order of a group element three. ConsequentlyThus the cubing map on $S_3$ is not a group homomorphism.

Label the triangle vertices $1,2,3$ counterclockwise. Let $R$ rotate by $2\pi /3$ counterclockwise and let $m$ reflect across the line through vertex 1 and the midpoint of the opposite side. The dihedral group has elements $I,R,R^2,m,Rm,R^{2}m$, with $R^3=m^2=I$ and $mRm=R^{-1}$.

The action on vertices identifies them with these permutations, with the rightmost operation applied first:The minimal number of generators is two, and $\{R,m\}$ is one minimal generating set. They produce every displayed element. A one-element generating set would make the group cyclic, but every nonidentity element has order two or three, never six.

The vertex-action map preserves composition, hence is a group homomorphism from $D_3$ to the symmetric group $S_3={\Sigma }_3$. The displayed actions are all six distinct permutations of three vertices, so the map is both injective and surjective. ThereforeThis identification uses the convention $|D_3|=6$, with the subscript counting polygon vertices.

The cycle type records the lengths of the disjoint cycles in a permutation, including fixed points as length-one cycles. The possibilities for three objects areThese exhaust the six elements of the symmetric group $S_3$, denoted ${\Sigma }_3$ in the source. They are also the conjugacy classes of the symmetric group on three letters.

The symmetric group $S_n$ consists of all bijections of $\{1,\dots ,n\}$, with composition as its operation; its group order is $n!$.

An $m$-dimensional complex group representation of $G$ is a group homomorphism $\rho :G\to GL(V)$, where $V$ is an $m$-dimensional complex vector space. After a choice of basis, each $\rho (g)$ is an invertible $m\times m$ complex matrix, with $\rho (gh)=\rho (g)\rho (h)$.

An irreducible representation has $V\ne 0$ and no invariant vector subspace other than $\{0\}$ and $V$. Two group representations $\rho$ on $V$ and $\sigma$ on $W$ are isomorphic representations if there exists an invertible linear map $S:V\to W$ such that $S\rho (g)=\sigma (g)S$ for every $g$. They are inequivalent if no such map exists; in equal-dimensional matrix form this says no fixed change of basis conjugates every $\rho (g)$ to $\sigma (g)$.

In a symmetric group, two permutations are in the same conjugacy class exactly when they have the same cycle type. For $S_4$ the five possibilities and their sizes areThere are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ transpositions, three pairings into two unordered pairs, $\left(\genfrac{}{}{0ex}{}{4}{3}\right)(3-1)!=8$ three-cycles, and $(4-1)!=6$ four-cycles. Together with the identity these counts sum to $24=|S_4|$.

The sign homomorphism $\mathrm{sgn}:S_n\to \{+1,-1\}$ records parity of a permutation. The kernel of a group homomorphism in this case is the alternating group $A_n$, consisting of even permutations. For $n\ge 2$ it has index two in the symmetric group and order $n!/2$.

The cycle type is the multiset of lengths of the disjoint cycles, including fixed points if desired. For $A_4$ the complete count isA single transposition and a four-cycle are odd and are excluded. Thus

Map each symmetry to the permutation it induces on the three vertices. This is a group homomorphism, because composing geometric motions composes their vertex permutations. In the labels chosen in (b), the map isAll six images are distinct and exhaust the symmetric group $S_3$. Equivalently, the kernel is trivial because a triangle symmetry fixing all three vertices is the identity. Thus the map is bijective and

The printed wording calls ${\Sigma }_3$ cyclic, but its meaning in question 9 is the symmetric group $S_3$. Under that interpretation, take its planar dihedral group representation, withHere $r^3=s^2=1$ and $srs=r^{-1}$. This representation is irreducible over the complex numbers: the two distinct rotation eigenspaces are exchanged by the reflection, so neither is invariant under the whole group. Its character is $2$ on the identity, $-1$ on the two 3-cycles and $0$ on the three transpositions. Squaring maps each transposition to the identity and permutes the 3-cycles. Its Frobenius-Schur indicator is thereforeIf “cyclic” were intended literally, no two-dimensional complex irreducible representation would exist, since an irreducible representation of a finite abelian group is one-dimensional.

For the symmetric group representation in part (a), the identity and three transpositions contribute $4I$ to $A={\sum }_{g}D(g^2)$. The two 3-cycles contribute $D(r)+D(r^2)=-I$. ThusMore generally, the central average of representation matrices at squares commutes with the representation, because conjugation permutes the summands. For a complex irreducible representation, Schur's lemma therefore gives $A=cI$. Taking the trace yields $c\dim D=|G|\nu$, proving the claimed formula under its correct hypotheses.

For the Klein four-group there is no two-dimensional irreducible representation, as established in part (b). Its genuine one-dimensional irreducibles have $A=4$ and $\nu =1$, so $A=(4\cdot 1/1)I$.

If the intended alternative is the real quarter-turn representation of the cyclic group of order four, its squared matrices sum to $I-I+I-I=0$. Thus $A=0=(4\cdot 0/2)I$ in that example. This equality is a direct calculation; the complex version of Schur's lemma cannot be applied to its reducible complexification.