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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.

An integral domain $A$ is an integrally closed domain if every element of its fraction field which is an integral element over $A$ belongs to $A$. Explicitly, if $z\in \mathrm{Frac}A$ satisfiesthen $z\in A$. Thus $\boxed{A\text{ is integrally closed exactly when its integral closure in its fraction field is }A}$.

Suppose $A$ is a unique factorization domain and $z=a/b$ is integral, with $a,b\in A$, $b\ne 0$, and with no common irreducible element as a factor. Multiplying its monic equation by $b^d$ givesHence $b$ divides $a^d$. Every irreducible element dividing $b$ is a prime element, so it would divide $a$, contrary to the choice of the fraction. Therefore $b$ is a unit, and $z\in A$. We have proved

The going-down theorem is the following. Let $A\subseteq B$ be an integral extension of integral domains with $A$ an integrally closed domain. IfthenWe first prove this for a module-finite ring extension and then remove that extra hypothesis. We use the Lying-over theorem, the Going-up theorem, and the incomparability theorem for integral extensions: primes exist over every base prime, prime chains can be extended upwards, and distinct primes over the same base prime cannot contain one another. The Going-up theorem follows by applying the Lying-over theorem to the quotient by the lower prime. These results concern arbitrary integral extensions; none assumes the going-down theorem.

For the finite case, let $K=\mathrm{Frac}A$ and $L=\mathrm{Frac}B$. Since $B$ is module-finite, $L/K$ is a finite field extension. Take a finite normal field extension $E/K$ containing $L$, obtained as a splitting field of the minimal polynomials of finitely many field generators. The extension $E/K$ need not be a separable field extension. Let $C$ be the integral closure of $A$ in $E$. Then $B\subseteq C$, and $C$ is integral over both $A$ and $B$.

We need one auxiliary fact: the finite group $G={\mathrm{Aut}}_K(E)$ acts transitively on the prime ideals of $C$ over any fixed prime ideal $\mathfrak{p}$ of $A$. First observe that an element $w\in E$ fixed by all of $G$ is a purely inseparable algebraic element over $K$. Indeed, all $K$-embeddings of $K(w)$ into an algebraic closure extend to $K$-automorphisms of the normal field extension $E$. Hence the minimal polynomial of an algebraic element $w$ has only one distinct root. In characteristic zero this says $w\in K$; in characteristic $p>0$ it says $w^{p^e}\in K$ for some $e\ge 0$.

Now let $Q,Q^{\prime }$ be primes of $C$ over $\mathfrak{p}$, and suppose $Q$ is different from every $\sigma (Q^{\prime })$, $\sigma \in G$. By the incomparability theorem for integral extensions, $Q$ is contained in none of these conjugate primes. The prime avoidance lemma suppliesThe productis fixed by $G$, is integral over $A$, and lies in $Q$ because one factor is $x$. Choose an exponent $e_0$ equal to $1$ in characteristic zero, or a power of $p$ in characteristic $p$, such that $h^{e_0}\in K$. Since $A$ is an integrally closed domain, $h^{e_0}\in A$. It also lies in $Q\cap A=\mathfrak{p}$, hence in $Q^{\prime }$. Because $Q^{\prime }$ is a prime ideal, some factor $\sigma (x)$ lies in $Q^{\prime }$. This means $x\in {\sigma }^{-1}(Q^{\prime })$, contradicting our choice. Thus $Q=\sigma (Q^{\prime })$ for some $\sigma$, proving transitivity. The power $e_0$ accounts for the possible purely inseparable field extension and makes the argument valid in every characteristic.

Lift the prescribed ${\mathfrak{q}}_2$ to a prime $Q_2$ of $C$ by the Lying-over theorem for $B\subseteq C$. Separately, choose a prime $Q_1^{\prime }$ over ${\mathfrak{p}}_1$ and use the Going-up theorem to find $Q_2^{\prime }\supseteq Q_1^{\prime }$ over ${\mathfrak{p}}_2$. Transitivity gives $\sigma \in G$ with $\sigma (Q_2^{\prime })=Q_2$. ThenContracting to $B$ gives ${\mathfrak{q}}_1=Q_1\cap B\subseteq {\mathfrak{q}}_2$ with the required contraction. This proves the going-down theorem for the module-finite case.

For a general integral extension $A\subseteq B$, consider $B_{{\mathfrak{q}}_2}$ and the idealWe claim that $J$ is disjoint from $A\setminus {\mathfrak{p}}_1$. Otherwise, clearing the finitely many denominators in an expression for an element $a\in J\cap (A\setminus {\mathfrak{p}}_1)$ givesThe subalgebra $B_0=A[s,b_1,\dots ,b_h]$ is a module-finite ring extension of $A$, since all its generators are integral elements. Apply the finite case to ${\mathfrak{q}}_2\cap B_0$: there is a prime $Q_1\subseteq {\mathfrak{q}}_2\cap B_0$ contracting to ${\mathfrak{p}}_1$. The displayed identity puts $sa$ in $Q_1$, but $a\notin Q_1$. Thus $s\in Q_1\subseteq {\mathfrak{q}}_2$, a contradiction.

Localize $B_{{\mathfrak{q}}_2}$ further at the multiplicative subset $A\setminus {\mathfrak{p}}_1$. The extension of $J$ is a proper ideal, by the disjointness just proved. Choose a maximal ideal containing it and contract back to a prime ideal of $B_{{\mathfrak{q}}_2}$. This prime contains $J$ and avoids $A\setminus {\mathfrak{p}}_1$. Its contraction ${\mathfrak{q}}_1$ to $B$ lies inside ${\mathfrak{q}}_2$ by the prime ideal correspondence for localization, and its contraction to $A$ is exactly ${\mathfrak{p}}_1$. This completes the proof of the going-down theorem without any finite generation assumption on $B$.

$\boxed{A_5\text{ is not a nontrivial free product}.}$ If $g\ne 1$ and $h\ne 1$ belong to different factors, the powers $(gh{)}^r$ have distinct normal forms of length $2r$. Every nontrivial free product is therefore an infinite group, whereas the alternating group $A_5$ is a finite group of order $60$.

The auxiliary Higman group $Q$ has no nontrivial finite quotients of a group. One can use this permitted property directly, but there is a short verification. In any finite quotient, let $n_i$ be the order of the image of $a_i$. Since conjugation sends $a_{i+1}$ to its square, $n_{i+1}$ must be odd. If some $n_i>1$, choose the least prime number $p$ dividing any $n_i$. It is odd. Conjugation by the preceding generator givesThe multiplicative order of $2$ modulo $p$ is therefore a divisor of $n_{i-1}$. It is greater than one and divides $p-1$ by the Fermat little theorem, so it has a prime divisor smaller than $p$. That prime also divides $n_{i-1}$, contradicting the choice of $p$. Hence every $n_i=1$ and the quotient is trivial.

Given any finitely presented group $P$, form $\hat{P}=P\ast ⟨v⟩$, and apply part (a) to the word $w=v$. This word is nontrivial by the normal form theorem for a free product, so the output $J=\hat{P}(v)$ contains $P$ by the embedding already proved.

To check its finite quotients, let $f:J\to F$ be any group homomorphism to a finite group. The image of the copy of $Q$ is trivial by the preceding property. The identifications $u_j=a_3^{j+1}{a}_1{a}_3^{-(j+1)}$ make all $f(u_j)$ trivial. Thus the equations $u_j^{-1}c{u}_j=g_j$ give, in the image,Since $f(c)=f(z)$, this forces $f(x_i)=1$ for all generators of $\hat{P}$, including $v$. The defining word $c=[v,z]$ then has trivial image, so $f(z)=f(d)=1$ as well. Every generator of $J$ has trivial image. ThereforeThis uses the concrete construction of part (a), not an assumption that arbitrary quotients preserve an embedding of $P$.

The Sylow theorems state that if a finite group has order $|G|=p^{a}m$ with $p$ prime and $p\nmid m$, it has a subgroup of order $p^a$, called a Sylow subgroup. Every p-group subgroup is contained in a Sylow subgroup, and all Sylow subgroups for the same prime are conjugate subgroups. Their number satisfiesMore precisely $n_p=[G:N_G(P)]$ for a Sylow subgroup $P$, where $N_G(P)$ is its normalizer.

Let $P$ be the averaging linear mapFor a unitary representation of a finite group, $P$ is the orthogonal projection onto the invariant subspace $V^G$, and $P\rho (h)=\rho (h)P=P$. Applying this to $\rho (g)v=\lambda v$ gives $Pv=\lambda Pv$. Since $\lambda \ne 1$, $Pv=0$. The vector $v$ is a cyclic vector for a group representation, so its translates span $V$, and every translate also has projection zero. Thus $P=0$ andThis is the cyclic eigenvector obstruction to invariant vectors. Unitarity also gives a direct proof by pairing $v$ with an invariant vector and then pairing every translate; the averaging proof makes the role of cyclicity especially transparent.

Write $x=g^{-1}h$ and $S(x)={\sum }_{\chi \in \hat{G}}\chi (x)$, where $\hat{G}$ is the character group of a finite abelian group. The Fundamental theorem of finitely generated abelian groups writes the finite group $G$ as a product of cyclic groups. A character of a cyclic group factor of order $r$ is determined by an arbitrary $r$th root of unity, so $|\hat{G}|=|G|$. This description also shows that the characters separate points: if $x\ne 1$, a nonzero coordinate of $x$ is detected by a character $\psi$ with $\psi (x)\ne 1$.

Multiplication by $\psi$ permutes $\hat{G}$, soFor $x\ne 1$ this forces $S(x)=0$. For $x=1$ every summand is one. Since $\chi (g{)}^{-1}\chi (h)=\chi (g^{-1}h)$, the character-sum cancellation lemma givesAll character values are roots of unity, so inversion here is also complex conjugation. This is the finite-abelian version of character orthogonality.

Write $q=p^r$, where $p=\mathrm{char}K$. Forthe derivative isso $h$ is a separable polynomial and all its roots in its splitting field $L$ are distinct.

Let $R$ be the set of these roots. The Frobenius endomorphism and the identitiesshow that $R$ is closed under addition and multiplication. It is also closed under additive inverses and, for $x\ne 0$, inversion. Thus $R$ is a finite field. Since $h$ splits into $q^n$ distinct linear factors, $|R|=q^n$. The field generated by all roots is therefore $R$ itself, so $L=R$ andHence the polynomial characterization of a finite field gives

Now let $f\in K[X]$ be irreducible of degree $n$, and let $\alpha$ be a root in an algebraic closure. The minimal polynomial of an algebraic element givesso $K(\alpha )$ has $q^n$ elements. Its nonzero elements form a finite group under multiplication, of order $q^n-1$; hence ${\alpha }^{q^n}=\alpha$. Therefore $\alpha$ is a root of $X^{q^n}-X$, and its minimal polynomial $f$ divides every polynomial over $K$ that vanishes at $\alpha$. Thus the irreducible factors of a finite-field Frobenius polynomial satisfy

Maschke's theorem says that if $H$ is a finite group and the characteristic of a field $F$ does not divide $|H|$, then every finite-dimensional $F$-representation of $H$ is semisimple.

To prove it, let $W$ be an invariant subspace of a representation $V$, and choose any linear projection $P:V\to W$. Average its conjugates:The scalar $|H{|}^{-1}$ exists by the characteristic assumption. Reindexing the sum shows that $\overline{P}$ commutes with every $\rho (g)$. Since each summand restricts to the identity on $W$, so does $\overline{P}$; hence $\overline{P}$ is an equivariant projection onto $W$. Its kernel is an invariant complement, and induction on $\dim V$ decomposes $V$ into irreducible representations.

The isometry group of $\mathbb{Z}$ is the infinite dihedral groupLet $\rho$ be a nonfaithful finite-dimensional complex representation. Its nontrivial normal kernel contains $t^m$ for some $m>0$, by the stated hint, so $\rho$ factors through the finite quotientMaschke's theorem decomposes this quotient representation into irreducibles. To bound their dimensions, choose a $t$-eigenvector $v$ with eigenvalue $\lambda$ in an irreducible constituent. The relation givesso $\mathrm{span}\{v,sv\}$ is invariant under both generators. Irreducibility makes it the whole constituent, whose dimension is therefore at most two. This proves the result on nonfaithful representations of the infinite dihedral group.

For the requested representation of $(\mathbb{Z},+)$, let its generator act on the polynomials of degree at most one by translation:In the basis $(1,x)$,This is a nontrivial Jordan block, so it is not diagonalizable. A direct sum of one-dimensional subrepresentations would make $T$ diagonalizable, proving that this nonsemisimple translation representation of the infinite cyclic group has the required property.

Finally, on $V=\mathbb{C}[x{]}_{\le 2}$ defineThen $S^2=I$ and $STS=T^{-1}$, so these operators define a quadratic-polynomial representation of the infinite dihedral group. In the basis $(1,x,x^2)$ they areHere $(T-I{)}^2\ne 0$ but $(T-I{)}^3=0$, so $T$ consists of one size-three Jordan block. If $V$ were a direct sum of subrepresentations of dimensions at most two, then $T$ would be block diagonal with Jordan blocks of sizes at most two, a contradiction. Thus this representation has the required failure of decomposition.

The product ${\prod }_j{G}_j$ of the finite groups with their discrete topology is a topological group under coordinatewise multiplication and inversion. The compatibility equations defining $G={\underleftarrow{\lim }}_j{G}_j$ are preserved by both operations, so $G$ is a subgroup. Their restrictions to the subspace topology on $G$ are continuous. Hence $G$ with its standard inverse-limit topology is a topological group.

Suppose first that $\Gamma$ is conjugacy separable. If $h\in \Gamma$ is not conjugate to $\gamma$ in $\Gamma$, some homomorphism to a finite group sends them to nonconjugate elements. This homomorphism factors through a finite quotient of $\hat{\Gamma }$, so $h$ cannot be conjugate to $\gamma$ in $\hat{\Gamma }$. Thus

Conversely, suppose this equality holds and $h$ is not conjugate to $\gamma$ in $\Gamma$. Then they are not conjugate in $\hat{\Gamma }$. By the finite-quotient criterion for conjugacy in a profinite group, their images fail to be conjugate in some finite quotient. This is precisely conjugacy separability.

The restriction form of the restriction branching rule for a symmetric group iswhere ${\lambda }^{-}$ contains the distinct partitions obtained by deleting one Removable node of a Young diagram. In particular, the restriction is multiplicity-free.

Restrict the alternating expressionThe supplied restriction formula for a Young permutation character, with $k=1$, says that each term restricts by subtracting one from each possible component. After collecting the alternating sums, this givesIf row $i$ has no removable node, part i straightens ${\psi }^{\lambda -{ϵ}_i}$ against the adjacent term with the opposite sign, or makes it zero when two shifted entries coincide. The surviving terms are exactly ${\psi }^{\mu }$ for $\mu \in {\lambda }^{-}$. Since $\lambda$ and each surviving $\mu$ are partitions, ${\psi }^{\lambda }={\chi }^{\lambda }$ and ${\psi }^{\mu }={\chi }^{\mu }$. We obtainComplex representations of a finite group are semisimple, so equality of characters proves the asserted module decomposition.

A group action of $G$ on $X$ is a map $(g,x)\mapsto gx$ such that $ex=x$ and $(gh)x=g(hx)$. The orbit and stabilizer of $x$ areFor a finite group, the mapis a well-defined bijection from the left cosets of $G_x$ to $Gx$. Each coset has $|G_x|$ elements, so the orbit-stabilizer theorem is

The Cauchy theorem for groups says that if a prime number $p$ divides $|G|$, then $G$ has an element of order $p$. To prove it, letThe first $p-1$ entries determine the last, so $|X|=|G{|}^{p-1}$, which is divisible by $p$. The cyclic group $C_p$ acts on $X$ by cyclically rotating the entries; rotation preserves the product condition becauseEvery orbit has size one or $p$. The fixed points are exactly the tuples $(g,\dots ,g)$ with $g^p=e$. Their number is therefore divisible by $p$. Since the identity gives one fixed point, there is another, and its entry has order $p$.

Now let $|G|=33$. Cauchy's theorem gives a subgroup $H$ of order $11$. Let $H$ act by the conjugation action on the set $\mathcal{X}$ of subgroups of order $11$. It fixes $H$. If it also fixed $K\ne H$, then $H$ would normalize $K$; since $H\cap K=\{e\}$, the product $HK$ would be a subgroup of order $121$, which is impossible. Every other $H$-orbit in $\mathcal{X}$ therefore has size $11$, soDistinct members of $\mathcal{X}$ share only the identity and each contributes ten nonidentity elements. Hence $1+10|\mathcal{X}|\le 33$, forcing $|\mathcal{X}|=1$. Thus $H$ is a normal subgroup.

Conjugation now defines a group homomorphismBecause $H\cong C_{11}$, its automorphism group has order $10$. By the Lagrange theorem, the image has order dividing both $33$ and $10$, so the image is trivial and $H$ lies in the center of $G$. Cauchy's theorem also supplies $x$ of order $3$. If $h$ generates $H$, then $h$ and $x$ commute and $hx$ has order $\mathrm{lcm}(11,3)=33$. Therefore

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

A group $G$ is residually finite when, for every $1\ne g\in G$, there are a finite group $Q$ and a group homomorphism $q:G\to Q$ such that $q(g)\ne 1$. Equivalently, the intersection of all finite-index normal subgroups of $G$ is trivial.

Use the injective group homomorphism $F_2\hookrightarrow {\mathrm{SL}}_2(\mathbb{Z})$ from part (a). If $1\ne g\in F_2$, its image is an integral matrix $M\ne I$. Choose a prime number $p$ that does not divide one nonzero entry of $M-I$. The reduction modulo a prime in an integral matrix group homomorphismthen sends $M$ to a nonidentity element. Its target is a finite group, so the composite map separates $g$ from the identity. Hence $F_2$ is a residually finite group.

Consider the principal congruence subgroupIt has finite index because $S{L}_2({\mathbb{F}}_3)$ is a finite group. It is torsion-free: by part (b), a finite-order element is conjugate to $A^i$ or $B^j$, while the reductions of $A$ and $B$ modulo $3$ still have orders $6$ and $4$, respectively. Hence no nonidentity power in either vertex group reduces to the identity.

It follows that $\Gamma (3)$ intersects every conjugate of the two vertex stabilizers trivially, so its action on the Bass-Serre tree is free. A group with a free group action on a tree is a free group. Consequently $S{L}_2(\mathbb{Z})$ contains the free subgroup $\Gamma (3)$ of finite index; equivalently, it is a virtually free group.

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

For a finite group with trivial centre of a group and a prime divisor $p$ of its order, some nonidentity conjugacy class has size coprime to $p$. Otherwise the class equation would give $|G|\equiv 1(\mathrm{mod}p)$. Primality is essential: in the symmetric group $S_3$, neither nonidentity class size is coprime to the composite divisor $6$.