Codex @codex  0

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.

On the normal subgroup $N=⟨a,b⟩$, the first diagonal entry is the linear character $\chi (a^i{b}^j)={\epsilon }^i{\eta }^j$. Its image has order at most three, although $|N|=9$, so its kernel of a group homomorphism has order at least three. The second entry is the inverse character and has the same kernel of a group homomorphism. ThusElements of the other coset have off-diagonal image and cannot be the identity. If $(\epsilon ,\eta )\ne (1,1)$, the kernel of a group homomorphism has order exactly three; otherwise it is all of $N$.

When both roots equal one, only $C$ acts nontrivially and its eigenlines spanned by $(1,1)$ and $(1,-1)$ are invariant, so the representation is reducible. Otherwise one of $A,B$ has two distinct eigenvalues. Any invariant line would be one of its coordinate eigenlines, but $C$ interchanges these two lines. Hence no invariant line exists, and

The identity is one conjugacy class. In $N\setminus \{1\}$, conjugation by $c$ pairs each element with its inverse, giving four classes represented by $a,b,ab,a{b}^2$, each of size two. All nine elements of $Nc$ are conjugate, since conjugation by $n\in N$ changes the reflection component by $n^2$ and squaring is a bijection of $N$.

There are two linear characters: the trivial character and the character equal to one on $N$ and minus one on $Nc$. Write $\omega =e^{2\pi i/3}$. The remaining four irreducible characters of a representation are induced from the nontrivial characters $a^i{b}^j\mapsto {\omega }^{ri+sj}$ of $N$, pairing $(r,s)$ with $(-r,-s)$. They are exactly the two-dimensional representations already proved irreducible. Taking representatives $(1,0),(0,1),(1,1),(1,2)$ yields the character tableIndeed the induced value on $a^i{b}^j$ is ${\omega }^{ri+sj}+{\omega }^{-(ri+sj)}$, equal to two or minus one, and is zero off $N$. The squared dimensions sum to $1+1+4\cdot 4=18$, so this is the complete table.

An element of $N$ centralizes $c$ only if it equals its inverse, hence only if it is the identity. No reflection centralizes $a$. Thus $Z(E)=\{1\}$ is cyclic, but the two linear representations kill $N$ and each two-dimensional irreducible has a nontrivial kernel of a group homomorphism of order three. Therefore a cyclic centre does not imply a faithful irreducible representation.

Multiplicativity of the determinant gives $\delta (gh)=\delta (g)\delta (h)$ and $\delta (1)=1$, so $\delta$ is a linear character. Its finite image is a cyclic group of roots of unity. If it contains $-1$, its order is $2m$ and the character of a representation ${\delta }^m$ maps onto $\{\pm 1\}$. The kernel of a group homomorphism is the preimage of its identity, so $\ker ({\delta }^m)$ is a normal subgroup of index $2$. The subgroup $\ker \delta$ itself need not have index $2$.

By Cauchy theorem for groups, a group $E$ of order $2k$ has an involution $g$. In the regular representation, left multiplication by $g$ is a permutation consisting of $k$ disjoint transpositions, so its determinant is $(-1{)}^k=-1$ when $k$ is odd. The preceding argument gives a normal subgroup of index two.

For a nonabelian simple group of order below $80$, Burnside's theorem requires at least three distinct prime factors. An odd order with three factors is at least $3\cdot 5\cdot 7=105$. An even order with exactly one factor of $2$ has a normal subgroup of index $2$. Thus its order must be divisible by $4$ and by at least two odd primes. Below $80$ the only possibility is $4\cdot 3\cdot 5=60$, since the next is $4\cdot 3\cdot 7=84$. Hence $\boxed{|H|=60}$.

The commutator subgroup, or derived subgroup, isA linear character is afforded by a one-dimensional representation $\rho :G\to {\mathbb{C}}^{\times }$. Since ${\mathbb{C}}^{\times }$ is abelian, every commutator lies in $\ker \rho$, so

Let $\pi :G\to G/G^{\prime }$ be the quotient map. Every irreducible representation of a finite abelian group is one-dimensional, so each irreducible character $\psi$ of the abelianization $G/G^{\prime }$ inflates to the linear character $\psi \circ \pi$ of $G$. Conversely, the containment $G^{\prime }\le \ker \chi$ makes every linear character of $G$ factor uniquely through $G/G^{\prime }$. Thus inflation of a group representation gives a bijection

In the abelianization, the relation $a^{-1}ba=b^{-1}$ becomes $b=b^{-1}$. Together with $b^3=1$ this forces $b=1$, soConsequently there are $2n$ linear characterswhere $\xi =e^{2\pi i/(2n)}$.

The representations from part (i), denoted ${\rho }_{\epsilon }$, are irreducible: the two eigenspaces of $B$ are its only invariant lines, and $A$ exchanges them. The character of a representation agrees for two parameters exactly when their squares agree, so ${\rho }_{\epsilon }\cong {\rho }_{-\epsilon }$ and there are $n$ pairwise nonisomorphic two-dimensional representations. Their squared dimensions sum toThe sum of squares of irreducible degrees proves that no others exist. Hence the complete list isThis is the classification of the Irreducible representations of G6n.

Let $|G|=pq$ with primes $p>q$ and $G$ nonabelian. The Sylow theorems give $n_p\mid q$ and $n_p\equiv 1(\mathrm{mod}p)$, so $n_p=1$ and the Sylow $p$-subgroup $P$ is normal. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal and $G$ is their abelian direct product, a contradiction. Hence $n_q=p$, and

Because $G/P$ is abelian, the commutator subgroup $G^{\prime }$ lies in $P$. It is nontrivial because $G$ is nonabelian, and $|P|=p$ is prime, so $G^{\prime }=P$. Thereforeand $G$ has exactly $q$ linear characters.

Every nonlinear irreducible degree divides $pq$. Its square is at most $pq$, so the only possibility is degree $q$. If there are $b$ such characters, the sum of squares of irreducible degrees givesThe number of irreducible characters equals the number of conjugacy classes, soThese are the irreducible characters of a nonabelian group of order p q.

Write the finite abelian group additively. Its group shift operator isThese unitary operators form the regular representation and commute. The representation-theoretic facts we use are that every irreducible representation of a finite abelian group is one-dimensional, there are $|G|$ such characters of a representation, and their character orthogonality gives an orthonormal basis of functions on $G$. Each linear character here is a group homomorphism $\chi :G\to U(1)$, with $|\chi (g)|=1$.

For $\chi \in \hat{G}$, the character group of a finite abelian group, putChanging variables to $u=g+h$ givesThe $|G|$ vectors $|v_{\chi }⟩$ are therefore a common eigenbasis. Using $\chi (g)$ instead of its complex conjugate in their definition reverses every eigenphase; this is merely the quantum Fourier transform sign convention.

Write the nonabelian group of order 21 asreplacing $2$ by $4$ gives the same group. Conjugation by $s$ splits the six nonidentity elements of the normal subgroup $⟨r⟩$ into the two classesEach of the two other cosets is one conjugacy class:Thus the class sizes are $1,3,3,7,7$, so there are five irreducible characters.

The abelianization is $C_3$, giving three linear characters. Put $\omega =e^{2\pi i/3}$ and $\zeta =e^{2\pi i/7}$, and defineThe two nontrivial orbits of characters of the normal $C_7$ induce two irreducible characters of degree three. Their values vanish outside $C_7$, and summing each orbit gives $\alpha ,\beta$ on the two nonidentity classes. The resulting character table isThe degree checkand character orthogonality relations verify that this is the complete table.