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For a finite group, the regular representation contains $d_{\rho }$ copies of each irreducible representation $\rho$. Its character of a representation is $|G|$ at the identity and zero elsewhere. Thus ${\sum }_{\rho }{d}_{\rho }{\chi }_{\rho }(x)=|G|1_{\{x=e\}}$, including the trivial representation. This formula converts independent uniform expectations into constrained configuration counts.

The Fourier transform on a finite group convention $\hat{f}(\rho )={\mathbb{E}}_{x}f(x)\rho (x)$ has the displayed inversion formula. The sum is over one representative from each equivalence class of unitary irreducible representations. It follows from the Schur orthogonality relations and the regular representation decomposition, and holds at every group element without a limiting argument.

The permutation matrices of the three displayed double transpositions, together with $I_4$, form a faithful representation of the Klein four-group. They satisfy $P^2=Q^2=I_4$ and $PQ=QP$. Every nonidentity element fixes no basis vector, so this is its four-dimensional regular representation.

The constant vector $(1,1,1,1{)}^T$ spans a nonzero proper invariant subspace, proving reducibility. More completely, the columns ofform a common orthonormal basis of eigenvectors. Direct multiplication gives $S^{T}PS=\mathrm{diag}(1,1,-1,-1)$ and $S^{T}QS=\mathrm{diag}(1,-1,1,-1)$. Thus the representation splits into the four distinct real one-dimensional characters, each once. The basis change is a normalized Hadamard matrix.

For a Frobenius complement $H$, a class function $\theta$ with $\theta (1)=0$ satisfies $({\theta }^G{)}_H=\theta$. For each nontrivial $\eta \in \mathrm{Irr}(H)$ of degree $d$, the virtual character $(\eta -d{1}_H{)}^G+d{1}_G$ has norm one and positive degree by Frobenius reciprocity. It is therefore irreducible, restricts to $\eta$ and is identically $d$ on the candidate kernel $N$. The intersection of the kernels of these characters is exactly $N$: outside $N$, separation follows from the character of the regular representation of $H$. This proves normality; $N\cap H=1$ and $|N|=[G:H]$ then give the semidirect product.

For a finite abelian group $G$, $U(h)|g⟩=|g+h⟩$ defines the regular representation on the basis labelled by $G$. The linear characters give the common eigenbasisCharacter orthogonality proves orthonormality, and changing variables by the translation in a group proves the eigenvalue formula.

Embed $C_4\times C_4$ and $Q_8\times C_2$ in $S_{16}$ by their regular representations. Both have one element of order one, three of order two, and twelve of order four. An order-$r$ element in a regular action has cycle type $r^{16/r}$. Their intersections with every conjugacy class of $S_{16}$ consequently have the same size, so they are Gassmann equivalent. The first group is abelian and the second is not, so their underlying groups are not isomorphic. Finite-cover constructions from the Sunada theorem therefore give isospectral manifolds with different fundamental groups.

The defining SU(3) representation restricts to the given matrix representation. All its restricted matrices are real, so the conjugate representation restricts identically. Hence the first two branchings areFor the adjoint representation of SU(3), use $3\otimes \overline{3}=1\oplus 8$. This follows by splitting endomorphisms into scalar and traceless matrices. Subtracting one trivial constituent from c givesAs an independent check, its restricted character is $|{\chi }_R{|}^2-1=(8,0,0,0,0)$, the regular character of $G$. Thus this is precisely the regular representation, whose representation multiplicities equal the corresponding dimensions.

Only the character ${\chi }_2$ of the two-dimensional irreducible representation remains unknown. Write it as $(2,a,b,c,d)$. The regular representation has character zero off the identity and decomposes with multiplicities equal to the irreducible representation dimensions. Thus, for a nonidentity element $g$,For the four nonidentity conjugacy classes this gives $a=0$, $b=2$, $c=-1$, $d=0$. The complete character table is consequentlyFor example, $⟨{\chi }_2,{\chi }_2⟩=(4+12+8)/24=1$, agreeing with character orthogonality.

For type $(1^n)$, all labels are distinct, so a semistandard tableau is precisely a standard Young tableau. Thus $K_{\lambda ,(1^n)}=f^{\lambda }$, and Young's rule givesThis is also the decomposition of the regular representation, since tabloids with one entry per row are naturally indexed by all permutations.

Let $f^{\lambda }$ be the number of standard tableaux of shape $\lambda$. First the associated polytabloids are linearly independent. Order tabloids lexicographically by reading each row after sorting its entries. For a column-increasing tableau $t$, the term $\{t\}$ in $e_t$ has coefficient one; every other tabloid term has larger row word. To see this, take the first row changed by a nontrivial column permutation. Earlier rows are fixed entry by entry, since each entry belongs to one original column. The changed row can only receive entries from lower positions of those increasing columns, so after sorting it is lexicographically larger. In a linear relation between standard polytabloids, choose the smallest row word with a nonzero coefficient. Its own tabloid coefficient cannot be cancelled by a larger standard polytabloid, a contradiction. Thus $\dim S^{\lambda }\ge f^{\lambda }$.

Work first over $\mathbb{C}$. The RSK bijection between permutations and pairs of standard tableaux of the same shape gives ${\sum }_{\lambda ⊢n}(f^{\lambda }{)}^2=n!$. The assumed pairwise nonisomorphic irreducible Specht modules, and the decomposition of the regular representation with irreducible multiplicities equal to their dimensions, give ${\sum }_{\lambda }(\dim S^{\lambda }{)}^2\le n!$. Combining the two inequalities forces equality term by term: $\dim S^{\lambda }=f^{\lambda }$. Hence the independent standard polytabloids are a basis.

The same conclusion holds over every characteristic-zero field. All polytabloid coordinates in the tabloid basis are integers, so the ranks of their coordinate matrices are the same over any characteristic-zero field as over $\mathbb{C}$. In particular the dimension equality and independence transfer without requiring an embedding of the whole field into $\mathbb{C}$.

The equality ${\sum }_{\lambda }(\dim S^{\lambda }{)}^2=n!$ exhausts the dimension of the regular representation. There is no room for another irreducible constituent, since it would contribute its positive squared dimension. ThusEach has the standard-polytabloid basis just proved. The usual split characteristic-zero statement follows over any characteristic-zero field because these integral Specht models remain irreducible after extension to an algebraic closure and already exhaust all dimensions there. In the literal wording of the source, polytabloids themselves are vectors; it is the Specht modules with those bases that form the family of representations.

If $A$ is empty, take the zero generalized character; if $B$ is empty, take the trivial character. Otherwise let $\pi$ be the set of primes dividing the orders of elements of $A$. The coprimality hypothesis makes every element of $B$ a ${\pi }^{\prime }$-element. Thus every nonidentity element of $G$ is either a pi-element in $A$ or a complementary-prime element in $B$. In particular these sets are unions of conjugacy classes, since conjugation preserves element order.

Put $m=|G{|}_{\pi }$ and $n=|G{|}_{{\pi }^{\prime }}$. Their greatest common divisor is one, so the Chinese remainder theorem supplies an integer $d$ such thatDefine the class function $f$ by $f(1)=d$, $f=1$ on $A$ and $f=0$ on $B$. We verify Brauer's characterization of characters directly.

For an elementary subgroup $E=P\times C$, the cyclic factor can be split into its $\pi$-part and ${\pi }^{\prime }$-part. Attach $P$ to the part containing its prime. This expresses $E=E_{\pi }\times E_{{\pi }^{\prime }}$. Both factors cannot be nontrivial: choosing a nonidentity element in each would give a commuting product of mixed prime order, whereas every element of $G$ has order involving only one of the two prime sets. Hence $E$ is wholly a $\pi$-group or wholly a ${\pi }^{\prime }$-group.

Let $r_E$ be the character of the regular representation of $E$, with value $|E|$ at the identity and zero elsewhere. In the two cases respectively,The coefficients are integers because $|E|$ divides $m$ or $n$. Thus both are generalized characters. The trivial subgroup causes no difficulty since $d$ is an integer. By Brauer's characterization of characters, $f$ is a generalized character of $G$.

This is a generalized character separating complementary prime elements; the freely chosen value at the identity is what makes all elementary restrictions integral.

Let $K$ be the intersection under consideration. The family is nonempty since it contains $G$, and is invariant under conjugation: conjugating an irreducible inducing pair $(H,\phi )$ gives another such pair. Hence $K$ is a normal subgroup of $G$.

For any $\chi \in \mathrm{Irr}(G)$, monomiality gives $\chi ={\lambda }^G$ with $\lambda$ a linear character of some $H$. This $H$ belongs to the family, so $K\le H$. Because $K$ is normal, $K\le tH{t}^{-1}$ for every $t\in G$. In the induced representation oneach coset line $t\otimes {\mathbb{C}}_{\lambda }$ is therefore invariant under $K$: for $k\in K$,Thus the restriction to $K$ is a direct sum of one-dimensional group representations. All representing matrices of $K$ commute, and the derived subgroup $K^{\prime }$ acts as the identity. Equivalently,The intersection of the character kernels is trivial: the regular representation is faithful and is a direct sum of irreducible representations. It follows that $K^{\prime }=1$.

This proves the abelian intersection of irreducible inducing subgroups without assuming that every subgroup of an M-group is itself monomial.

The translation subgroup $U$ is normal and $G/U\cong {\mathbb{F}}_p^{\times }$, via the upper-left entry $a$. Thus the coset representation is the regular representation of that cyclic quotient. If $\chi$ runs through its $p-1$ complex one-dimensional characters, inflated to $G$, thenEach occurs once: the regular character is $p-1$ at the quotient identity and zero elsewhere, which is also the sum of all quotient characters by finite character orthogonality.

Put $w_j=|C_j|/|G|$. The character orthogonality for the five supplied rows say ${\sum }_j{w}_j{\chi }_r(C_j){\chi }_s(C_j)={\delta }_{rs}$. Solving these linear equations givesFor example orthogonality of the two degree-four rows gives $16w_1+w_3+w_4-4w_5-w_6=0$, and their norms give $16w_1+w_3+w_4+4w_5+w_6=1$; the remaining row products determine the other weights. Since $C_1=\{e\}$, $w_1=1/|G|$, so $|G|=120$ and the class sizes are the displayed numerators.

Tensoring the fifth irreducible representation with the second, one-dimensional one gives another irreducible character, with row $(5,1,-1,0,-1,-1,1)$. It differs from all five known rows. There are seven conjugacy classes, hence seven irreducible characters. The sum of squared degrees is $|G|$, so the final degree is $\sqrt{120-(1+1+16+16+25+25)}=6$.

The regular representation has character zero off the identity and is the sum of irreducible characters weighted by their degrees. Applying this column by column determines the last row. The completed character table, in the supplied class order, isNo identification of the group with a familiar group is needed.

For a complex irreducible representation, every central element acts as a scalar by Schur lemma. If the representation is faithful, these scalars embed $Z$ in ${\mathbb{C}}^{\ast }$. Every finite subgroup of ${\mathbb{C}}^{\ast }$ is cyclic: if its order is $m$, all its elements satisfy $z^m=1$, and the $m$th roots of unity form a cyclic group. Thus $Z$ is cyclic.

The trivial group satisfies the conclusion immediately. Otherwise let $G$ be a finite $p$-group with cyclic centre, and decompose a faithful representation as a direct sum of irreducibles using Maschke's theorem. The nontrivial cyclic $p$-group $Z$ has a unique subgroup $C$ of order $p$. Some component does not kill $C$, since otherwise the whole representation would kill it. If that component had a nontrivial kernel $N$, the kernel would be a normal subgroup and the allowed intersection theorem gives $N\cap Z\ne 1$. Every nontrivial subgroup of cyclic $Z$ contains $C$, contradicting the choice of component. That irreducible component is faithful.

Finally the regular representation is faithful, since left multiplication by a nonidentity element does not fix the basis vector at the identity. A cyclic centre therefore supplies a faithful irreducible component of it. The first argument proves the converse, giving a finite $p$-group has a faithful irreducible complex representation exactly when its centre is cyclic.

Take a finite presentation $T=⟨x_1,\dots ,x_r\mid R_1,\dots ,R_s⟩$. For a closed-manifold realization of a finitely presented fundamental group, begin with a five-dimensional ball and attach $r$ one-handles, giving a compact oriented five-manifold $W_1$ with fundamental group the free group on the $x_i$. Represent the relators by disjoint embedded circles in its four-dimensional boundary. General position allows them to be embedded and disjoint, and their oriented normal bundles admit framings. Attach a two-handle along each such circle. The resulting compact five-manifold $W$ hasby the Seifert-van Kampen theorem. Now set $B=\partial W$. It is a compact connected smooth four-manifold without boundary. Turning this handle decomposition upside down gives a relative handle decomposition of $(W,B)$ with handles of indices three, four and five only. Attaching such handles does not change the fundamental group: there are no new one-handles or two-handle relations. Thus ${\pi }_1(B)\to {\pi }_1(W)$ is an isomorphism andThese are the differential-topological handle facts used in the construction; the dimension-five thickening ensures the boundary retains the prescribed fundamental group.

Now assume $T$ finite. Choose any smooth Riemannian metric on $B$ and lift it to its universal cover $\tilde{B}$. This cover is compact, since its degree is $|T|$, and simply connected. Its deck transformations give a free isometric $T$-action. For the two Gassmann equivalent subgroups putEach is a closed Riemannian manifold, and $\tilde{B}$ is its simply connected universal cover, with deck transformation group $U_i$. Consequently ${\pi }_1(M_i)\cong U_i$.

Here is the complete spectral argument behind the Sunada theorem in this situation. Each finite-dimensional upstairs eigenspace $E_{\lambda }$ is a representation of $T$, because the action is isometric and commutes with the Laplace-Beltrami operator. Functions downstairs correspond exactly to invariant functions upstairs, so their multiplicities are $\dim E_{\lambda }^{U_i}$. Averaging the representation matrices over a subgroup is a projection onto its invariant subspace. Taking its trace givesThe character ${\chi }_{\lambda }$ is constant on each conjugacy class of $T$. Gassmann equivalence means $|U_1\cap C|=|U_2\cap C|$ for every such class $C$; summing these equalities also gives $|U_1|=|U_2|$. The two averages are therefore equal for every $\lambda$. The quotients are isospectral, with ${\pi }_1(M_i)\cong U_i$.

If $U_2=t{U}_1{t}^{-1}$, the map $[x{]}_{U_1}\mapsto [tx{]}_{U_2}$ is well defined, invertible and isometric because $t$ acts by an isometry upstairs. Conjugate subgroups give isometric quotients.

For an explicit example with different fundamental groups, use the order-sixteen Gassmann pairEmbed each in $T=S_{16}$ by left multiplication on its own sixteen elements, identified with the same sixteen-point set. These regular representations are faithful. Both groups have one identity element, three elements of order two, and twelve of order four: in $C_4^2$ the elements of order dividing two are precisely its four even-coordinate elements; in $Q_8\times C_2$ they are the four elements in $\{1,-1\}\times C_2$. Every remaining element has order four.

In a regular action, an order-$k$ element is a product of $16/k$ disjoint $k$-cycles. Thus each subgroup intersects the identity conjugacy class of $S_{16}$ in one element, the class of type $2^8$ in three, the class of type $4^4$ in twelve, and every other class in zero. They are Gassmann equivalent. However $H_1$ is abelian and $H_2$ is nonabelian, so they are not isomorphic. Apply the preceding construction with fundamental group $S_{16}$ and these subgroups. It producesIn particular they cannot even be homeomorphic, and therefore cannot be isometric.

We prove closure of $N$ through character kernels, rather than assuming that a conjugacy-invariant set is a subgroup. Let $\eta \in \mathrm{Irr}(H)\setminus \{1_H\}$, write $d=\eta (1)$, and define the generalised characterHere $1_H,1_G$ are the trivial characters. Since $(\eta -d{1}_H)(1)=0$, part (i) gives $({\Psi }_{\eta }{)}_H=\eta$ and ${\Psi }_{\eta }(1)=d$. Put $\vartheta =\eta -d{1}_H$. By Frobenius reciprocity and character orthogonality,ConsequentlyTo justify the resulting irreducibility, write any generalised character as an integer combination ${\sum }_{\chi }{a}_{\chi }\chi$ of irreducible characters. Its squared norm is $\sum a_{\chi }^2$, so norm one makes it $\chi$ or $-\chi$ for one irreducible $\chi$. Positive degree rules out the negative sign. Thus ${\Psi }_{\eta }$ is an irreducible character, an instance of the fact that positive-degree norm-one virtual characters are irreducible.

If $x\in N\setminus \{1\}$, no conjugate of $x$ lies in $H$, so the induction formula gives ${\vartheta }^G(x)=0$. At $1$ it is also zero. Hence ${\Psi }_{\eta }(x)=d$ for all $x\in N$. For a finite-group representation we may choose an invariant positive Hermitian form, making every representing matrix unitary. Its trace equals its dimension only when all its unit-modulus eigenvalues are $1$. Thus the kernel of a character is $\{g:{\Psi }_{\eta }(g)={\Psi }_{\eta }(1)\}$, and $N\subseteq \ker {\Psi }_{\eta }$.

Conversely, if $g\notin N$, then $g$ is conjugate to some $h\in H\setminus \{1\}$. The character of the regular representation of $H$ isIf every nontrivial $\eta$ satisfied $\eta (h)=\eta (1)$, the sum at $h$ would instead equal $\sum \eta (1{)}^2=|H|$, a contradiction. Choose an $\eta$ for which these values differ. Since ${\Psi }_{\eta }$ restricts to $\eta$, it has ${\Psi }_{\eta }(g)=\eta (h)\ne d$, so $g\notin \ker {\Psi }_{\eta }$. ThereforeThis proves the Frobenius kernel theorem's subgroup assertion. The definition also gives $N\cap H=\{1\}$. Using the permitted cardinality $|N|=[G:H]$, normality ensures $NH$ is a subgroup andThus $N$ is the Frobenius kernel, and $H$ supplies the complementary factor.

In character theory, a p-elementary group is a finite group isomorphic to $P\times C$, where $P$ is a $p$-group and $C$ is cyclic of order coprime to $p$. Either factor may be trivial. An elementary group is p-elementary for at least one prime $p$; it need not be an elementary abelian group.

A generalised character means an integer linear combination of irreducible complex characters. Brauer's characterisation of characters states that a complex class function $f$ on a finite group $G$ is a generalised character if and only if $f_E$ is a generalised character for every elementary subgroup $E\le G$. Equivalently,An ordinary character additionally requires nonnegative coefficients in its global irreducible decomposition. We use the virtual-character criterion just stated.

For the extension construction, set $d=\theta (1)$. The determinant character $\det \theta$ is linear. Since its extension $\mu$ has the same value $1$ at the identity, $\mu$ is also a linear character. For any $g\in G$, the subgroup $J_g=N⟨g⟩$ has cyclic quotient $J_g/N$, hence a solvable quotient. The invariant character $\theta$ has degree coprime to $[J_g:N]$, since this index divides $[G:N]$, and ${\mu }_{J_g}$ extends its determinant. The permitted coprime-degree determinant extension theorem therefore gives a unique character ${\chi }_g\in \mathrm{Irr}(J_g)$ withDefine a function on $G$ by $f(g)={\chi }_g(g)$. We next verify that these individually defined values form a compatible class function.

If $x\in G$, conjugation carries $J_g$ to $J_{xg{x}^{-1}}$. Transporting ${\chi }_g$ by this conjugation gives a character extending $\theta$, because $\theta$ is $G$-invariant, and with determinant $\mu$ restricted to the conjugate subgroup, because $\mu$ is a class function. Uniqueness identifies the transported character with ${\chi }_{xg{x}^{-1}}$. Consequently $f(xg{x}^{-1})=f(g)$.

Now take an elementary subgroup $E\le G$ and let $J=NE$. A finite $p$-group is solvable, a cyclic group is solvable, and a direct product of solvable groups is solvable. Thus $E$ is solvable, and so is $J/N\cong E/(E\cap N)$. Again the index $[J:N]$ divides $[G:N]$, so the same extension theorem gives ${\chi }_J\in \mathrm{Irr}(J)$ extending $\theta$ with determinant ${\mu }_J$.

For $g\in E$, we have $J_g\le J$. The restriction $({\chi }_J{)}_{J_g}$ is irreducible: an invariant subspace for $J_g$ would be invariant for $N$, whose restriction already affords the irreducible $\theta$. Its determinant is ${\mu }_{J_g}$. By uniqueness on $J_g$, it equals ${\chi }_g$. Thereforean ordinary character and hence a generalised character of $E$. Brauer's characterisation of characters now makes $f$ a generalised character of $G$. If $n\in N$, then $J_n=N$ and ${\chi }_n=\theta$, so $f(n)=\theta (n)$. We have proved, by gluing determinant-normalized character extensions,Only the intermediate quotients $J_g/N$ and $NE/N$ were required to be solvable; no solvability assumption on $G/N$ has been added.

For the prime-set assertion, let $E=P\times C$ be p-elementary, and split the cyclic group as $C=C_{\pi }\times C_{{\pi }^{\prime }}$, where each factor has the indicated prime divisors in its order. If $p\in \pi$, put $E_{\pi }=P\times C_{\pi }$ and $E_{{\pi }^{\prime }}=C_{{\pi }^{\prime }}$. If $p\notin \pi$, put $E_{\pi }=C_{\pi }$ and $E_{{\pi }^{\prime }}=P\times C_{{\pi }^{\prime }}$. In both cases the factors commute, have trivial intersection and have orders with disjoint prime spectra. This proves the prime-set decomposition of elementary groupsHere a pi-group has order divisible only by primes in $\pi$, and a pi-element has such an element order; the identity qualifies for both complementary prime sets.

Finally, suppose $K$ satisfies the given element-order condition. The sets $A,B$ are unions of conjugacy classes and are disjoint, because membership in both would force every prime divisor of the element order into $\pi \cap {\pi }^{\prime }=\varnothing$, hence force order $1$. Let $E\le K$ be elementary and use $E=E_{\pi }\times E_{{\pi }^{\prime }}$. If both factors were nontrivial, choose $a\ne 1$ in the first and $b\ne 1$ in the second. They commute and have coprime orders, so $ab$ has order $|a||b|$, containing primes from both $\pi$ and ${\pi }^{\prime }$. It would belong to neither $A$ nor $B$ nor $\{1\}$, contrary to the hypothesis. Thus every elementary subgroup is entirely a pi-group or entirely a complementary-prime group.

Write $m=|K{|}_{\pi }$ and $r=|K{|}_{{\pi }^{\prime }}$ for the two prime parts of $|K|$, so $|K|=mr$ and $\gcd (m,r)=1$. The Chinese remainder theorem supplies an integer $D$ satisfyingDefine the class functionFor an elementary $\pi$-subgroup $E$, its order divides $m$. Let ${\rho }_E$ be the character of its regular representation, equal to $|E|$ at $1$ and zero elsewhere. Thenwith integral coefficient because $D\equiv 1(\mathrm{mod}|E|)$. For an elementary ${\pi }^{\prime }$-subgroup, its order divides $r$, andagain has integral coefficient. The trivial subgroup satisfies either formula. Every elementary restriction is therefore a generalised character. A second application of Brauer's characterisation of characters provesThe freely chosen identity value, fixed by the two congruences, makes the restrictions integral. This construction also covers the cases $\pi =\varnothing$, ${\pi }^{\prime }=\varnothing$, or $K=\{1\}$, using modulus-one congruences where appropriate.

The three quadratic subfields are $\mathbb{Q}(\sqrt{-2})$, $\mathbb{Q}(\sqrt{-3})$ and $\mathbb{Q}(\sqrt{6})$, with respective fundamental discriminants $-8,-3,24$. Their associated quadratic Dirichlet characters are the Kronecker symbolsThe regular representation of $C_2\times C_2$ is the direct sum of the trivial character and these three nontrivial characters. At an unramified prime the character is the sign by which Frobenius automorphism acts on the square root defining its quadratic subfield; at a ramified prime the linear character has no inertia group invariants. The local factor of an Artin L-function consequently givesThe zero values of a quadratic Dirichlet character at conductor primes mean that its local factor is one, not that the whole Artin L-function vanishes there.

Only the rational primes $2,3,5,7$ can contribute to coefficients with index at most ten. Their character values areFor the odd unramified primes these entries follow by checking squares: modulo five, both $-8$ and $-3$ are the nonsquare $2$, while $24$ is the square $4$; modulo seven they are respectively $6,4,3$. At $3$, $-8\equiv 1$ is a square. At $2$, the $-3$ character is $-1$ because the integral quadratic generator has polynomial $X^2+X+1$, irreducible modulo two. The other entries at $2$ and $3$ are zero exactly where the corresponding fundamental discriminant is divisible by that prime. This verifies every local factor, including the ramified ones.

The Euler product makes $a_n$ a multiplicative arithmetic function. Expanding the first two factors givesTherefore $a_2=0$, $a_4=1$, $a_8=0$, $a_3=2$ and $a_9=3$. Both factors for $5$ and $7$ have zero coefficient of $T$, so $a_5=a_7=0$. Multiplicativity gives $a_6=a_2{a}_3=0$ and $a_{10}=a_2{a}_5=0$, while $a_1=1$. The answer isAs an arithmetic check, the regular representation Artin L-function is the Dedekind zeta function of $F$. The local factors say there is one prime of norm four above two and two primes of norm three above three. Their powers give respectively one ideal of norm four and three ideals of norm nine, matching the nonzero coefficients above.