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The complex regular representation has basis $(e_h{)}_{h\in G}$ and left actionIf $\rho (g)$ is the identity, thenso $g=1$. Thus $\mathbb{C}G$ is a faithful representation.

By Maschke's theorem, the regular representation decomposes as a direct sum of irreducibles,Its kernel is the intersection of the kernels of the constituent representations:Each constituent kernel is a normal subgroup of $G$. If $G$ is simple, every such kernel is either $\{1\}$ or $G$. They cannot all be $G$, because then their intersection would be $G$, contradicting faithfulness of the regular representation. Hence some $V_j$ has trivial kernel. Therefore every finite simple group has a faithful irreducible representation of a finite simple group.

A representation is completely reducible, or semisimple, when it is a direct sum of irreducible representation. Equivalently, every invariant subspace has an invariant complement.

Maschke's theorem says that every finite-dimensional representation of a finite group $G$ over a field of characteristic zero is completely reducible. More generally, it is enough that the field characteristic does not divide $|G|$.

Schur lemma states the following.

Indeed, $\ker T\subseteq V$ and $\mathrm{im}T\subseteq W$ are invariant subspaces. If $T\ne 0$, irreducibility forcesso $T$ is an isomorphism. For an endomorphism of a finite-dimensional complex representation, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has nonzero kernel, so the first part forcesThis proves the Proof of Schur lemma.

Now let $\rho :G\to \mathrm{GL}(V)$ be faithful and irreducible over $\mathbb{C}$. For every $z\in Z(G)$, the operator $\rho (z)$ commutes with every $\rho (g)$. Schur's lemma therefore givesfor some ${\lambda }_z\in {\mathbb{C}}^{\times }$. Faithfulness makes $z\mapsto {\lambda }_z$ an embeddingSince every finite subgroup of the multiplicative complex numbers is cyclic,This is the cyclic-center obstruction to a faithful irreducible representation.

The seventh cyclotomic polynomial isIt is irreducible over $\mathbb{Q}$, and all its roots ${\zeta }_7^a$, $a=1,\dots ,6$, lie in $L=\mathbb{Q}({\zeta }_7)$. Hence $L/\mathbb{Q}$ is a degree-six Finite Galois extension, withThe group $(\mathbb{Z}/7\mathbb{Z}{)}^{\times }$ is cyclic of order six; for example,

By the Galois correspondence, the subfields are the fixed fields of the four subgroups of $C_6$:Thus these are all the subfields.

For the Quadratic Gaussian period in the seventh cyclotomic field, putThe cyclotomic relation gives $\eta +\bar{\eta }=-1$, while direct multiplication gives $\eta \bar{\eta }=2$. HenceIts discriminant is $-7$, so

For the real cubic subfield of the seventh cyclotomic field, divideby ${\zeta }_7^3$. With $\theta ={\zeta }_7+{\zeta }_7^{-1}$,soThis cubic has no rational root, and therefore

The requested primitive elements, minimal polynomials, and automorphism groups are consequentlyMore explicitly, the nontrivial automorphism of $\mathbb{Q}(\eta )$ sendsand the three automorphisms of $\mathbb{Q}(\theta )$ cyclically permute

Finally, $G$ is abelian, so every subgroup is normal. The subextensions of an abelian Galois extension theorem shows that

A field extension $L/K$ is a splitting field for $f\in K[X]$ when $f$ splits into linear factors over $L$ and $L$ is generated over $K$ by the roots of $f$. Equivalently, it is the smallest extension inside an algebraic closure over which $f$ splits.

To prove existence, choose an irreducible factor of $f$, adjoin one of its roots, and factor $f$ over the resulting finite extension. If a nonlinear factor remains, adjoin one of its roots and repeat. Each step lowers the total degree of the unsplit factors, so after finitely many finite extensions the polynomial splits. The field generated by the roots obtained is therefore a splitting field.

The precise uniqueness theorem says that if $L$ and $L^{\prime }$ are splitting fields of the same polynomial $f$ over $K$, then there is a field embedding isomorphismwhose restriction to $K$ is the identity. More generally, an isomorphism of base fields carrying one polynomial to another extends to an isomorphism of their splitting fields. Thus splitting fields are unique up to base-field isomorphism; inside a fixed algebraic closure, the field generated by all the roots is literally unique. These facts are summarized by existence and uniqueness of splitting fields.

LetThe roots of $X^3-2$ are $\alpha ,\omega \alpha ,{\omega }^2\alpha$, sois the splitting field of x cubed minus two. As a subfield of $\mathbb{C}$ it is unique, because any splitting field in $\mathbb{C}$ must contain all three roots and hence must contain both $\alpha$ andminimality then forces it to equal $\mathbb{Q}(\alpha ,\omega )$.

Partition the vertex set into disjoint sets $A$ and $B$ of sizes $\lfloor n/2\rfloor$ and $\lceil n/2\rceil$. The induced subgraph on $A$ has distributionFor $p=n^{-0.9}$,so part (b) shows that $A$ contains a triangle with probability tending to one.

On that event, choose a triangle using only the edges internal to $A$. The edges from its three vertices to $B$ remain independent of this choice. The probability that all $3|B|$ such edges are absent isbecause $p|B|\sim n^{0.1}/2$. Thus, with probability tending to one, some vertex of $B$ is adjacent to a vertex of the triangle. Those four vertices and the four required edges form the given graph $H$, a triangle with an attached leaf in a binomial random graph. Therefore

Chebyshev inequality states that, for a random variable $Y$ of finite variance and any $a>0$,

Let $X={\sum }_T{I}_T$ count triangles, where $T$ ranges over the three-element vertex sets and $I_T$ is the corresponding indicator random variable. ThenTwo distinct triangle indicators are independent unless their triangles share an edge. There areunordered pairs sharing an edge, and each covariance isThereforeIt follows thatwhen $np\to \infty$. Chebyshev's inequality now givesHenceas summarized by the triangle count in a binomial random graph.

The binomial random graph $G(n,p)$ has vertex set $[n]=\{1,\dots ,n\}$, and each unordered pair of distinct vertices is included as an edge independently with probability $p$.

Let $X_t$ count copies of the complete graph $K_t$. Each $t$-element vertex set forms a copy with probability $p^{\left(\genfrac{}{}{0ex}{}{t}{2}\right)}$, so linearity of expectation givesThe event $E_t$ is $\{X_t\ge 1\}$. By the first moment method,This is the clique count in a binomial random graph.

If some $T_i$ were inconsistent, it would prove every formula by the classical principle of explosion. Because $T_i$ is a deductively closed set of formulae, it would then contain every formula and could have no proper extension. This contradicts $T_i⊊T_{i+1}$. Thus every $T_i$ is consistent.

LetIf $T$ were inconsistent, the finite character of formal proofs would give a finite inconsistent subset $R\subseteq T$. The chain is increasing, so all finitely many members of $R$ lie in some common $T_N$. Then $T_N$ would be inconsistent, a contradiction. Hence $T$ is consistent.

If $T⊢\phi$, a proof again uses only finitely many assumptions from $T$, all lying in some $T_N$. ThusSince $T_N$ is deductively closed, $\phi \in T_N\subseteq T$. Therefore $T$ is deductively closed.

Finally, suppose that $T$ were finitary. Part (b) would provide a finite $R\subseteq T$ such that $R⊢T$. Choose $N$ with $R\subseteq T_N$. Then $T_N⊢T$; since $T_N$ is deductively closed, this implies $T\subseteq T_N$. Buta contradiction. Consequently $T$ is not finitary, proving the increasing union of deductively closed sets.

Suppose the finitary set of formulae $S$ is equivalent to the finite setFor each $f_j$, the relation $S⊢f_j$ has a finite proof, so it uses a finite set $R_j\subseteq S$ of assumptions. PutThis is a finite subset of $S$, and $R⊢f_j$ for every $j$, hence $R⊢F$. Since equivalence also gives $F⊢S$, transitivity of deduction yields

A set $S$ is a deductively closed set of formulae when every consequence of $S$ already belongs to it:It is a consistent set of formulae when it does not prove a contradiction,

If $S$ is inconsistent, there is a formal derivation of $\perp$ from assumptions in $S$. By the finite character of formal proofs, that derivation mentions only finitely many assumptions $s_1,\dots ,s_m\in S$. Henceso this finite subset of $S$ is already inconsistent.

In the far past, $|ck\tau |\gg 1$, the term $2/{\tau }^2$ is negligible and the rescaled mode is a quantum harmonic oscillator of angular frequency $ck$. The stated choice of its ground state is the Bunch-Davies vacuum.

At late times, after the mode is far outside the Hubble scale, $|ck\tau |\ll 1$. Its variance becomesSince $P_k$ has dimensions of length cubed, the dimensionless power is proportional to $k^3{P}_k$. More precisely, with the standard isotropic Fourier convention,which is independent of the wavenumber $k$. Thus equal logarithmic intervals in $k$ carry equal late-time power. This is the sense in which inflation naturally generates a scale-invariant inflationary power spectrum.

Conformal time is defined byFor $a=a_0{e}^{H_{\inf }t}$, choose $\tau \to 0$ as $t\to +\infty$. Integration givesas recorded by Conformal time during de Sitter expansion.

Writing a prime for $d/d\tau$, the relationsturn the Fourier-mode equation intoSetA direct substitution givesHere $a^{\prime \prime }/a=2/{\tau }^2$, so the Canonically rescaled de Sitter scalar mode obeys

When $V$ is approximately constant, $dV/d\varphi \simeq 0$, and exponential expansion gives the constant Hubble parameter $H=H_{\inf }$. Take the spatial Fourier transformSinceeach independent inflationary scalar Fourier mode satisfies

Write the action density asUnder a compactly supported variation $\varphi \mapsto \varphi +ϵ\eta$,Integrating the first term by parts in time and the second in space, and discarding the boundary terms, givesThe principle of stationary action requires this to vanish for every $\eta$. Sincedivision by $-a^3$ yields the field Euler-Lagrange equationThis is the equation generated by the inflaton action in an expanding universe.

If $\mathrm{Im}z>0$, the same upper-half-plane contour calculation givesIf $\mathrm{Im}z<0$, the pole lies below the real integration contour. Closing upward encloses no pole, so Jordan lemma and the Cauchy integral theorem giveHence at a real point $x$ the upper and lower boundary values satisfyThus crossing from the upper half-plane to the lower half-plane produces the negative of this jump.

An analytic continuation must be holomorphic, hence continuous, across every point where it is defined. The undeformed real-axis formula for $G$ has the nonzero jump above, while the continuation constructed in part (c) retains $2\pi i^n{e}^{iz}/(n-1)!$ below the axis. Therefore $G$ cannot be the analytic continuation of $F$.

For $\mathrm{Im}z>0$, close the $t$-contour in the upper half-plane. Jordan lemma removes the large semicircle, and the only enclosed pole is the order-$n$ pole at $t=z$. By the residue at a pole of order n,The residue theorem therefore gives

To continue the integral itself, move the $t$-contour downward locally as $z$ approaches and crosses the real axis, always keeping the pole above the contour. This analytic continuation by contour deformation defineswhere ${\mathcal{C}}_z$ passes below $z$. Closing ${\mathcal{C}}_z$ upward continues to enclose the pole, even when $\mathrm{Im}z\le 0$, sothere as well. The right-hand side is an entire function, so it is the unique analytic continuation of $F$ to the whole complex plane.

An analytic continuation of $f_1$ from ${\mathcal{D}}_1$ to ${\mathcal{D}}_2$ is a holomorphic function $f_2$ on ${\mathcal{D}}_2$ whose restriction to the nonempty overlap ${\mathcal{D}}_1\cap {\mathcal{D}}_2$ agrees with $f_1$.

Suppose $f_2$ and ${\tilde{f}}_2$ are two such continuations. Both equal $f_1$ on ${\mathcal{D}}_1\cap {\mathcal{D}}_2$, soon that nonempty open set. Since ${\mathcal{D}}_2$ is connected, the identity theorem givesThus analytic continuation to ${\mathcal{D}}_2$ is unique.

The identity theorem states that if two holomorphic functions on a connected open set $\mathcal{D}$ agree on a subset having an accumulation point in $\mathcal{D}$, then they agree throughout $\mathcal{D}$. In particular, agreement on any nonempty open subset is sufficient.