We prove the stronger ordinal identityby transfinite induction. It holds at zero. At a successor stage an ordinal $\gamma$ belongs to $\mathcal{P}(V_{\beta })$ exactly when every $\delta <\gamma$ belongs to $V_{\beta }$. By induction this means every $\delta <\gamma$ is less than $\beta$, equivalently $\gamma \le \beta$. Thus the ordinals in $V_{\beta +1}$ are precisely those less than $\beta +1$. At a limit the identity follows by taking unions.

Every $\beta <\alpha$ consequently belongs to $V_{\alpha }$, so $\alpha \subseteq V_{\alpha }$ and $\alpha \in \mathcal{P}(V_{\alpha })=V_{\alpha +1}$. But $\alpha \notin V_{\alpha }$, since an ordinal is not less than itself. HenceThe only well-foundedness used is that of the ordinal order, available independently of Foundation for arbitrary ambient sets.

The preceding inclusion is exactly $V_{\alpha }\subseteq \mathcal{P}(V_{\alpha })$. Therefore every $x\in V_{\alpha }$ is a subset of $V_{\alpha }$. Equivalently,which says that $V_{\alpha }$ is a transitive set. This uses the power set definition and the proved inclusion, without any ambient use of Foundation.

Use transfinite induction. The inclusion at zero is immediate. If $V_{\beta }\subseteq V_{\beta +1}$, monotonicity of the power set givesIf $\alpha$ is a limit and $x\in V_{\alpha }$, choose $\beta <\alpha$ with $x\in V_{\beta }$. The inductive inclusion $V_{\beta }\subseteq \mathcal{P}(V_{\beta })$ means that $x\subseteq V_{\beta }\subseteq V_{\alpha }$. Thus $x\in \mathcal{P}(V_{\alpha })=V_{\alpha +1}$. This proves $V_{\alpha }\subseteq V_{\alpha +1}$ for every ordinal. These successive inclusions and the union definition also give $V_{\beta }\subseteq V_{\alpha }$ whenever $\beta \le \alpha$.

Let ${\omega }_G=h_G(T)dT$. The pullback through the formal group homomorphism isThe identity $f(F(X,T))=G(f(X),f(T))$ and the chain rule show that this pullback is invariant under $F$: translation by $T$ before $f$ becomes translation by $f(T)$ afterward. By part (i) it is a scalar multiple of ${\omega }_F$. Its coefficient at zero is $h_G(0)f^{\prime }(0)=f^{\prime }(0)$, so

Apply this to the multiplication-by-$p$ endomorphism $q(T)=[p](T)$. Since the linear coefficient of $F$ is $X+Y$, repeated addition gives $q^{\prime }(0)=p$. If ${\omega }_F=h(T)dT$ with $h(0)=1$, the pullback formula givesbecause $h(q(T))$ is a unit in a ring. Write $q(T)={\sum }_{m\ge 1}{a}_m{T}^m$. Then $m{a}_m\in pR$ for every $m$. If $p\nmid m$, choose integers $u,v$ with $um+vp=1$. It follows thatChoose $b_m\in R$ with $a_m=p{b}_m$ for those $m$ not divisible by $p$, and defineCoefficient comparison now provesNo cancellation of $p$ was used, so this prime-multiplication coefficient splitting for formal groups remains valid when $p$ is a zero divisor in $R$.

Put $U(T)=F_X(0,T)$. The formal group identity $F(X,0)=X$ gives $U(0)=1$, so $U$ is a unit in a ring in $R[[T]]$, even if $R$ has zero divisors.

A differential $h(T)dT$ is invariant under formal translation ifas an identity of formal power series. Setting $X=0$ gives $h(T)U(T)=h(0)$. Consequently every invariant differential of a formal group law, if it exists, must have the form $h(0)U(T{)}^{-1}dT$. In particular there can be only one with constant coefficient one.

Existence follows directly from associativity. Differentiatewith respect to $Z$ at zero, obtainingEvery displayed $U$ is a unit in a ring, soThis is precisely invariance of $U(T{)}^{-1}dT$. ThusNormalization means its constant coefficient is one. This proves both existence and uniqueness over an arbitrary commutative ring, without division by any integer.

The preceding bound at $m=2$ gives $|h_x(2Q)-4h_x(Q)|\le C$ for every point $Q$. For $a_n=4^{-n}{h}_x(2^{n}P)$, take $Q=2^{n}P$ to obtainFor $m>n$, summing the geometric tail givesThus the sequence is Cauchy and converges by completeness of the real numbers. The estimate also gives a quantitative bound on its distance from the limit.

Fix $P$ and put $e_m=h_x(mP)-m^2{h}_x(P)$ for $m\ge 0$. Then $e_0=e_1=0$. Apply the given approximate parallelogram identity with its two arguments $mP$ and $P$. The quadratic terms cancel exactly, leavingSet $d_m=e_m-e_{m-1}$. Starting from $d_1=0$, the inequality gives $|d_m|\le (m-1)C$. Summing from one to $m$ therefore yieldsThus $C_m=m(m-1)C/2$ works independently of $P$, including $m=1$.

Apply continuity of the inner product to the convergent expansion of $x$:The series is absolutely convergent because the Cauchy-Schwarz inequality for scalar sequences and Bessel's inequality giveThis is the polarized Parseval identity.

For $a_n=⟨x,e_n⟩$, the finite orthogonal projection $P_{N}x={\sum }_{|n|\le N}{a}_n{e}_n$ satisfiesThis is Bessel's inequality, so the coefficients are square summable. The preceding result gives a limit $s={\sum }_n{a}_n{e}_n$. For each fixed $k$, continuity of the inner product gives $⟨x-s,e_k⟩=0$. The completeness assumption on the orthonormal system forces $x-s=0$. HenceTaking squared norms also gives Parseval's identity, $\Vert x{\Vert }^2={\sum }_n|⟨x,e_n⟩{|}^2$.

For finite subsets $F\subset \mathbb{Z}$, orthonormality givesFor $S_N={\sum }_{|n|\le N}{\alpha }_n{e}_n$, the squared norm of $S_N-S_M$ is the corresponding finite tail sum. Since ${\sum }_{n\in \mathbb{Z}}|{\alpha }_n{|}^2<\infty$, these tails tend to zero. Thus $(S_N)$ is a Cauchy sequence in the complete Hilbert space and converges.

The same estimate for arbitrary finite tails proves that the sum is unconditional, so the bilateral orthogonal series converges in $X$ independently of its ordering. This is the Riesz-Fischer theorem for an orthonormal system.

For a fixed second argument the norm formula makes $P$ continuous in its first argument. Rational homogeneity therefore extends to real homogeneity by rational approximation; additivity and $P(iu,v)=iP(u,v)$ then give full complex linearity in that argument.

The real pairing $B$ is symmetric, and $B(v,iu)=B(iu,v)=-B(u,iv)$. Hence $P(v,u)=\overline{P(u,v)}$, giving conjugate linearity in the second argument. Finally $B(u,u)=Q(u)$ and $B(u,iu)=0$, because $\Vert (1+i)u\Vert =\Vert (1-i)u\Vert$. Thereforewhich is positive for $u\ne 0$. This proves $P$ is a complex inner product inducing exactly the original norm. The assumed Banach completeness makes the resulting inner product space Hilbert. This is the Jordan-von Neumann theorem.

Additivity applied to $u+(-u)=0$ gives $P(-u,v)=-P(u,v)$. Multiplication by $i$ preserves the norm, soFor example the first follows by multiplying $iu\pm v$ by $-i$ inside each norm. ConsequentlyThis proves the two required identities, with the convention of linearity in the first argument.

Repeated additivity gives $P(mu,v)=mP(u,v)$ for every positive integer $m$. Since $u$ is the sum of $n$ copies of $u/n$, $P(u/n,v)=P(u,v)/n$. Combining these provesThe case $m=0$ also holds by additivity at zero.

Put $Q(u)=\Vert u{\Vert }^2$ and $B(u,w)=\frac{1}{4}[Q(u+w)-Q(u-w)]$. Then $B(0,w)=0$ and the parallelogram identity, applied to $u+w,v$ and to $u-w,v$ and subtracted, givesTaking $v=u$ shows $B(2u,w)=2B(u,w)$. Replacing $u,v$ in the identity by $(u+v)/2,(u-v)/2$ now givesThe proposed complex pairing is $P(u,w)=B(u,w)+iB(u,iw)$. Applying these same identities with $w$ and $iw$ proves both asserted midpoint and additivity equalities for $P$.

Take the free Abelian groupIts coordinate flag $0<\mathbb{Z}<{\mathbb{Z}}^2<\cdots <{\mathbb{Z}}^m$ is a subnormal series of length $m$, with every factor isomorphic to the infinite cyclic group $\mathbb{Z}$. Hence it has exactly $m$ infinite cyclic factors.

TakeStart with $1<A_3<S_3$, whose factors are $C_3$ and $C_2$, and then append the $m-2$ coordinate factors of the elementary Abelian group $C_2^{m-2}$. More explicitly, append $S_3\times C_2^j$ for $1\le j\le m-2$. Each subgroup is normal in the next, and the factors consist of one $C_3$ and $m-1$ copies of $C_2$. Thus there are exactly $m$ finite cyclic factors. The group is non-Abelian because it contains $S_3$.

First establish the two general claims about polycyclic groups. Intersect a subnormal cyclic-factor series $(G_i)$ with $H\le G$. Then $H\cap G_i⊴H\cap G_{i+1}$, and the inclusion induces an injective homomorphismA subgroup of a cyclic group is cyclic, so this is a cyclic-factor subnormal series for $H$, after omitting repetitions. This proves subgroups of polycyclic groups are polycyclic.

To prove the generator bound, choose $g_i\in G_i$ whose coset generates $G_i/G_{i-1}$ whenever that factor is nontrivial. Inductively $G_i=⟨G_{i-1},g_i⟩$, so $G=⟨g_1,\dots ,g_n⟩$. This is the generator bound from a cyclic-factor series: at most $n$ generators suffice.

For the first example, take $G=C_2^m$ with its coordinate flagIt has length exactly $m$ and each factor is $C_2$, so all factors are cyclic.

True; there are exactly five. On a simple module, the central element $Y$ acts as zero: its image is a submodule, and surjectivity would contradict $Y^6=0$. The nilpotent central elements annihilate simple modules argument shows that the action factors throughby the Chinese remainder theorem. Therefore the simple modules are one-dimensional withThis also rules out hypothetical infinite-dimensional simple modules, rather than counting only finite-dimensional ones.

True. The polynomial $X^7-1$ has seven distinct roots ${\zeta }_7^j$ in $\mathbb{C}$. The Chinese remainder theorem givesThe simple modules are exactly the seven one-dimensional evaluation modules, on which $X$ acts by ${\zeta }_7^j$. Thus there are seven up to isomorphism.

True. If every simple module over $\mathbb{C}G$ were one-dimensional, the Maschke theorem and the Artin–Wedderburn theorem would make $\mathbb{C}G$ a product of copies of $\mathbb{C}$, hence commutative. But its basis elements multiply as in $G$, so commutativity of $\mathbb{C}G$ implies commutativity of $G$. This contradicts the assumption. Therefore a non-Abelian finite group has a simple module of dimension greater than one.