For fixed-endpoint variations, the second variation iswith $p=f_{vv}$, $r=f_{uv}$, $s=f_{uu}$ evaluated along $(x,\bar{u},{\bar{u}}^{\prime })$. The Jacobi equation, written without differentiating $r$ separately, isWhen $r$ is differentiable this is $-(p{z}^{\prime }{)}^{\prime }+(s-r^{\prime })z=0$.

Positive definiteness means $q(\phi )>0$ for every nonzero admissible variation. Uniform or coercive positivity of a second variation is the stronger assertion that $q(\phi )\ge c\int {\phi }^{\prime 2}$ for some fixed $c>0$. On a fixed-endpoint interval this is equivalent to a positive lower bound in the $W^{1,2}$ norm.

The strict Legendre implication uses the uniform interpretation of positivity. If $q\ge c\int {\phi }^{\prime 2}$, localized rapidly oscillating variations show $p(x)\ge c$: otherwise support them in an interval where $p-c<0$, so the derivative term of $q-c\int {\phi }^{\prime 2}$ is negative of order frequency squared, overwhelming its mixed and zero-order terms. Continuity extends the bound to the endpoints, giving $p>0$ on $[a,b]$. Under mere positive definiteness this strict conclusion is false: $f=u^2$ at $\bar{u}=0$ gives $q=2\int {\phi }^2>0$ but $p=0$. In that weaker convention only the ordinary Legendre necessary condition $p\ge 0$ follows. This distinction makes the implication precise.

Under the stated regular condition $p>0$, a conjugate point $c\in (a,b]$ to $a$ is an endpoint admitting a nonzero Jacobi solution with $z(a)=z(c)=0$. The disconjugacy form of the Jacobi positivity criterion says that no such point in $(a,b]$ exists if and only if there is a Jacobi solution strictly positive on all of $[a,b]$.

For the required identity, let that positive solution be $z$ and write $\phi =zh$. Set $A=pz{z}^{\prime }+r{z}^2$. The Jacobi equation givesExpanding the second-variation integrand therefore yieldsFor compactly supported $\phi$, the boundary term vanishes, provingThis is the Jacobi identity for the second variation. The divergence form is also appropriate under the original $C^1$ regularity; with $p>0$, the Euler-Lagrange equation gives the usual further regularity along the extremal.

Let $L$ be the subgroup of $H\mathrm{wr}{A}_m$ generated by all its elements of order $3$. The top group $A_m$ is generated by 3-cycles, also when $m=3$, so $A_m\le L$.

Take $\sigma =(123)$. For any $h\in H$, the conjugate $h_1\sigma h_1^{-1}$ has order $3$, as does ${\sigma }^{-1}$. Thus their product belongs to $L$ and equalsUsing ${\sigma }^{-1}$ instead supplies $d_{13}(h)=h_1{h}_3^{-1}$. Since distinct coordinate groups commute,The hypothesis that $H$ is perfect says that products of its commutators generate $H$, so $H_1\le L$. Transitivity of $A_m$ now gives $H_i\le L$ for every coordinate. Hence $H^m\le L$ and $A_m\le L$, provingThe proof does not require $H$ itself to be finitely generated, nor does it require $A_m$ to be perfect for $m=3,4$.

Put $A=A_m$ and let $T$ be the rooted tree of finite words over $\{1,\dots ,m\}$. View $a\in A$ as a rooted automorphism: it permutes the first letter by $a$ and fixes the suffix. Denote this rooted copy by $A$ again.

For $a\in A$, define a directed tree automorphism $b(a)$ by putting local permutation $a$ at every vertexand identity local permutation at every other vertex. These marked vertices are pairwise incomparable, so the prescribed actions are on disjoint rooted subtrees and define a unique automorphism. Equivalently its root permutation is trivial and its sections areThe map $a\mapsto b(a)$ is an injective homomorphism: it is the same copy of $a$ acting on each of those disjoint subtrees. Let $B=b(A)$ and defineBoth $A$ and $B$ are finite, so $H$ is finitely generated. For $m\ge 5$, $A_m$ is a nonabelian simple group and hence perfect. Thus $A,B\le H^{\prime }$, and because they generate $H$, $H$ is perfect.

Every section of either defining generator belongs to $H$, so sections of every element of $H$ belong to $H$. Its root permutation belongs to $A$. For a tree automorphism $g$, write $g(iw)=\sigma (i)g_i(w)$, where $g_i$ is its section at input letter $i$. In the semidirect-product convention of part (a), defineComposition of tree automorphisms gives exactly the wreath-product multiplication: this is why $\theta$ is a homomorphism. It is injective because an automorphism is determined by its first-level permutation and all its first-level sections. On the generators,It remains to prove surjectivity; merely writing these recursions would not establish it. Choose $\sigma =(234)\in A$, which fixes $1$ and sends $2$ to $3$. For $a,c\in A$, the two base tuples $\theta (b(a))$ and $\theta (\sigma b(c){\sigma }^{-1})$ overlap only in their first coordinate. Their commutator is thereforeSince $A$ is perfect, products of these commutators give every element of the copy of $B$ localized in coordinate $1$. Canceling this first coordinate from $\theta (b(a))$ then gives $(1,a,1,\dots ,1;1)$, the rooted copy of $A$ localized in coordinate $2$. Conjugating by the transitive rooted group $A_m$ supplies both localized $A$ and localized $B$ in every coordinate. Since $H=⟨A,B⟩$, the image contains the full base $H^m$. It also contains the top $A_m$. Therefore it is the entire wreath product andFinally $H$ is infinite: it contains the nontrivial rooted $A_m$, and a finite order $h=|H|$ would satisfy $h=h^m|A_m|$, which is impossible for $m\ge 5$.

Let $B=H_1\times \cdots \times H_m$, with each $H_i$ a copy of $H$. The permutation group $F\le S_m$ acts on $B$ byDefine the permutation wreath product to be the semidirect productExplicitly,Writing $h_i$ for the element with entry $h$ in coordinate $i$ and identity elsewhere, and writing $\sigma$ for $(1,\sigma )$, conjugation satisfiesThus the top permutation group permutes the coordinate subgroups in exactly its given action. Distinct coordinate subgroups commute, including when $H$ itself is nonabelian.

First establish a stronger form of the coset-counting argument. If $K$ is a finitely generated infinite-index subgroup of a finitely generated group $L$ of growth at most $q$, choose a finite generating set $T$ for $K$, and let $M$ bound its lengths in a generating set $S$ of $L$. The Schreier graph supplies $r+1$ distinct left-coset representatives of $K$ of $S$-length at most $r$. The corresponding translates of $B_T(r)$ are disjoint and lie in $B_S((M+1)r)$. HenceThus an infinite-index finitely generated subgroup has growth at most $q-1$ whenever the ambient upper bound is $q$.

Every finitely generated $K\le H$ has infinite index in $G$. Applying this result with $L=G$ gives growth at most $n-1$. Minimality of $n$ among the supposed counterexamples therefore makes each such $K$ virtually nilpotent. Here $n\ge 1$, since a group with a uniform bound on its finite word balls is finite, and finite groups are virtually nilpotent.

For each finitely generated $K\le H$, let $d(K)$ be the least nonnegative integer giving a polynomial growth upper bound. It exists and belongs to $\{0,\dots ,n-1\}$. Choose $K$ for which this integer is maximal. If $K\le L\le H$ and $L$ is finitely generated, then $d(L)\le d(K)$. Were $[L:K]$ infinite, the strengthened bound would make $K$ have growth at most $d(L)-1\le d(K)-1$, contradicting the definition of $d(K)$. If $d(K)=0$, the same inequality would force the nonempty balls of $K$ to have cardinality tending to zero, also impossible. Hence every such overgroup has $[L:K]<\infty$.

Since $K$ is virtually nilpotent, choose a nilpotent finite-index subgroup $N\le K$. It is finitely generated by Schreier's lemma. For any $h\in H$, the subgroup $L=⟨K,h⟩$ is finitely generated and $[L:K]<\infty$. Pigeonholing the cosets $h^{j}K$ gives a positive $a$ with $h^a\in K$. Pigeonholing the cosets of $N$ in $K$ then gives a positive $b$ with $(h^a{)}^b\in N$. ConsequentlyNo normality of $K$ in $L$ or of $N$ in $K$ is required. Using least integer upper bounds also avoids needing any further structural theorem about the exact growth degree of a virtually nilpotent group.

Choose finite symmetric generating sets $S_i$ for all $G_i$, and let $L\ge 1$ bound the $S$-length of their elements. If $[G_{i-1}:G_i]=\infty$, the connected locally finite Schreier graph of left cosets of $G_i$ in $G_{i-1}$ is infinite. Its radius-$r$ ball has at least $r+1$ vertices: otherwise a finite-diameter locally finite connected graph would be finite. Equivalently, we can choose $r+1$ distinct left-coset representatives $t_{i,0},\dots ,t_{i,r}$ of $G_i$, each of $S_{i-1}$-length at most $r$, and hence $S$-length at most $Lr$.

The $(r+1{)}^n$ productsare distinct. Indeed, equality first forces equality of the $G_1$-cosets of the first factors, and hence their indices; cancellation then forces equality of the $G_2$-cosets of the second factors, and so on. Each product has $S$-length at most $nLr$. ThereforeLetting $r\to \infty$ yieldsFor $n=0$ the assertion is automatic.