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For a finite group $G$ with a symmetric generating set $S$, its diameter isEquivalently, it is the diameter of the Cayley graph ${\mathrm{Cay}}_S(G)$.

A finitely generated abelian group has a finite generating set. Equivalently, it is the cokernel of a homomorphism between finitely generated free abelian groups.

Suppose inductively that $G=H{\gamma }_i(G)$. In the quotient $G/{\gamma }_{i+1}(G)$, the image of ${\gamma }_i(G)$ is central. Hence this quotient is generated by the image of $H$ together with a central subgroup, so its commutator subgroup lies in the image of $H$. Pulling this back givesThe original assumption now gives $G=H{\gamma }_2(G)\le H{\gamma }_{i+1}(G)\le G$. Starting with $i=2$, induction proves $G=H{\gamma }_i(G)$ for every $i\ge 2$. These products are subgroups because each ${\gamma }_i(G)$ is normal; no normality of $H$ is needed.

If $G$ is finitely generated, every quotient, including its abelianization $G/{\gamma }_2(G)$, is finitely generated. Conversely, lift a finite generating set of that quotient to $g_1,\dots ,g_t\in G$, and set $H=⟨g_1,\dots ,g_t⟩$. Then $G=H{\gamma }_2(G)$. For a nilpotent group, ${\gamma }_{c+1}(G)=1$ for some $c$, and the established identity gives $G=H$. Thus

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.

The hidden subgroup problem for a group $G$ is specified by an oracle $f:G\to S$ that hides an unknown subgroup $H\le G$: it is constant on each left coset of $H$ and takes distinct values on distinct cosets. The task is to determine $H$, usually by finding a generating set for it.

A deck transformation of a covering $p:\tilde{X}\to X$ is a homeomorphism $d:\tilde{X}\to \tilde{X}$ with $p\circ d=p$. A connected covering is a regular covering map if its deck transformation group acts transitively on each fiber; for a path-connected base, transitivity on the fiber over the chosen basepoint is equivalent.

Write $H=p_{\ast }{\pi }_1(\tilde{X},\tilde{b})$ and $G={\pi }_1(X,b)$. A based loop represents an element of $H$ exactly when its lift starting at $\tilde{b}$ is closed. Indeed a closed lift itself projects to a representative of an image element; conversely the image of a closed path lifts to that path, and the closed-endpoint property is unchanged by based homotopy.

Suppose the covering is regular, let $h\in H$, and let $g\in G$. Lift a representative of $g$ from $\tilde{b}$ to an endpoint $v$. By regularity some deck transformation sends $\tilde{b}$ to $v$. Applying it to the closed lift of $h$ gives a closed lift of $h$ at $v$, by uniqueness in the path lifting theorem. The concatenation of the lift of $g$, that closed lift of $h$, and the reverse lift of $g$ is therefore a closed lift of $gh{g}^{-1}$. Thus $gH{g}^{-1}\subseteq H$; applying the same argument to $g^{-1}$ gives equality. Hence a regular covering has a normal image subgroup.

For the explicit cover, label the bottom vertex $B=\tilde{b}$, the top $T$, the left $L$, the right $C=c$, and the rear/front equatorial vertices $U,V$, respectively. Reading the arrows gives the complete monodromy table:

Thus the $x$ edges form the cycle $B,V,L,T,C,U,B$, and the $y$ edges form the cycle $B,C,V,T,U,L,B$. In particular every vertex has one incoming and one outgoing edge of each label, as a covering of the two-circle wedge requires. All word paths below are traversed from left to right.

Choose the spanning tree consisting of the first five $x$ edges from $B$ to $U$. The tree paths to $B,V,L,T,C,U$ have labels $1,x,x^2,x^3,x^4,x^5$. For every directed non-tree edge $v\stackrel{z}{\to }w$, the path from $B$ along the tree to $v$, then along the edge, then back along the tree from $w$, has label $t_{v}z{t}_w^{-1}$. Collapsing a spanning tree turns a connected graph into a wedge of circles, so these paths form a free basis of its fundamental group. There are seven non-tree edges, since $12-6+1=7$. Their projected labels give the explicit generating setThe covering map induces an injection on fundamental groups, so these seven generators are also a free basis for the image subgroup. The first comes from the remaining $x$ edge $U\to B$; the other six come from the outgoing $y$ edge at each vertex in the table.

Now the word $h=y{x}^2$ lifts at $B$ around the triangleso $h\in H$. Starting at $C$, however, its lift iswhich is not closed. A deck transformation sending $B$ to $C$ would take the first closed path to a closed lift of the same word at $C$, contradicting uniqueness of lifts and the displayed nonclosed path. No such deck transformation exists. This is also the closed-lift obstruction expressed by the deck transformation group as a monodromy centralizer.

For a connected covering of a connected, locally path-connected, semilocally simply connected space, the classification of connected covering spaces says that the image subgroup is normal if and only if the covering is regular. The two-circle wedge and this connected finite graph satisfy those hypotheses. Since regularity fails, $H$ is not normal in the base group ${\pi }_1(X,b)\cong F(x,y)$. All normality comparisons here are in that base group.

Finally takeThe lift of $gh{g}^{-1}=y^2{x}^2{y}^{-1}$ from $B$ followsIt ends at $V\ne B$, so $gh{g}^{-1}\notin H$. This is the required explicit witness, and independently proves the failure of normality.