Solution

ID: past-exam-of-the-building-infinite-groups-course-of-the-university-of-oxford/2011/2/c/solution

First establish a stronger form of the coset-counting argument. If is a finitely generated infinite-index subgroup of a finitely generated group of growth at most , choose a finite generating set for , and let bound its lengths in a generating set of . The Schreier graph supplies distinct left-coset representatives of of -length at most . The corresponding translates of are disjoint and lie in . Hence
Thus an infinite-index finitely generated subgroup has growth at most whenever the ambient upper bound is .
Every finitely generated has infinite index in . Applying this result with gives growth at most . Minimality of among the supposed counterexamples therefore makes each such virtually nilpotent. Here , since a group with a uniform bound on its finite word balls is finite, and finite groups are virtually nilpotent.
For each finitely generated , let be the least nonnegative integer giving a polynomial growth upper bound. It exists and belongs to . Choose for which this integer is maximal. If and is finitely generated, then . Were infinite, the strengthened bound would make have growth at most , contradicting the definition of . If , the same inequality would force the nonempty balls of to have cardinality tending to zero, also impossible. Hence every such overgroup has .
Since is virtually nilpotent, choose a nilpotent finite-index subgroup . It is finitely generated by Schreier's lemma. For any , the subgroup is finitely generated and . Pigeonholing the cosets gives a positive with . Pigeonholing the cosets of in then gives a positive with . Consequently
No normality of in or of in is required. Using least integer upper bounds also avoids needing any further structural theorem about the exact growth degree of a virtually nilpotent group.

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