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For a group presentation $\mathcal{P}=⟨X\mid R⟩$ with $X$ finite, let $F(X)$ be its free group. For a nontrivial relator defineThe p-deficiency in the unshifted convention used here isIf the weighted sum diverges the value is $-\infty$; identity relators may be omitted or assigned weight zero. Roots are taken in the free group, not in the presented quotient. Some authors subtract one from this definition; here the requested threshold is $1$.

Two elementary bounds explain why p-deficiency detects infinitude. The p-rank of a group isHere $G^p$ denotes the subgroup generated by all $p$th powers. Each relator that is not a $p$th power imposes at most one linear relation in this vector space, and a $p$th-power relator imposes none. If there are $s$ relators of the first type, then

The second bound is the index-p rewriting bound for p-deficiency. Suppose $H⊴G$ has index $p$, and its preimage in $F(X)$ is $V$. The Nielsen–Schreier formula gives $V$ rank $1+p(|X|-1)$. For a relator $r=w^{p^k}$, there are two cases in the Reidemeister–Schreier theorem. If $w\in V$, its $p$ coset-conjugates are all $p^k$th powers in $V$, with total weight at most $p{p}^{-k}$. If $w\notin V$, then $k\ge 1$, since $r\in V$. Its cosets generate $F(X)/V$, so representatives $1,w,\dots ,w^{p-1}$ show that the $p$ rewritten conjugates of $r$ are redundant up to conjugation in $V$. One relator suffices, and $r=(w^p{)}^{p^{k-1}}$ has weight at most $p^{-(k-1)}$. In both cases the total weight is at most $p$ times the old weight. Thus the induced group presentation $\mathcal{Q}$ of $H$ satisfiesThe argument applies termwise to infinitely many relators whenever the weighted sum converges.

If ${\mathrm{def}}_p(\mathcal{P})\ge 1$, the p-rank of a group bound gives a surjection to $C_p$, hence a normal subgroup of index $p$. The rewriting bound gives that subgroup another presentation of p-deficiency at least one. Iterating produces subgroups of index $p^j$ for every $j$. p-deficiency at least one implies infinitude.

Now enumerate the nonidentity elements $w_1,w_2,\dots$ of $F(x,y)$ and choose the presentationIts p-deficiency obeysThe infinitude criterion shows that $T$ is infinite. It is generated by two elements, and every element is represented by some $w_i$ or is the identity; the imposed relation makes its order a power of $p$. Thus $\boxed{T\text{ is an infinite finitely generated torsion group}}$. This is a torsion group construction by p-power relators; the presentation intentionally has infinitely many relators.

A group is a torsion group when every element is a torsion element, meaning that each element has finite order. The orders need not have a common finite bound. A torsion group has no nontrivial torsion-free group as a subgroup; in particular it cannot contain a nonabelian free group. Infinite finitely generated groups of this kind can be constructed using a torsion group construction by p-power relators.