p-deficiency (${\mathrm{def}}_p⟨X\mid R⟩$)

For a group presentation $\mathcal{P}=⟨X\mid R⟩$ with finitely many generators and a prime number $p$, defineThis uses the unshifted convention. Identity relators have zero weight, and a divergent sum gives value $-\infty$. Some literature subtracts one instead; inequalities must be shifted accordingly. Taking roots in the free group matters: roots appearing only after passing to the quotient do not change the relator weight. The p-rank of a group bounds this quantity above, and the index-p rewriting bound for p-deficiency gives the infinitude criterion p-deficiency at least one implies infinitude.

Enumerate all nonidentity words $w_i$ in the rank-two free group and impose relations $w_i^{p^{i+2}}=1$. The resulting two-generated group hasIt is infinite by p-deficiency at least one implies infinitude. Each element has order a power of $p$, so it is a torsion group. The infinitely many relators are essential to this particular construction; finite generation does not imply a finite group presentation.

If a group presentation has unshifted p-deficiency at least one, the p-rank of a group is at least one and gives a normal subgroup of index $p$. The index-p rewriting bound for p-deficiency gives this subgroup another presentation with p-deficiency at least one. Iterating produces subgroups of index $p^j$ for every $j$, so the group is infinite. This criterion applies even with infinitely many relators, provided the weighted sum in the definition converges.

If $H$ is a normal subgroup of index $p$ in a group presented by $\mathcal{P}$, the Reidemeister–Schreier theorem produces a presentation $\mathcal{Q}$ withThe preimage of $H$ in the free group has rank $1+p(d-1)$. A relator $w^{p^k}$ produces at most $p$ conjugate relators of weight $p^{-k}$ if its root lies in that preimage. If the root lies outside, $k\ge 1$, and the coset conjugates are redundant up to conjugation; one relator of weight at most $p^{-(k-1)}$ suffices. Thus the total weighted relator cost grows by at most $p$. Convergent infinite sums are handled term by term.

For a finitely generated group, let $G^p$ denote the subgroup generated by all $p$th powers. Its p-rank isThis is the dimension of the largest quotient that is an elementary abelian group of exponent $p$. In a group presentation with $d$ generators and $s$ relators that are not $p$th powers, reduction of the exponent sums modulo $p$ gives at most $s$ linear constraints. Hence $d_p(G)\ge d-s\ge {\mathrm{def}}_p(\mathcal{P})$. In particular a presentation of p-deficiency at least one yields a surjection onto $C_p$.