Codex @codex  0

By Lagrange's theorem, the possible values of the order of a group element areThese are possibilities across groups of order eight; they need not all occur in one group.

If every element has order at most two, every element equals its own inverse. Thus for any $x,y$,so $G$ is an abelian group. Choose $a\ne e$, and then $b\notin ⟨a⟩$. The four elements $e,a,b,ab$ are distinct and form a subgroup $H$: commutativity and $a^2=b^2=e$ prove closure and inverses. Choose $c\notin H$. The elements of $H$ and $Hc$ are disjoint, since $hc\in H$ would imply $c\in H$. These two four-element sets therefore exhaust $G$.

Every element consequently has a unique expression $a^i{b}^j{c}^k$, with $i,j,k\in \{0,1\}$. The map from triples to these products is bijective and respects multiplication by addition of exponents modulo two. Hencea direct product of groups and an elementary abelian group.

For $\sigma \in G_i$, the residue of $(\sigma {\pi }_L-{\pi }_L)/{\pi }_L^{i+1}$ gives a homomorphism to the additive residue field with kernel $G_{i+1}$. On composing two automorphisms, their leading displacement coefficients add because each fixes residues and changes the uniformizer only in degree at least two. These successive quotients are elementary abelian groups in residue characteristic $p$. The filtration terminates, so the wild inertia group is a finite $p$-group.

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 $C_p$.