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A subgroup $H$ of a finitely generated group $G$ is benign if some finitely presented group $P$ contains $G$ and some finitely generated subgroup $L\le P$ satisfies $H=G\cap L$. The subgroup $H$ itself need not be finitely generated. A centralizing HNN extension of $P$ along $L$ is finitely presented even when the corresponding intersection subgroup $H$ is infinitely generated.

The Ohnishi orderability criterion states that, for each finite list $g_1,\dots ,g_n\ne 1$, there are signs ${\epsilon }_i\in \{-1,1\}$ such that the semigroup of nonempty products of conjugates of $g_i^{{\epsilon }_i}$ does not contain $1$. The conjugates are essential for a two-sided invariant total order. Necessity follows by choosing all signed elements positive: their conjugates and nonempty products are still positive.

For sufficiency, consider the compact product $\{+,-{\}}^{G\setminus \{1\}}$. Impose the closed conditions that an inverse has the opposite sign, that conjugation preserves sign, and that a product of two positive elements is positive whenever it is nonidentity. A positive pair cannot multiply to the identity because of the inverse condition. Each condition uses finitely many coordinates. Given finitely many conditions, collect every nonidentity element appearing in them into a finite list and choose signs using the criterion. These signs satisfy the conditions: an incorrectly negative product would contribute its positive inverse to the generated semigroup, producing identity with the positive factors; an incorrect conjugate or inverse gives the same contradiction. Extend the signs arbitrarily to the remaining coordinates. Thus the finite intersection property and compactness give a global sign assignment. Its positive elements form a conjugation-invariant semigroup $S$ with $S\cap S^{-1}=\varnothing$ and $S\cup S^{-1}=G\setminus \{1\}$. Defining $a<b$ by $a^{-1}b\in S$ gives the required linearly ordered group.

Restriction shows that every finitely generated subgroup of an orderable group is orderable. Conversely, suppose every such subgroup is orderable. If the criterion failed on $g_1,\dots ,g_n$, each of the finitely many sign choices would have an identity witness made of finitely many conjugates. Collect all $g_i$ and all conjugating elements in these witnesses into one finitely generated subgroup $H$. An order on $H$ supplies one positive sign choice, but its corresponding witness is a nonempty product of positive elements in $H$, a contradiction. Hence the criterion holds in $G$.

Finally, a finitely generated subgroup of a torsion-free abelian group is ${\mathbb{Z}}^r$ by the Fundamental theorem of finitely generated abelian groups. Its lexicographic order is two-sided invariant. Applying the local-to-global conclusion proves that every torsion-free abelian group admits the required order.

A subgroup $H\le G$ is a benign subgroup when there are an embedding of $G$ into a finitely presented group $E$ and a finitely generated subgroup $L\le E$ with $G\cap L=H$. Finite generation is required of $L$, not of $H$ itself. We use Britton's lemma, the normal form theorem for an amalgamated free product, and, for the final quotient assertion, the Higman embedding theorem.

Choose witnesses $E_H,L_H$ and $E_K,L_K$ for $H,K$. The amalgamated free product $P=E_H{\ast }_G{E}_K$ embeds both witnesses and their common copy of $G$. It is finitely presented: take their two finite presentations and add the finitely many equalities identifying the images of a finite generating set of $G$. These equalities identify all elements of $G$, so they present precisely the amalgam. The original intersections $G\cap L_H=H$ and $G\cap L_K=K$ are preserved by the embeddings.

Now form the finitely presented groupOnly finitely many commutation relators, one for each of the chosen subgroup generators, are needed. The subgroup $Q=⟨G,t,s⟩$ is exactly the multiple HNN extension of $G$ centralizing $H$ and $K$. To prove this claim, take a word in $G,t,s$ reduced for those associated subgroups. It remains reduced in $T$: a $t$-pinch with a coefficient in $G$ occurs precisely when that coefficient lies in $L_H\cap G=H$, and an $s$-pinch precisely when it lies in $K$. Britton's lemma therefore prevents any extra relation, and it also embeds $G$.

Write $G^u=u^{-1}Gu$. The subgroup $L_{\cap }=G^{ts}$ is finitely generated. If $s^{-1}{t}^{-1}gts\in G$, Britton reduction must first pinch $t^{-1}gt$, requiring $g\in H$; its value is then $g$. The remaining $s^{-1}gs$ can represent a base element only if $g\in K$, and then its value is again $g$. Conversely every $g\in H\cap K$ commutes with both letters. Henceso the intersection is benign.

For the join, put $J=⟨H,K⟩$ and $L_{\vee }=⟨G^t,G^s⟩$, again finitely generated. The subgroup generated by $G,G^t,G^s$ in $Q$ isIndeed, expanding conjugate-copy syllables as $t^{-1}gt$ or $s^{-1}gs$ shows that the only possible Britton reductions are exactly the identifications in the indicated amalgamations; a reduced amalgam word therefore has no additional way to become trivial.

We use this elementary consequence of amalgam normal forms: if $C\le D\le A$, then inside $A{\ast }_{C}B$ the subgroup $⟨D,B⟩$ is $D{\ast }_{C}B$ and intersects $A$ exactly in $D$. A reduced expression with a $B\setminus C$ syllable cannot lie in $A$, because multiplying it by an inverse base element still leaves a reduced nontrivial amalgam word. Apply this first to $G{\ast }_H{G}^t$ with $D=J$, obtaining $D_1=⟨J,G^t⟩$ and $D_1\cap G=J$. Since $K\le J\le D_1$, apply it again in $U=(G{\ast }_H{G}^t){\ast }_K{G}^s$ to obtain $⟨D_1,G^s⟩\cap (G{\ast }_H{G}^t)=D_1$.

Each element of $H$ lies in $G^t$, and each element of $K$ lies in $G^s$, since the corresponding stable letter centralizes it. Thus $J\le L_{\vee }$, and $L_{\vee }=⟨D_1,G^s⟩$. Intersecting the previous identities with $G$ givesThis proves the intersection and join of benign subgroups construction with actual finitely generated witnesses inside $T$.

Finally let $N◃G$ be benign, witnessed by $N=G\cap L$ in a finitely presented group $E$. Choose finite generators $X$ for $G$ and express both $X$ and a finite generating set for $L$ as words in $E$. Enumerate every word $w$ on $X$, every word $v$ on those generators of $L$, and every finite proof in the presentation of $E$ that their substituted words are equal. Dovetail these searches and list $w$ when such a proof is found. Exactly the words representing elements of $G\cap L=N$ are listed: membership has a finite expression in the generators of $L$, and every valid equality has a finite relator proof. Since $N$ is normal, these are exactly the null words of $G/N$.

Use this enumerable list as relators on the finite alphabet $X$. Its normal closure is precisely the kernel of $F(X)\to G/N$, so it is a recursive presentation of a group for $G/N$. The Higman embedding theorem now yieldsThis is normal benign subgroup quotient embedding; no word algorithm for $E$ or membership algorithm for $L$ was assumed.