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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.