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The Higman embedding theorem says that a finitely generated group embeds in a finitely presented group if and only if it has a recursive presentation of a group, that is, a presentation on finitely many generators with a recursively enumerable set of relators. We use its embedding direction as an assumed theorem. The reverse direction follows by expressing subgroup generators as words in the larger group and enumerating finite products of conjugates of its relators and their inverses; comparison after free reduction enumerates all null words in the subgroup. This is the enumerable identities of finitely generated subgroups argument. The theorem does not assume that the recursively presented group's word problem for a group is decidable.

False. We construct a two-generated countable group that cannot embed in any finitely presented group. Choose a subset $S\subseteq {\mathbb{N}}_{>0}$ which is not a recursively enumerable set, for example a positive-integer coding of the complement of the halting set. Such sets exist because there are only countably many finite programs but uncountably many subsets of the positive integers. On generators $b_i$ indexed by $i\in \mathbb{Z}$, form the possibly infinite-graph Right-angled Artin groupThe shift $\sigma (b_i)=b_{i+1}$ preserves the defining relations and has inverse shift $i\mapsto i-1$, so it is an automorphism. Form the semidirect productIts base embeds by the definition of a semidirect product. Since $b_i=t^i{b}_0{t}^{-i}$, the group is generated by just $b_0,t$ and is countable.

The shift-invariant graph groups encode arbitrary sets construction has the exact testThe forward direction for $n\in S$ is a defining edge relation. If $n\notin S$, kill all vertex generators except $b_0,b_n$. Every defining commutator then either contains a killed generator or is trivial; no relation is imposed between these two retained generators. Thus there is a retraction of $B_S$ onto the rank-two free group on $b_0,b_n$. Their commutator is a nonempty reduced word there, so it is nonidentity in $B_S$ and remains so in $G_S$.

Now suppose $G_S$ embedded in a finitely presented group $P$. Choose fixed words $u,v$ in its finite generators representing the images of $b_0,t$. The null words of $P$ are recursively enumerable: enumerate all finite products of conjugates of its relators and their inverses, then freely reduce. Substituting $u,v$ into $[b_0,t^n{b}_0{t}^{-n}]$ and dovetailing these enumerations for all $n>0$ would enumerate exactly $S$, contradicting its choice. These finitely many image words may be hard-coded; the argument does not require an algorithm finding an embedding. This proves the enumerable identities of finitely generated subgroups obstruction and supplies a concrete construction of the required countable counterexample.

For $S\subseteq {\mathbb{N}}_{>0}$, form the Right-angled Artin group on generators $b_i$ indexed by $i\in \mathbb{Z}$, imposing $[b_i,b_j]=1$ precisely when $|i-j|\in S$. Translation $b_i\mapsto b_{i+1}$ is an automorphism. Its semidirect product with an infinite cyclic group generated by $t$ is countable and generated by $b_0,t$. If $n\notin S$, killing all $b_i$ except $b_0,b_n$ retracts the graph group onto the rank-two free group, so their commutator is nonidentity. This proves the displayed equivalence. For non-recursively-enumerable $S$, enumerable identities of finitely generated subgroups rules out an embedding of this two-generated group in any finitely presented group.