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Ignore this: Higman's theorem

It is about finitely generatedknown [Ould Houcine, Abderezak. Embeddings in finitely presented groups which preserve the center. J. Algebra 307 (2007), no. 1, and 1--23. MR2278040 (2007i:20043)] that there is a finitely presented group which has $\mathbb Q$ as center, which is not a very nice result! Indeed, the paper shows that (i) every countable group $G$ embeds into a finitely generated ! group $K$ such that $Z(G)=Z(K)$ and (ii) Every finitely generated recursively presented group $G$ embeds into a finitely presented group $K$ such that $Z(G)=Z(K)$.

I had originally started the answer with the following:

It is known [Ould Houcine, Abderezak. Embeddings in

but then retracted it because Rotman actually states (and proves) Higman's theorem for finitely generated finitely presented groupswhich preserve the center. J. Algebra 307 (2007), no. 1, 1--23. MR2278040 (2007i:20043)] that there is a finitely presented group which has and $\mathbb Q$ as center, which is a very nice resultnot finitely generated! Indeed

Later, though, Jack Schmidt observed than in fact Higman does deal with the paper shows that (i) every countable group $G$ embeds into a finitely countably generated group $K$ such that $Z(G)=Z(K)$ and case in the original paper (ii) Every finitely generated recursively presented group $G$ embeds into a finitely presented group $K$ such that $Z(G)=Z(K)$.please refer to his comments below for details) so the retracted text should be unretracted.

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Ignore this: Higman's theorem is about finitely generated finitely presented groups, and $\mathbb Q$ is not finitely generated!

The statement to which you are referring is a consequence of G. Higman's theorem that states that every group having a recursive presentation can be embedded in a finitely presented group. See [Higman, G. Subgroups of finitely presented groups. Proc. Roy. Soc. Ser. A 262 1961 455--475. MR0130286 (24 #A152)]

Since $\mathbb Q$ is plainly recursively presentable, there is a group like the one you want... You can follow the construction given by Rotman in the last chapter of his Introduction to the Theory of Groups to obtain a presentation---the end result is not going to be pretty, though... (It is the proof of this result that makes use of the unsettling folded plates that come with the book)

It is known [Ould Houcine, Abderezak. Embeddings in finitely presented groups which preserve the center. J. Algebra 307 (2007), no. 1, 1--23. MR2278040 (2007i:20043)] that there is a finitely presented group which has $\mathbb Q$ as center, which is a very nice result! Indeed, the paper shows that (i) every countable group $G$ embeds into a finitely generated group $K$ such that $Z(G)=Z(K)$ and (ii) Every finitely generated recursively presented group $G$ embeds into a finitely presented group $K$ such that $Z(G)=Z(K)$.
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Ignore this: Higman's theorem is about finitely generated finitely presented groups, and $\mathbb Q$ is not finitely generated!

The statement to which you are referring is a consequence of G. Higman's theorem that states that every group having a recursive presentation can be embedded in a finitely presented group. See [Higman, G. Subgroups of finitely presented groups. Proc. Roy. Soc. Ser. A 262 1961 455--475. MR0130286 (24 #A152)]

Since $\mathbb Q$ is plainly recursively presentable, there is a group like the one you want... You can follow the construction given by Rotman in the last chapter of his Introduction to the Theory of Groups to obtain a presentation---the end result is not going to be pretty, though... (It is the proof of this result that makes use of the unsettling folded plates that come with the book)

In fact, it is know [Ould Houcine, Abderezak. Embeddings in finitely presented groups which preserve the center. J. Algebra 307 (2007), no. 1, 1--23. MR2278040 (2007i:20043)] that there is a finitely presented group which has $\mathbb Q$ as center, which is a very nice result!
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