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I have a group generated by 3 elements: let it be $G=\langle a,b,c\rangle$. The concrete form is given by some matrices, but never mind; my question is purely group theoretic.

Let us assume the generating elements are pairwise free(i.e., each of $\langle a,b\rangle$, $\langle b,c\rangle$, $\langle c,a\rangle$ is isomorphic to $F_2$). The question is: Can it be possible that there exists a subgroup $H\le G$ such that $\langle a\rangle \lneq H$ and $Z(H)=\langle a\rangle$?

The simplest case: if $G$ is itself free, then clearly it's impossible, since every subgroup of a free group is again free.

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Yes, that would be the case in the group defined by the presentation $$\langle a,b,c \mid [a,bc] = [a,cb]=1 \rangle.$$ with $H = \langle a,bc,cb \rangle$, which is the direct product of $\langle a \rangle$ with the free group $\langle bc,cb \rangle$.

This is an HNN-extension, and all of the properties you need follow from the basic theory of HNN-extensions, such as Britton's lemma, and the fact that the subgroup $\langle bc,cb \rangle$ has trivial intersection with $\langle b \rangle$ and with $\langle c \rangle.$

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