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Given a group $G$ denote by $G^\ast=\mathrm{Hom}(G,\Bbb Z)$ its dual and by $j\colon G\to G^{\ast\ast}$ the canonical homomorphism $g\mapsto (f\mapsto f(g))$. A group is reflexive iff $j$ is an isomorphism.

There are known examples of groups such that $G\simeq G^{\ast\ast}$ but $G$ is not reflexive, the one I'm aware of is $G=\mathcal C((\omega_1+1)\setminus E,\Bbb Z)$, where $E\subseteq\omega_1$ is a stationary costationary set of limit ordinals. In this case $G\simeq G^{\ast\ast}$, but not through $j$, in fact $j(G)$ has codimension $1$ in $G^{\ast\ast}$.

However the existence of a stationary costationary subset of $\omega_1$ cannot be proved in $\mathsf{ZF}$ alone, hence my question: are there known examples of groups $G$ such that $G$ is not reflexive, $G\simeq G^{\ast\ast}$ and the construction of $G$ (and the proofs of those properties) can be carried out without using $\mathsf{AC}$?

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