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Let me give half an answer by pointing out that it is relatively consistent with ZFC that there is such a group. Indeed, it is relatively consistent with ZFC that there are numerous such groups, and indeed, that the continuum is very large and every dense subgroup of size less than the continuum (which would include many uncountable subgroups) has the property you mention. This is a consequence of Martin's Axiom plus $\neg$CH.

The reason is that the classical Vitali argument generalizes to uncountable subgroups, provided that they have size less than the additivity number $\text{add}(\mathcal{N})$ of the null ideal $\mathcal{N}$, and actually, it suffices to be less than the covering number $\text{cov}(\mathcal{N})$. The additivity number is the largest cardinal such that the union of fewer than $\text{add}(\mathcal{N})$ many measure zero sets still has measure zero (see this MO question for further information). We all know that the union of countably many measure zero sets has measure zero, and so $\aleph_1\leq\text{add}(\mathcal{N})\leq 2^{\aleph_0}$. But it is also known to be relatively consistent with $\text{ZFC}+\neg\text{CH}$ that one may take the union of any $\aleph_1$ many (or more) measure zero sets and still have a measure zero set. In other words, it is relatively consistent that $\text{add}(\mathcal{N})$ is much larger than $\aleph_1$. Indeed, for any ordinal $\alpha$, one can arrange that $\aleph_\alpha\leq\text{add}(\mathcal{N})=2^{\aleph_0}$. Indeed, $\text{add}(\mathcal{N})=2^{\aleph_0}$ is a consequence of Martin's Axiom MA, which is consistent with very large values of the continuum.

The point now is that the classical Vitali argument shows that if $\Gamma$ is any subgroup of $\mathbb{R}$ of size less than the additivity number $\text{add}(\mathcal{N})$, and $V$ is a selector with respect to translation by $\Gamma$, selecting one element from each equivalence class, then $V$ will be non-measurable. To see this, observe that since $\mathbb{R}$ is the union of $|\Gamma|$ many translates of $V$, it follows that $V$ cannot have measure zero, since the union of fewer than $\text{add}(\mathcal{N})$ many measure zero sets still has measure zero. And $V$ cannot have positive measure, since then it will have positive measure on an finite interval, and one can proceed just as in the Vitali case, finding infinitely many disjoint positive measure sets in a bounded interval, a contradiction.

The argument can be improved to the case of $\Gamma$ of size less than the covering number of the null ideal $\text{cov}(\mathcal{N})$, the smallest number of measure zero sets that cover $\mathbb{R}$, since we had covered $\mathbb{R}$ with the $\Gamma$-translates of $V$. This is an improvement, since it is consistent that the covering number is strictly larger than the additivity number.

In summary, what the argument shows is that it is consistent with ZFC that the continuum is very large, but every dense subgroup of $\mathbb{R}$ of size less than the continuum, and this includes many uncountable subgroups since the continuum is large, has all their selectors being non-measurable. This situation is a consequence of $\text{MA}+\neg\text{CH}$.

Let me give half an answer by pointing out that it is relatively consistent with ZFC that there is such a group. Indeed, it is relatively consistent with ZFC that there are numerous such groups, and indeed, that the continuum is very large and every dense subgroup of size less than the continuum (which would include many uncountable subgroups) has the property you mention. This is a consequence of Martin's Axiom plus $\neg$CH.

The reason is that the classical Vitali argument generalizes to uncountable subgroups, provided that they have size less than the additivity number $\text{add}(\mathcal{N})$ of the null ideal $\mathcal{N}$, and actually, it suffices to be less than the covering number $\text{cov}(\mathcal{N})$. The additivity number is the largest cardinal such that the union of fewer than $\text{add}(\mathcal{N})$ many measure zero sets still has measure zero (see this MO question for further information). We all know that the union of countably many measure zero sets has measure zero, and so $\aleph_1\leq\text{add}(\mathcal{N})\leq 2^{\aleph_0}$. But it is also known to be relatively consistent with $\text{ZFC}+\neg\text{CH}$ that one may take the union of any $\aleph_1$ many (or more) measure zero sets and still have a measure zero set. In other words, it is relatively consistent that $\text{add}(\mathcal{N})$ is much larger than $\aleph_1$. Indeed, for any ordinal $\alpha$, one can arrange that $\aleph_\alpha\leq\text{add}(\mathcal{N})=2^{\aleph_0}$. Indeed, $\text{add}(\mathcal{N})=2^{\aleph_0}$ is a consequence of Martin's Axiom MA, which is consistent with very large values of the continuum.

The point now is that the classical Vitali argument shows that if $\Gamma$ is any subgroup of $\mathbb{R}$ of size less than the additivity number $\text{add}(\mathcal{N})$, and $V$ is a selector with respect to translation by $\Gamma$, selecting one element from each equivalence class, then $V$ will be non-measurable. To see this, observe that since $\mathbb{R}$ is the union of $|\Gamma|$ many translates of $V$, it follows that $V$ cannot have measure zero, since the union of fewer than $\text{add}(\mathcal{N})$ many measure zero sets still has measure zero. And $V$ cannot have positive measure, since then it will have positive measure on an finite interval, and one can proceed just as in the Vitali case, finding infinitely many disjoint positive measure sets in a bounded interval, a contradiction.

The argument can be improved to the case of $\Gamma$ of size less than the covering number of the null ideal $\text{cov}(\mathcal{N})$, the smallest number of measure zero sets that cover $\mathbb{R}$, since we had covered $\mathbb{R}$ with the $\Gamma$-translates of $V$. This is an improvement, since it is consistent that the covering number is strictly larger than the additivity number.

In summary, what the argument shows is that it is consistent with ZFC that the continuum is very large, but every dense subgroup of $\mathbb{R}$ of size less than the continuum, and this includes many uncountable subgroups since the continuum is large, has all their selectors being non-measurable. This situation is a consequence of $\text{MA}+\neg\text{CH}$.

Let me give half an answer by pointing out that it is relatively consistent with ZFC that there is such a group. Indeed, it is relatively consistent with ZFC that there are numerous such groups, and indeed, that the continuum is very large and every dense subgroup of size less than the continuum (which would include many uncountable subgroups) has the property you mention. This is a consequence of Martin's Axiom plus $\neg$CH.

The reason is that the classical Vitali argument generalizes to uncountable subgroups, provided that they have size less than $\mathfrak{a}$, the additivity number of the null ideal. This is the largest cardinal such that the union of fewer than $\mathfrak{a}$ many measure zero sets still has measure zero (see this MO question for further information). We all know that the union of countably many measure zero sets has measure zero, and so $\aleph_1\leq\mathfrak{a}\leq 2^{\aleph_0}$. But it is also known to be relatively consistent with $\text{ZFC}+\neg\text{CH}$ that one may take the union of any $\aleph_1$ many (or more) measure zero sets and still have a measure zero set. In other words, it is relatively consistent that $\mathfrak{a}$ is much larger than $\aleph_1$. Indeed, for any ordinal $\alpha$, one can arrange that $\aleph_\alpha\leq\mathfrak{a}=2^{\aleph_0}$. Indeed, $\mathfrak{a}=2^{\aleph_0}$ is a consequence of Martin's Axiom MA, which is consistent with very large values of the continuum.

The point now is that the classical Vitali argument shows that if $\Gamma$ is any subgroup of $\mathbb{R}$ of size less than the additivity number $\mathfrak{a}$, and $V$ is a selector with respect to translation by $\Gamma$, selecting one element from each equivalence class, then $V$ will be non-measurable. To see this, observe that since $\mathbb{R}$ is the union of $|\Gamma|$ many translates of $V$, it follows that $V$ cannot have measure zero, since the union of fewer than $\mathfrak{a}$ many measure zero sets still has measure zero. And $V$ cannot have positive measure, since then it will have positive measure on an finite interval, and one can proceed just as in the Vitali case, finding infinitely many disjoint positive measure sets in a bounded interval, a contradiction.

What the argument shows is that it is consistent with ZFC that the continuum is very large, but every dense subgroup of $\mathbb{R}$ of size less than the continuum, and this includes many uncountable subgroups since the continuum is large, has all their selectors being non-measurable. This situation is a consequence of $\text{MA}+\neg\text{CH}$.

Let me give half an answer by pointing out that it is relatively consistent with ZFC that there is such a group. Indeed, it is relatively consistent with ZFC that there are numerous such groups, and indeed, that the continuum is very large and every dense subgroup of size less than the continuum (which would include many uncountable subgroups) has the property you mention. This is a consequence of Martin's Axiom plus $\neg$CH.

The reason is that the classical Vitali argument generalizes to uncountable subgroups, provided that they have size less than the additivity number $\text{add}(\mathcal{N})$ of the null ideal $\mathcal{N}$, and actually, it suffices to be less than the covering number $\text{cov}(\mathcal{N})$. The additivity number is the largest cardinal such that the union of fewer than $\text{add}(\mathcal{N})$ many measure zero sets still has measure zero (see this MO question for further information). We all know that the union of countably many measure zero sets has measure zero, and so $\aleph_1\leq\text{add}(\mathcal{N})\leq 2^{\aleph_0}$. But it is also known to be relatively consistent with $\text{ZFC}+\neg\text{CH}$ that one may take the union of any $\aleph_1$ many (or more) measure zero sets and still have a measure zero set. In other words, it is relatively consistent that $\text{add}(\mathcal{N})$ is much larger than $\aleph_1$. Indeed, for any ordinal $\alpha$, one can arrange that $\aleph_\alpha\leq\text{add}(\mathcal{N})=2^{\aleph_0}$. Indeed, $\text{add}(\mathcal{N})=2^{\aleph_0}$ is a consequence of Martin's Axiom MA, which is consistent with very large values of the continuum.

The point now is that the classical Vitali argument shows that if $\Gamma$ is any subgroup of $\mathbb{R}$ of size less than the additivity number $\text{add}(\mathcal{N})$, and $V$ is a selector with respect to translation by $\Gamma$, selecting one element from each equivalence class, then $V$ will be non-measurable. To see this, observe that since $\mathbb{R}$ is the union of $|\Gamma|$ many translates of $V$, it follows that $V$ cannot have measure zero, since the union of fewer than $\text{add}(\mathcal{N})$ many measure zero sets still has measure zero. And $V$ cannot have positive measure, since then it will have positive measure on an finite interval, and one can proceed just as in the Vitali case, finding infinitely many disjoint positive measure sets in a bounded interval, a contradiction.

The argument can be improved to the case of $\Gamma$ of size less than the covering number of the null ideal $\text{cov}(\mathcal{N})$, the smallest number of measure zero sets that cover $\mathbb{R}$, since we had covered $\mathbb{R}$ with the $\Gamma$-translates of $V$. This is an improvement, since it is consistent that the covering number is strictly larger than the additivity number.

In summary, what the argument shows is that it is consistent with ZFC that the continuum is very large, but every dense subgroup of $\mathbb{R}$ of size less than the continuum, and this includes many uncountable subgroups since the continuum is large, has all their selectors being non-measurable. This situation is a consequence of $\text{MA}+\neg\text{CH}$.

Let me point out at least that it is relatively consistent with ZFC that there is such a group. Indeed, it is a consequence of Martin's Axiom plus $\neg$CH.

This is because any dense subgroup of size less than the additivity $\mathfrak{a}$ of the null ideal will have all non-measurable selectors (see this MO question for more information about what this means). This follows by essentially the same reasons as for the Vitali set. Namely, suppose that $\Gamma$ is a dense subgroup of $\mathbb{R}$ of size less than $\mathfrak{a}$, and suppose that $V$ selects one element from each equivalence class by $\Gamma$-translation. Thus,  $\mathbb{R}$ is the disjoint union of $|\Gamma|$ many translations of $V$. So $V$ cannot have measure $0$, since by assumption the union of $|\Gamma|$ many measure zero sets still has measure zero, and $V$ cannot have positive measure, since then it will have positive measure on an finite interval, and one can proceed just as in the Vitali case, finding infinitely many disjoint positive measure sets in a bounded interval, contradiction.

Finally, it is known to be consistent with ZFC that the additivity number can be strictly larger than $\omega_1$. This is a consequence, for example, of Martin's Axiom plus $\neg$CH.

Let me give half an answer by pointing out that it is relatively consistent with ZFC that there is such a group. Indeed, it is relatively consistent with ZFC that there are numerous such groups, and indeed, that the continuum is very large and every dense subgroup of size less than the continuum (which would include many uncountable subgroups) has the property you mention. This is a consequence of Martin's Axiom plus $\neg$CH.

The reason is that the classical Vitali argument generalizes to uncountable subgroups, provided that they have size less than $\mathfrak{a}$, the additivity number of the null ideal. This is the largest cardinal such that the union of fewer than $\mathfrak{a}$ many measure zero sets still has measure zero (see this MO question for further information). We all know that the union of countably many measure zero sets has measure zero, and so $\aleph_1\leq\mathfrak{a}\leq 2^{\aleph_0}$. But it is also known to be relatively consistent with $\text{ZFC}+\neg\text{CH}$ that one may take the union of any $\aleph_1$ many (or more) measure zero sets and still have a measure zero set. In other words, it is relatively consistent that $\mathfrak{a}$ is much larger than $\aleph_1$. Indeed, for any ordinal $\alpha$, one can arrange that $\aleph_\alpha\leq\mathfrak{a}=2^{\aleph_0}$. Indeed, $\mathfrak{a}=2^{\aleph_0}$ is a consequence of Martin's Axiom MA, which is consistent with very large values of the continuum.

The point now is that the classical Vitali argument shows that if $\Gamma$ is any subgroup of $\mathbb{R}$ of size less than the additivity number $\mathfrak{a}$, and $V$ is a selector with respect to translation by $\Gamma$, selecting one element from each equivalence class, then $V$ will be non-measurable. To see this, observe that since $\mathbb{R}$ is the union of $|\Gamma|$ many translates of $V$, it follows that $V$ cannot have measure zero, since the union of fewer than $\mathfrak{a}$ many measure zero sets still has measure zero. And $V$ cannot have positive measure, since then it will have positive measure on an finite interval, and one can proceed just as in the Vitali case, finding infinitely many disjoint positive measure sets in a bounded interval, a contradiction.

What the argument shows is that it is consistent with ZFC that the continuum is very large, but every dense subgroup of $\mathbb{R}$ of size less than the continuum, and this includes many uncountable subgroups since the continuum is large, has all their selectors being non-measurable. This situation is a consequence of $\text{MA}+\neg\text{CH}$.