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Your guess is correct: if $H$ is a closed subgroup containing $\operatorname{SL}_n({\mathbb Z})$,then it is a Lie subgroup. If $\mathfrak h$ is its Lie algebra, then $\mathfrak h$ is stable under the adjoint action of $SL_n(\mathbb Z)$ and hence under all of $\operatorname{SL}_n(\mathbb R)$ (since $\operatorname{SL}_n(\mathbb Z)$ is Zariski dense in $\operatorname{SL}_n(\mathbb R)$). If $\mathfrak h$ is non-zero, that means, by the simplicity of $\operatorname{SL}_n(\mathbb R)$, that $\mathfrak h$ is the Lie algebra of $\operatorname{SL}_n(\mathbb R)$ and hence $H$ is all of $\operatorname{SL}_n(\mathbb R)$.

If $\mathfrak h =0$, then $H$ is discrete and you have accepted the result in this case.

In the generality that you have asked, $G(\mathbb Z)$ need not be maximal in $G(\mathbb R)$; if $G$ is a simply connected semi-simple algebraic group over $\mathbb Q$ (and is $\mathbb Q$-simple), with $G(\mathbb R)$ non-compact, and $K_p$ is a maximal compact open subgroup of $G(\mathbb Q_p)$ for each prime $p$, then the intersection of $G(\mathbb Q)$ with all the $K_p$ may be "called" $G(\mathbb Z)$ and the same argument goes through to say that $G(\mathbb Z)$ is maximal among closed subgroups of $G(\mathbb R)$.

Finally, your statement that integer points are dense in $\operatorname{SO}(3)$ is incorrect. In this case, integer points are finite, since $\operatorname{SO}(3)$ is compact.

Your guess is correct: if $H$ is a closed subgroup containing $\operatorname{SL}_n({\mathbb Z})$,then it is a Lie subgroup. If $\mathfrak h$ is its Lie algebra, then $\mathfrak h$ is stable under the adjoint action of $SL_n(\mathbb Z)$ and hence under all of $\operatorname{SL}_n(\mathbb R)$ (since $\operatorname{SL}_n(\mathbb Z)$ is Zariski dense in $\operatorname{SL}_n(\mathbb R)$). If $\mathfrak h$ is non-zero, that means, by the simplicity of $\operatorname{SL}_n(\mathbb R)$, that $\mathfrak h$ is the Lie algebra of $\operatorname{SL}_n(\mathbb R)$ and hence $H$ is all of $\operatorname{SL}_n(\mathbb R)$.

If $\mathfrak h =0$, then $H$ is discrete and you have accepted the result in this case.

In the generality that you have asked, $G(\mathbb Z)$ need not be maximal in $G(\mathbb R)$; if $G$ is a simply connected semi-simple algebraic group over $\mathbb Q$ (and is $\mathbb Q$-simple), with $G(\mathbb R)$ non-compact, and $K_p$ is a maximal compact open subgroup of $G(\mathbb Q_p)$ for each prime $p$, then the intersection of $G(\mathbb Q)$ with all the $K_p$ may be "called" $G(\mathbb Z)$ and the same argument goes through to say that $G(\mathbb Z)$ is maximal among closed subgroups of $G(\mathbb R)$.

Your guess is correct: if $H$ is a closed subgroup containing $SL_n({\mathbb Z})$,then it is a Lie subgroup. If $\mathfrak h$ is its Lie algebra, then $\mathfrak h$ is stable under the adjoint action of $SL_n(\mathbb Z)$ and hence under all of $SL_n(\mathbb R)$ (since $SL_n(\mathbb Z)$ is Zariski dense in $SL_n(\mathbb R)$). If $\mathfrak h$ is non-zero, that means, by the simplicity of $SL_n(\mathbb R)$, that $\mathfrak h$ is the Lie algebra of $SL_n(\mathbb R)$ and hence $H$ is all of $SL_n(\mathbb R)$.

If $\mathfrak h =0$, then $H$ is discrete and you have accepted the result in this case.

In the generality that you have asked, $G(\mathbb Z)$ need not be maximal in $G(\mathbb R)$; if $G$ is a simply connected semi-simple algebraic group over $\mathbb Q$ ( and is $\mathbb Q$  -simple), with $G(\mathbb R)$ non-compact, and $K_p$ is a maximal compact open subgroup of $G(\mathbb Q_p)$ for each prime $p$, then the intersection of $G(\mathbb Q)$ with all the $K_p$ may be "called" $G(\mathbb Z)$ and the same argument goes through to say that $G(\mathbb Z)$ is maximal among closed subgroups of $G(\mathbb R)$.

Finally, your statement that integer points are dense in SO(3) is incorrect. In this case, integer points are finite, since SO(3) is compact.

Your guess is correct: if $H$ is a closed subgroup containing $\operatorname{SL}_n({\mathbb Z})$,then it is a Lie subgroup. If $\mathfrak h$ is its Lie algebra, then $\mathfrak h$ is stable under the adjoint action of $SL_n(\mathbb Z)$ and hence under all of $\operatorname{SL}_n(\mathbb R)$ (since $\operatorname{SL}_n(\mathbb Z)$ is Zariski dense in $\operatorname{SL}_n(\mathbb R)$). If $\mathfrak h$ is non-zero, that means, by the simplicity of $\operatorname{SL}_n(\mathbb R)$, that $\mathfrak h$ is the Lie algebra of $\operatorname{SL}_n(\mathbb R)$ and hence $H$ is all of $\operatorname{SL}_n(\mathbb R)$.

If $\mathfrak h =0$, then $H$ is discrete and you have accepted the result in this case.

In the generality that you have asked, $G(\mathbb Z)$ need not be maximal in $G(\mathbb R)$; if $G$ is a simply connected semi-simple algebraic group over $\mathbb Q$ (and is $\mathbb Q$-simple), with $G(\mathbb R)$ non-compact, and $K_p$ is a maximal compact open subgroup of $G(\mathbb Q_p)$ for each prime $p$, then the intersection of $G(\mathbb Q)$ with all the $K_p$ may be "called" $G(\mathbb Z)$ and the same argument goes through to say that $G(\mathbb Z)$ is maximal among closed subgroups of $G(\mathbb R)$.

Finally, your statement that integer points are dense in $\operatorname{SO}(3)$ is incorrect. In this case, integer points are finite, since $\operatorname{SO}(3)$ is compact.

Your guess is correct: if $H$ is a closed subgroup containing $SL_n({\mathbb Z})$,then it is a Lie subgroup. If $\mathfrak h$ is its Lie algebra, then $\mathfrak h$ is stable under the adjoint action of $SL_n(\mathbb Z)$ and hence under all of $SL_n(\mathbb R)$ (since $SL_n(\mathbb Z)$ is Zariski dense in $SL_n(\mathbb R)$). If $\mathfrak h$ is non-zero, that means, by the simplicity of $SL_n(\mathbb R)$, that $\mathfrak h$ is the Lie algebra of $SL_n(\mathbb R)$ and hence $H$ is all of $SL_n(\mathbb R)$.

If $\mathfrak h =0$, then $H$ is discrete and you have accepted the result in this case.

In the generality that you have asked, $G(\mathbb Z)$ need not be maximal in $G(\mathbb R)$; if $G$ is a simply connected semi-simple algebraic group over $\mathbb Q$ ( and is $\mathbb Q$ -simple), with $G(\mathbb R)$ non-compact, and $K_p$ is a maximal compact open subgroup of $G(\mathbb Q_p)$ for each prime $p$, then the intersection of $G(\mathbb Q)$ with all the $K_p$ may be "called" $G(\mathbb Z)$ and the same argument goes through to say that $G(\mathbb Z)$ is maximal among closed subgroups of $G(\mathbb R)$.

Your guess is correct: if $H$ is a closed subgroup containing $SL_n({\mathbb Z})$,then it is a Lie subgroup. If $\mathfrak h$ is its Lie algebra, then $\mathfrak h$ is stable under the adjoint action of $SL_n(\mathbb Z)$ and hence under all of $SL_n(\mathbb R)$ (since $SL_n(\mathbb Z)$ is Zariski dense in $SL_n(\mathbb R)$). If $\mathfrak h$ is non-zero, that means, by the simplicity of $SL_n(\mathbb R)$, that $\mathfrak h$ is the Lie algebra of $SL_n(\mathbb R)$ and hence $H$ is all of $SL_n(\mathbb R)$.

If $\mathfrak h =0$, then $H$ is discrete and you have accepted the result in this case.

In the generality that you have asked, $G(\mathbb Z)$ need not be maximal in $G(\mathbb R)$; if $G$ is a simply connected semi-simple algebraic group over $\mathbb Q$ ( and is $\mathbb Q$ -simple), with $G(\mathbb R)$ non-compact, and $K_p$ is a maximal compact open subgroup of $G(\mathbb Q_p)$ for each prime $p$, then the intersection of $G(\mathbb Q)$ with all the $K_p$ may be "called" $G(\mathbb Z)$ and the same argument goes through to say that $G(\mathbb Z)$ is maximal among closed subgroups of $G(\mathbb R)$.

Finally, your statement that integer points are dense in SO(3) is incorrect. In this case, integer points are finite, since SO(3) is compact.