A certain finite group of rotations There are twenty four $\ \ 3$-dimensional  rotations matrices where entries are from the values $\{-1,0,1\}$,    see question (*).
Name  the set of all such matrices $T$ and matrices from this set "orthogonally regular".
They form a finite group as every inverse and product of any two matrices from the set $T$ belongs to this set. 
Suppose that we add to the set $T$ a single other rotation matrix which has some entries beyond values $\{ -1,0,1\}$. 


*

*How to  prove that the set generated by all possible matrices from $T$
and this additional matrix can't be a finite group?
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I'm suspect that the claim is true. Some experimental results support this.   
For example suppose that the added to $T$ is the matrix of rotation  about axis $Z$ by $\pi/4$ i.e    
$\begin{bmatrix}  \dfrac {\sqrt{2}}{2}   & -\dfrac {\sqrt{2}}{2} & 0 \\  \dfrac {\sqrt{2}}{2} &  \dfrac {\sqrt{2}}{2} & 0 \\ 0 & 0 & 1 \end{bmatrix}$.      
If such matrix belongs to the hypothetical finite group also similar rotations about axes $X$ and $Y$ belong to this group.     
For example:    
$ \begin{bmatrix}  \  1   & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix}  \ \dfrac {\sqrt{2}}{2}   & -\dfrac {\sqrt{2}}{2} & 0 \\ \dfrac {\sqrt{2}}{2} & \dfrac {\sqrt{2}}{2} & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}  \  1   & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{bmatrix}= \begin{bmatrix}   \dfrac {\sqrt{2}}{2} & 0  & -\dfrac {\sqrt{2}}{2}   \\   0 & 1 & 0 \\  \dfrac {\sqrt{2}}{2} & 0 & \dfrac {\sqrt{2}}{2}   \\  \end{bmatrix}$      
what is rotation by $(-\pi/4)$ about $Y$.  Inverse also belongs to the group. In similar way we can prove that rotation by $ \pi/4 $ about $X$ also belongs to the group.
But then
$A=\begin{bmatrix}  \ \dfrac {\sqrt{2}}{2}   & -\dfrac {\sqrt{2}}{2} & 0 \\ \dfrac {\sqrt{2}}{2}& \dfrac {\sqrt{2}}{2} & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}    1 & 0 & 0 \\ 0 & \dfrac {\sqrt{2}}{2}   & -\dfrac {\sqrt{2}}{2}   \\ 0 & \dfrac {\sqrt{2}}{2} & \dfrac {\sqrt{2}}{2}   \end{bmatrix}=\begin{bmatrix}  \dfrac{1}{\sqrt{2}} & -\dfrac{ 1}{2} & \dfrac{1}{2}  \\  \dfrac {1}{\sqrt{2}} & \dfrac {1}{2} & -\dfrac{ 1}{2} \\  0 & \dfrac{1}{\sqrt{2}} & \dfrac{1}{ \sqrt{2} } \end{bmatrix} $  should  belong also to this group.    
However it seems that a sequence $A,A^2,A^3, ... $ is not cyclic so the group can't be finite. But how to prove a general case ? Is it impossible a finite group which includes all "orthogonally regular" matrices which has more elements than $T$?
 A: If you view the diagonal matrices with entries $\pm1$ as a copy of $\mathbb{Z}_2^3$, then only half of them are rotations (the ones with determinant $+1$), and form a subgroup $V$ isomorphic to the Klein four group, $\mathbb{Z}_2^2$, which is invariant under permuting the diagonal entries. Note that permutations of the diagonal entries are group automorphisms of $V$ (and $\mathbb{Z}_2^3$) and can be achieved by conjugating the diagonal matrix by the appropriate permutation matrix. Moreover, if you multiply such a diagonal matrix by a permutation matrix, you get a "signed permutation matrix" which is a permutation matrix but with possible minus signs. Altogether, this means your group is an internal semidirect product isomorphic to $\mathbb{Z}_2^2\rtimes S_3$.
(Aside: since $S_3=\mathrm{Aut}(\mathbb{Z}_2^2)$, which by the way is a fun fact to observe, this means your group is the holomorph of the Klein four group $\mathrm{Hol}(V)=V\rtimes\mathrm{Aut}(V)$. It may also be interpreted as an index $2$ subgroup (or quotient) of the full wreath product $\mathbb{Z}_2\wr S_3=\mathbb{Z}_2^3\rtimes S_3$ which can be achieved as the full group of signed permutation matrices, including those with determinant $-1$.)
Finite subgroups of $\mathrm{SO}(3)$ are classified: they are all cyclic, dihedral, or one of the three permutation groups $A_4,S_4,A_5$, up to isomorphism. (This looks to be an excellent exercise in group actions.) Being nonabelian and of order $24$, the only possibilities are that it is a copy of $S_4$, and in fact $S_4\cong V_4\rtimes S_3$ is a semidirect product. Moreover, checking orders in this classification shows it cannot be a subgroup of a bigger finite group.
