Stimulated by some Physics backgrounds, consider the following two sets of matrices.
Notations and definitions:Let $A,B$ be two complex $n\times n$ matrices, then $\left [ A,B \right ]\overset{\underset{\mathrm{def}}{}}{=}AB-BA$ and $ e^A\overset{\underset{\mathrm{def}}{}}{=} \sum_{m=0}^{\infty }\frac{A^m}{m!}. $
Let $M_1,M_2,M_3$ be three $n\times n$ Hermitian matrices, and they satisfy $[M_1,M_2]=iM_3,[M_2,M_3]=iM_1,[M_3,M_1]=iM_2$ identities, where $i=\sqrt{-1}$.
Now define a set $X$ of unitary matrices:$X=\left \{ e^{i\alpha M_3}e^{i\beta M_2}e^{i\gamma M_3} :\alpha,\beta,\gamma \in \mathbb{R}\right \},$ and another set $Y$ of unitary matrices:$Y=\left \{ e^{i(\alpha M_1+\beta M_2+\gamma M_3)} :\alpha,\beta,\gamma \in \mathbb{R}\right \},$ where $i=\sqrt{-1}$ ( Note that the number indices of the Hermitian matrices $M$ are different in $X$ and $Y$ ).
And my questions are as follows:
(1) Is $X=Y$ ?
(2) If (1) is true, is the set $X$ a group ?
(3) If both (1) and (2) are true, is $X\cong SU(2)$ true ?
Supplements: For concreteness, let's take a look at the following simple example. Consider the Physically called spin-$\frac{1}{2}$ "Pauli" matrices $M_1=\frac{1}{2}\bigl(\begin{smallmatrix} 0& 1\\ 1&0 \end{smallmatrix}\bigr),M_2=\frac{1}{2}\bigl(\begin{smallmatrix} 0& -i\\ i&0 \end{smallmatrix}\bigr)$ and $M_3=\frac{1}{2}\bigl(\begin{smallmatrix} 1& 0\\ 0&-1 \end{smallmatrix}\bigr),$ and it's easy to find that $M_1^2+M_2^2+M_3^2=\frac{1}{2}(\frac{1}{2}+1)\mathbb{I}$, where $\mathbb{I}$ is a $2\times2$ identity matrix.
In the above example, direct calculation of matrices $e^{i\alpha M_3}e^{i\beta M_2}e^{i\gamma M_3}$ in $X$ shows that $X=SU(2)$ (then $X$ is a group), and it's also easy to verify that $Y\subseteq SU(2)$. So now the question is, is $SU(2)\subseteq Y$ too ?
Thanks in advance.
Comments to the question (v3):
(We will from now on assume that the $n\times n$ matrices $M_1$, $M_2$ and $M_3$ are linearly independent, and that $n\geq 2$.)
OP defines two sets:
$$\tag{1} X_n~:=~ \{ e^{i\alpha M_3}e^{i\beta M_2}e^{i\gamma M_3}\in {\rm Mat}_{n\times n}(\mathbb{C}) \mid \alpha,\beta,\gamma \in \mathbb{R} \}, $$
$$\tag{2} Y_n~:=~ \{ e^{i(\alpha M_1+\beta M_2+\gamma M_3)} \in {\rm Mat}_{n\times n}(\mathbb{C}) \mid\alpha,\beta,\gamma \in \mathbb{R} \}.$$
The set $Y_n$ is the image $Y_n=\rho(SU(2))$ of an $n$-dimensional $SU(2)$ group representation $\rho:SU(2) \to GL(n,\mathbb{C})$, also known in physics as a spin $\frac{n-1}{2}$ representation. In particular, the set $Y$ is itself a group.
$$\tag{3} Y_n~\cong~\left\{ \begin{array}{rcl} SU(2)/ \mathbb{Z}_2 ~\cong~SO(3) &\text{for}&n&\text{odd}, \\ SU(2) &\text{for}&n&\text{even}.\end{array} \right. $$
The set $X_n\subseteq Y_n$ is a subset of $Y_n$; due, among other things, to the Baker-Campbell-Hausdorff formula.
In fact, the set $X_n$ is a generalized Euler angle realization of $SU(2)$, as mentioned by OP himself. One may check by inspection that the $X_n$ hit every element in $Y_n$, so that $X_n=Y_n$.