Questions about the subgroups of $SU(2)$ and relevant problems?

Question

This question is based on the invariant gauge groups in condensed matter physics( https://physics.stackexchange.com/questions/76644/different-invariant-gauge-groups-igg-on-different-lattices-with-the-same-form ).

As we know, the group $SU(2)$ can be viewed as the set of all the $2\times2$ unitary matrices with determinant $1$. And let the groups $U(1)=\left \{\begin{pmatrix} e^{i\theta}&0 \\ 0 & e^{-i\theta} \end{pmatrix} \mid 0\leqslant\theta<2\pi \right \}$, $\mathbb{Z}_2=\left \{I,-I \right \}$, where $I$ is the $2\times2$ identity matrix. Then $U(1)$ and $\mathbb{Z}_2$ are both the subgroups of $SU(2)$. My questions are as follows:

(1) In addition to $U(1)$ and $\mathbb{Z}_2$(of course $SU(2)$ itself and its subgroup { $I$ }), are there any other subgroup in $SU(2)$?

(2) Does $SU(2)$ have two subgroups called $A$ and $B$ such that: $U(1)$ is a subgroup of $A$ , $B$ is a subgroup of $U(1)$, and $\mathbb{Z}_2$ is a subgroup of $B$ ?

(3) Also as we know, $SO(3)\cong SU(2)/\mathbb{Z}_2$, and does $SU(2)$ have a subgroup called $C$ such that $C\cong SO(3)$? More generally, if $K$ is a normal subgroup of group $G$, then does $G$ have a subgroup called $H$ such that $H\cong G/K$?

Thanks in advance.

Answer 1

1. There are plenty of interesting subgroups of $SU(2)$. For instance the rotational symmetry group of the icosahedron embeds (in many ways) into $SO(3)$, and the inverse image of that subgroup under the projection $SU(2)\to SO(3)$ is a subgroup of $SU(2)$ of order $120$. More prosaically the indicated copy of $U(1)$ has many conjugate subgroups in $SU(2)$.
2. The existence of $A$ and $B$ are independent. An interesting choice for $A$ is the normaliser of $U(1)$ in $SU(2)$, which apart from the matrices $D$ of $U(1)$ itself also contains matrices $\binom{0~-1}{1~\phantom+0}D$. For $B$ you could take the subset of matrices $D$ with $\theta$ a multiple of $\frac\pi{n}$ for some integer $n>1$. Explicitly one can take $$ A = \left \{\begin{pmatrix}e^{i\theta}&0 \\ 0 & e^{-i\theta} \end{pmatrix} \mid 0\leq\theta<2\pi \right \} \cup \left \{\begin{pmatrix} 0 & -e^{-i\theta}\\ e^{i\theta}&0\end{pmatrix} \mid 0\leq\theta<2\pi \right \} $$ and $$ B= \left \{\begin{pmatrix}e^{\frac{k\pi}7i}&0 \\ 0 & e^{-\frac{ k\pi}7i} \end{pmatrix} \mid k=0,1,\ldots,13 \right \} $$
3. No, $SO(3,\Bbb R)$ does not embed into $SU(2)$ (which is an example of the fact that in general quotient subgroups do not always embed as subgroups). It would be somewhat technical to show that there is no embedding as abstract subgroup (where one would have to exclude highly discontinuous embeddings as well), so I'll just show that such an embedding as a Lie subgroup (so compatible with the differential structure of the groups) is impossible. In such an embedding there would be some Lie algebra element $X$ such that the one parameter subgroup $\{\,\exp tX\mid t\in\Bbb R\,\}$ lies in the subgroup. But that set always contains the element $-I$ of $SU(2)$, whereas $SO(3,\Bbb R)$ contains no central elements of order$~2$.

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Added much later: There is actually an easy argument that $SO(3,\Bbb R)$ does not embed into $SU(2)$ even as plain groups that I did not think of at the time: $SU(2)$ just has one element of order$~2$, which is central, while $SO(3,\Bbb R)$ has infinitely many elements of order$~2$, none of which are central, so they have nowhere to go in $SU(2)$ (and their pre-images under the projection from $SU(2)$ all have order$~4$).

Answer 2

1) What about any discrete subgroup in $U(1)$?

2) $B=A=U(1)$ and $Z_2≅±1⊂U(1)$

3) The representation theory of $SO(3)$ is wellknown, so you only have to look up whether there are two-dim'l complex representations. For the general question, no! Consider $G=\mathbb{Z}$.

Answer 3

Referring to my second question, let $\mathbb{Z}_N=\left \{ \begin{pmatrix} e^{i\theta}&0 \\ 0& e^{-i\theta} \end{pmatrix} \mid \theta=\frac{2\pi}{N},2\cdot\frac{2\pi}{N},...,2\pi\right \}$ with positive integer $N$, then $\mathbb{Z}_N$ is a subgroup of $U(1)$. And $\mathbb{Z}_N=B$ for even $N$.

Answer 4

Concerning my question (3). $SO(3)$ can not be viewed as a subgroup of $SU(2)$. Since there is only one element $-1$ of order 2 in $SU(2)$, while there are more than one elements of order 2 in $SO(3)$.