Good evening,
I don't manage to solve the following problem : Is the cyclic group of order 5, Z5, a subgroup of the group Su(2) ? And for SU(3) ?
Is it possible to solve this using representation of Z5 or have I to show it "by hand " ? Because I don't really see a theorem that give a link between the representation of a subgroup abd these from the group himself.
Thank you very much and excuse my english because it is not my native language
Let $\zeta$ be a $n-$th root of unity. Then, the group generated by $$A = \begin{pmatrix} \zeta & 0 \\ 0 & \zeta^{-1} \end{pmatrix}$$
is cyclic of order $n$. Also notice that you can embedd $SU_2$ in $SU_3$ so it is also true for $SU_3$.
In fact, the finite subgroups of $SU_2$ corresponds to finite subgroups of $SL_2(\Bbb C)$ : up to conjugacy, the list is $C_n$, the dihedral groups $D_n$ (for all $n \geq 1$), $\mathfrak S_4$, $\mathfrak A_4$ and $\mathfrak A_5$. It corresponds to ADE diagrams and there are many interesting connexions with algebraic geometry, representation theory and Lie algebras.