Why the bosonic part of the superconformal group $SU(2,2|1)$ is $SO(4,2) \times U(1)_R$? 
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*Why in $d=4$ $\mathcal{N}=1$ SCFT the bosonic part of the superconformal group $SU(2,2|1)$ is $SO(4,2) \times U(1)_R$?

*More generally how can I determine the such a thing in other theories? Is there some specific way to think about how to find such a subgroup? Say in $d=4$ $\mathcal{N}=2$ theory. I know this is known, I want to know how one finds it though.
 A: OP's questions are quite broad. Here we will focus on OP's first question, but hopefully the reader gets some idea how this can be generalized.

*

*Consider the super inner product space
$$\begin{align}V~:=~&\mathbb{C}^{2,2|1}~=~V_0\oplus V_1, \cr V_0~:=~&\mathbb{C}^{2,2|0}, \cr V_1~:=~&\mathbb{C}^{0|1},\end{align} $$
which has 2+2=4 bosonic and 1 fermionic dimensions,
and which is endowed with the standard metric
$$ \eta ~=~ {\rm diag}(1,1,-1,-1|1) ~\in~ 
{\rm End}(\mathbb{C}^{2,2|1}).$$


*Supermatrices
$$ m~=~\begin{pmatrix} m_{00} & m_{01} \cr m_{10} & m_{11} \end{pmatrix}, $$
corresponding to the super matrix algebra
$$\begin{align}A~:=~&{\rm End}(V)~=~A_0 \oplus A_1, \cr A_0~:=~&{\rm End}(V_0)\oplus {\rm End}(V_1), \cr A_1~:=~&{\cal L}(V_0;V_1)\oplus {\cal L}(V_1;V_0), \end{align} $$ of endomorphisms can be decomposed in two diagonal bosonic blocks $$m_{00}\in{\rm End}(V_0)\qquad\text{and}\qquad m_{11}\in{\rm End}(V_1),$$
and two off-diagonal fermionic blocks $m_{01}$ and $m_{10}$. The fermionic sector $A_1$ contains linear maps between $V_0$ and $V_1$.


*The super Lie group
$$ U(2,2|1) ~~:=~~ \{U\in {\rm End}(\mathbb{C}^{2,2|1}) \mid  U^{\dagger}\eta U = \eta \}$$
has corresponding super Lie algebra
$$ u(2,2|1) ~~:=~~ \{m\in {\rm End}(\mathbb{C}^{2,2|1}) \mid  m^{\dagger} =-\eta m \eta^{-1}  \}.$$
(Warning: The super-Hermitian conjugation "$\dagger$" involves appropriate sign-factors.) The bosonic part of the super Lie algebra is
$$\begin{align} A_0\cap u(2,2|1)
~\cong~& u(2,2) \oplus u(1)_R \cr
 ~\cong~& su(2,2) \oplus u(1) \oplus u(1)_R \cr
~\cong~& so(4,2) \oplus u(1) \oplus u(1)_R.\end{align}$$
Here subscript $R$ stands for the $R$-charge in the fermionic sector.
The bosonic part of the super Lie group is
$$\begin{align} A_0 \cap U(2,2|1)
~\cong~&  U(2,2) \times U(1)_R \cr
~\cong~&\frac{ SU(2,2)\times U(1)}{\mathbb{Z}_4} \times U(1)_R \cr
~\cong~&\frac{ SPIN(4,2)\times U(1)}{\mathbb{Z}_4} \times U(1)_R, \end{align}$$
cf. e.g. this Phys.SE post and this Math.SE post.


*Now let us return to OP's first question. The super Lie group
$$ SU(2,2|1) ~~:=~~ \{U\in U(2,2|1) \mid  {\rm sdet} (U) =1\}$$
has corresponding super Lie algebra
$$ su(2,2|1) ~~:=~~ \{m\in u(2,2|1) \mid {\rm str} (m) =0
  \}.$$
The bosonic part of the super Lie algebra becomes
$$A_0\cap su(2,2|1)
~\cong~ su(2,2) \oplus u(1)_R  
~\cong~ so(4,2) \oplus u(1)_R.$$
This is the answer to OP's first question at the Lie algebra level.


*Concerning conformal groups without SUSY, see e.g. this Phys.SE post.
In 3+1D the connected component that contains the identity element is
$$ {\rm Conf}_0(3,1)~\cong~ SO^+(4,2)/\mathbb{Z}_2~\cong~SU(2,2)/[\mathbb{Z}_2\times \mathbb{Z}_2]. $$
For superconformal groups, see e.g. Wikipedia and nLab.
