Let $\alpha\in H^2(B\operatorname{PSU}(N) ; \mathbb{Z}\_N)$ be the obstruction class for lifting a $\operatorname{PSU}(N)$-bundle to an $\mathrm{SU}(N)$-bundle. Note that $\operatorname{PSU}(N)\cong \operatorname{SU}(N)/\mathbb{Z}\_N$ is the quotient group of $\operatorname{SU}(N)$ by its center, and $B\operatorname{PSU}(N)$ is the classifying space.


The short exact sequence
$$
0 \rightarrow \mathbb{Z}\_q \rightarrow \mathbb{Z}\_{q \cdot N} \rightarrow \mathbb{Z}\_N \rightarrow 0
$$
invokes the Bockstein map $\textrm{Bock}:H^2(B \operatorname{PSU}(N) ; \mathbb{Z}\_N) \rightarrow H^3\left(B \operatorname{PSU}(N) ; \mathbb{Z}\_q\right)$.
The image $\textrm{Bock}(\alpha)\in H^3\left(B \operatorname{PSU}(N) ; \mathbb{Z}\_q\right)$ of the obstruction class under the Bockstein map is a Postnokov class. My question is under which condition (on $q$ and $N$) the Postnikov class $\textrm{Bock}(\alpha)$ is non-trivial. Is there a classification or a study to tell whether the Postnikov class in $H^3\left(B \operatorname{PSU}(N) ; \mathbb{Z}\_q\right)$ is non-trivial?


**Background on physics**. The non-triviality of the Postnikov class will tell us the 2-group structure of 1-form symmetry and flavor symmetry in three dimensions, and this will tell equivalence classes of line defects in three dimensions.



Non-triviality of a Postnikov class in $H^3\left(B \operatorname{PSU}(N) ; \mathbb{Z}_q\right)$
Consider the map of short exact sequences


$$\begin{array}{ccccccccc} 0 & \rightarrow & \mathbb{Z} & \xrightarrow{\cdot N} & \mathbb{Z} & \xrightarrow{mod N} & \mathbb{Z}\_N & \rightarrow & 0\\
\downarrow & & \downarrow & & \downarrow & & \downarrow{=} & & \downarrow\\
0 & \rightarrow & \mathbb{Z}\_q & \xrightarrow{1 \mapsto N} & \mathbb{Z}\_{Nq} & \xrightarrow{mod N} & \mathbb{Z}\_N & \rightarrow & 0\end{array}$$


and the stretch of long exact sequence in cohomology from the top row:


$$H^2(BPSU(N); \mathbb{Z}) \rightarrow H^2(BPSU(N); \mathbb{Z}\_N) \xrightarrow{\beta} H^3(BPSU(N); \mathbb{Z})$$


Now, from the fibration $B\mathbb{Z}\_N \to BSU(N) \to BPSU(N)$ we have that $BPSU(N)$ is simply connected and $H\_2(BPSU(N);\mathbb{Z}) \cong \pi\_2(BPSU(N)) \cong \mathbb{Z}\_N$. Then from universal coefficients we get $H^2(BPSU(N); \mathbb{Z})= 0$ and $H^2(BPSU(N); \mathbb{Z}\_N) \cong \mathbb{Z}\_N$. Furthermore, we have that $\pi\_3(BPSU(N)) = \pi\_2(SU(N)) = 0$ surjects onto $H\_3(BPSU(N);\mathbb{Z})$
by Hurewicz, giving us $H^3(BPSU(N);\mathbb{Z}) \cong \mathbb{Z}\_N$ again by universal coefficients.


Therefore the integral Bockstein above is an injection of $\mathbb{Z}\_N$ into itself and hence an isomorphism, sending 1 to an invertible element of $\mathbb{Z}\_N$. Now we use the commutativity of


$$\begin{array}{ccc} H^2(BPSU(N);\mathbb{Z}\_N) & \xrightarrow{\beta} & H^3(BPSU(N);\mathbb{Z}) \\ \downarrow{=} & & \downarrow \\ H^2(BPSU(N); \mathbb{Z}\_N) &\xrightarrow{\beta} & H^3(BPSU(N);\mathbb{Z}\_q) \end{array}$$


to see that $1 \in H^2(BPSU(N);\mathbb{Z}\_N)$ is in the kernel of the lower Bockstein (the one you are interested in) if and only if $1$ (or equivalently any invertible element in $\mathbb{Z}\_N$) gets sent to 0 under mod $q$ reduction, i.e. if and only if $1$ is $q$ times another integral class, i.e. if and only if $q$ is invertible modulo $N$.


That is, the Bockstein is non-trivial if and only if $q$ and $N$ have a prime factor in common (or $q = 0$).


