Cohomology groups $H^*(B(\mathbb{Z}_2\ltimes PSU(4)),\mathbb{Z}_2)$ for $*\le3$

I'm trying to compute $$H^*(B(\mathbb{Z}_2\ltimes PSU(4)),\mathbb{Z}_2)$$ for $$*\le3$$. Here the semi-direct product is given by the outer automorphism of $$PSU(4)$$.

By Serre spectral sequence, we have $$H^p(B\mathbb{Z}_2,H^q(BPSU(4),\mathbb{Z}_2))\Rightarrow H^{p+q}(B(\mathbb{Z}_2\ltimes PSU(4)),\mathbb{Z}_2).$$

The relevant piece is

There could be several differentials, I'm not sure which of them actually exist.

Thank you!

• There are no differentials to the bottom row, which survives to $E_\infty$ by comparison to $H^*(\Bbb Z/2; H^*(pt;\Bbb Z/2))$. The only interesting one is $d_2: E^{0,3} \to E^{2,2}$. I don't know what that one is. – user98602 Nov 6 '18 at 16:10
• Are you sure your SSeq is correct? I get $H^3(PSU(4);\Bbb Z/2) = H^3(B^2\Bbb Z/4; \Bbb Z/2) = 0$ and $H^2(B^2\Bbb Z/4; \Bbb Z/2) = (\Bbb Z/2)^2$ from table C.3 here. If this is the case, then with my observation above, you see that everything in the relevant range survives to $E_\infty$. – user98602 Nov 6 '18 at 16:35
• I derive my SSeq from that table exactly, I think mine is correct. – Borromean Nov 7 '18 at 1:24
• Oops! I misread $H_2(B^2\Bbb Z/4) = \Bbb Z/2^2 = \Bbb Z/4$ as $(\Bbb Z/2)^2$. My apologies. – user98602 Nov 7 '18 at 1:53