Computing $\pi_4(SU(3))$ using Serre spectral sequence I'm trying to show that $SU(3)$ and $S^3\times S^5$ are not homotopy equivalent, by computing their fourth homotopy groups. I know that of $S^3\times S^5$ is $\mathbb Z_2$, and I want to show that $\pi_4(SU(3))=0$.
My idea for this purpose is to kill its first non trivial homotopy group by taking the homotopy fiber of $SU(3)\to K(\mathbb Z, 3)$, say $X$, then up to homotopy we have a fibration $K(\mathbb Z,2)\to X\to SU(3)$ which yields a quite simple spectral sequence. 
Let's call $x$ a generator of $H^2(K(\mathbb Z,2))$, then $x_3:=d_3(x)$ is a generator of $H^3(SU(3))$ by dimension arguments, and there is also some generator $x_5\in H^5(SU(3))$ such that $H^*(SU(3))=\Lambda(x_3, x_5)$. Let's take $x^n\cdot x_3$ as a generator of $E_2^{3,2n}$, then by Leibniz rule we know that $d_3:H^{2n}(K(\mathbb Z,2))\to E^{3,2n-2}$ acts as the multiplication by $n$, so it is injective and the three first columns of $E_4$ page are trivial. The fourth is $E_4^{4,2n}=\mathbb Z_{n+1}$ because $d_3^{3,k}=0$. The only differential which can touch $E_2^{5,0}=H^5(K(\mathbb Z,2))$ is $d_5$, starting from $E_5^{0,3}$ which has vanished since page 4, so everything below $p+q=5$ has converged after page 4.
From this, we can get $H^4(X)=0$ (which I don't use to get a contradiction) and $H^5(X)=\mathbb Z\oplus \mathbb Z_2$, and this is where problems begin. From universal coefficients theorem, as $\text{Hom}(H^5(X),\mathbb Z)=\mathbb Z$, we get that $\text{Ext}(H_4(X))=\mathbb Z_2$ and so by Hurewicz theorem, $H_4(X)=\pi_4(X)=\pi_4(SU(3))\neq 0$, which I know is not true.
Could someone tell me where I'm wrong ? Thank you very much.
 A: The problem is your conclusion that $H^5X$ is isomorphic to $\mathbb{Z}\oplus\mathbb{Z}_2$. The spectral sequence converges to the the associated graded module. To extract $H^5X$ from the resulting information leaves you with an extension problem, which in this case is the short exact sequence
$$0\rightarrow \mathbb{Z}\rightarrow H^5X\rightarrow \mathbb{Z}_2\rightarrow 0.$$
Here we have identified 
$$\mathbb{Z}\cong E^{5,0}_\infty\cong F^5H^{5}X/F^6H^5X=F^5H^{5}X$$
and 
$$\mathbb{Z}_2\cong E^{3,2}_\infty\cong F^3H^5X/F^4H^5X\cong F^0H^5X/F^5H^5X=H^5X/F^5H^5X=H^5X/\mathbb{Z}.$$ 
The point is that there are clearly two solutions to the extension above, and you have chosen the wrong one. We know that the correct solution should be $H^5X\cong\mathbb{Z}$, and using the methods you describe we check that everything goes through correctly.
As it turns out, such extension problems arise frequently, and it is not always clear how to solve them. Some kind of extra topological or algebraic input is most often needed. For instance you can check that if you replace $SU_3$ with $S^3\times S^5$, then all of the above goes through identically until it comes to solving the extension problem. Clearly your idea needs a little bit of extra input to be realised fully.
One way obtain such an input is to note the existence of a non-trivial Steenrod square $Sq^2:H^3(SU_3;\mathbb{Z}_2)\rightarrow H^5(SU_3;\mathbb{Z}_2)$. Clearly this already indicates that $SU_3\not\simeq S^3\times S^5$, but you can check your method by computing the mod 2 LSSS of your fibration as a module over the Steenrod algebra. You can squeeze out just enough information to solve the previous extension problem.
To see the non-trivial $Sq^2$ note the existence of a map $\Sigma\mathbb{C}P^2\rightarrow SU_3$. This takes a complex line to the linear map which is a generalised reflection through it. This map induces isomorphisms on $H^3$ and $H^5$. In particular $\Sigma\mathbb{C}P^2$ is a $7$-skeleton of $SU_3=\Sigma\mathbb{C}P^2\cup e^8$.
By the way, a really quick way to see that $SU_3\not\simeq S^3\times S^5$ is as follows: a retract of an H-space is an H-space, so if the homotopy equivalence were true, $S^5$ would be an H-space. But it is classically known that the only spheres which are H-spaces are $S^1,S^3$ and $S^7$.
