Does there exists a simply connected closed 5-manifold with $H_2=\mathbb{Z}_p$ for p an odd prime? Let $M$ be a simply connected closed 5-manifold. If we suppose that $H_2(M)=\mathbb{Z}_p$ for some odd prime $p$ then one can show that the other homology groups of $M$ must be trivial with the exceptions of $H_0(M)=H_5(M)=\mathbb{Z}$. The question is if this leads to some contradiction. I know that if $p$ instead was equal to two then the $SU(3)/SO(3)$ satisfies this criterion, but don't know of any examples when $p\neq 2$. Any tips on how to proceed would be greatly appreciated.
 A: This cannot occur.  That is, there is no simply connected $5$-manifold $M$ with $H_2(M;\mathbb{Z})\cong \mathbb{Z}_p$ where $p$ is odd (regardless of whether it is prime or not.)
The obstruction comes from the Linking Form.
Recall that the linking form on an oriented $n$-dimensional manifold is a bilinear pairing $L:TH_i(M;\mathbb{Z})\times TH_{n-i-1}(M;\mathbb{Z})\rightarrow \mathbb{Q}/\mathbb{Z}$ where, for an abelian group $G$, $TG$ denotes the torsion subgroup.
The key properties of the linking form are:

*

*It is non-degenerate

*$L(x,y) = (-1)^{i(n-i)} L(y,x)$.

Let's taylor this to the case of $5$-manifolds.  Because $M$ is simply connected, it is orientable, so let's fix some orientation on it.
Then, taking $i=2$, we see that $(-1)^{5(5-2)} = -1$, so $L$ is an antisymmetric non-degenerate form.  The non-existence of the a $5$-manifold with homology as you describe is then a simple consequence of the following algebra fact:
Proposition:  Suppose $L:\mathbb{Z}_p\times \mathbb{Z}_p\rightarrow \mathbb{Q}/\mathbb{Z}$ is non-degenerate and antisymmetric.  Then $p$ is even.
Proof:  Consider $1\in \mathbb{Z}_p$.  Non-degeneracy implies that there is a $x\in \mathbb{Z}_p$ with $L(1,x)\neq 0$.  Writing $x$ as a multiple of $1$, $x= n\cdot 1$, bilinearity now implies that $0\neq L(1,n\cdot 1) = nL(1,1)$, so $L(1,1)\neq 0$.
But  then then antisymmetry gives $L(1,1) = -L(1,1)$, so $2L(1,1) = 0$.  Since $L(1,1)\neq 0$, but is $2$-torsion in $\mathbb{Q}/\mathbb{Z}$, we conclude that $L(1,1) = 1/2$.
But, again using bilinearity, we have $$0 = L(0,1) = L(p\cdot 1, 1) = pL(1,1) = \frac{p}{2}.$$  Since $\frac{p}{2} = 0 \in \mathbb{Q}/\mathbb{Z}$, $\frac{p}{2}\in \mathbb{Z}$.  That is, $p$ is even.
