This is an example from a question sheet (non-assessed) of a university class. If M is a non-orientable, closed, connected 3 manifold, prove $H_1(M;\mathbb{Z})$ is an infinite group. I know that since $M$ is non-orientable it follows that $H_3(M;\mathbb{Z}) = 0$ and since $M$ is connected $H_0(M;\mathbb{Z}) = \mathbb{Z}$. I want to apply a combination of Poincare duality and Universal coefficient theorem to try and find a contradiction if $H_1(M;\mathbb{Z})$ has no free part. However poincare duality doesn't apply here so I'm stuck.
I would appreciate any hints as to how to proceed.
We can use the fact that the Euler characteristic of any odd dimensional closed manifold is $0$.
Given this, we have $0= \mbox{rk}H_0 - \mbox{rk}H_1 + \mbox{rk}H_2 - \mbox{rk}H_3$.
$M$ is connected so $\mbox{rk}H_0 = 1$. $M$ is non-orientable so $\mbox{rk}H_3 = 0$. If $H_1$ is finite, then $\mbox{rk}H_1 = 0$, so we get $$0 = 1-0+\mbox{rk}H_2 - 0$$ which implies that $\mbox{rk}H_2$ is negative which is impossible.
So $H_1$ must be infinite.