# Non-orientable 3-manifold has infinite fundamental group

I'm doing past papers for a first course in algebraic topology.

The question is:

Let $M$ be a 3-dimensional, closed, connected, non-orientable manifold. Show that $M$ has infinite fundamental group.

Is there any way of answering this question without simply quoting a classification theorem for 3-manifolds with finite fundamental group?

• What definition are you using for orientability? Jun 15, 2013 at 18:34
• @dfeuer Good question. I guess I can assume that top homology vanishes. Jun 15, 2013 at 23:07
• @Earthliŋ, do you know a reference for such classification theorem? Jun 8, 2015 at 19:42

$\def\QQ{\mathbb Q}$If $\pi_1(M)$ is finite, $H_1(M;\QQ)=0$. If $M$ is non-orientable, $H_3(M;\QQ)=0$. So $\chi(M)=h_0(M;\QQ)-h_1(M;\QQ)+h_2(M;\QQ)-h_3(M;\QQ)=1+h_2(M;\QQ)>0$.
• A little remark: the Euler characteristic of any closed odd-dimensional manifold is 0, but in this case it doesn't follow by Poincaré duality, since $M$ is supposed to be non-orientable. Jun 23, 2014 at 7:09
• @Dario just use $\mathbb Z_2$ coefficients. Nov 7, 2014 at 19:21
• @Dario Or use the two-fold orientable cover $\tilde M\to M$. $\chi(\tilde M)=0$, and $\chi(\tilde M)=2\chi(M)$, hence $\chi(M)=0$. Apr 21, 2015 at 23:15