# What is a K(Z/nZ, 1)?

I am currently writing parts of Farb-Margalit's Primer on Mapping Class Groups. However, I don't have a strong background in geometry, and I'm bit stumped on an argument.

Let $S$ be a compact surface, and $f$ an element of the mapping class group $Mod(S)$. We assume that $f$ is an element of torsion of order $n$. We want to prove that $f$ fixes some point of $Teich(S)$. It is written, p.391, that:

Since $Teich(S)$ is contractible, the finite cyclic group $\langle f \rangle$ [NB: isomorphic to $\mathbb{Z}/n\mathbb{Z}$] cannot act freely on $Teich(S)$, for otherwise the quotient would be a finite dimensional $K(\mathbb{Z}/n\mathbb{Z}, 1)$.

What is a $K(\mathbb{Z}/n\mathbb{Z}, 1)$, and why can't it be finite dimensional?

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As it answered by others, I only want to point out a concrete model for this particular space is infinite lens space. See here mathoverflow.net/questions/20176/… – Hua Feb 22 '14 at 19:54

$K(G,n)$ is an Eilenberg-MacLane space, a space that has $\pi_i(K(G,n))=\begin{cases}G &i=n\\0&\text{else}\end{cases}$. In the case $n=1$ it is the same as the so-called classifying space $BG$. There exists a nice cell structure for $BG$ and you can compute that the homology of $K(\mathbb Z/\mathbb nZ,1)$ is infinite. See also mathoverflow.
A $K(G,n)$ is an Eilenberg-MacLane space or classifying space. See Proposition 2.45 (p. 149) of Hatcher for an explanation of why a $K(G,1)$ can't be finite dimensional if $G$ has any torsion elements.