Is an Eilenberg-MacLane space $K(G,1)$ the same as the classifying space $BG$ for a group $G$ ?
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$\begingroup$Let $G$ be a discrete group (i.e. a group equipped with the discrete topology), $BG$ its classifying space, and $EG$ the universal $G$-bundle. Then we have a fibration $$G \hookrightarrow EG \rightarrow BG,$$ and the long exact sequence of this fibration reads $$\cdots \to \pi_{n+1}(BG) \to \pi_n(G) \to \pi_n(EG) \to \pi_n(BG) \to \cdots \to \pi_0(G) \to \pi_0(EG) \to 0.$$ The total space of the universal $G$-bundle is contractible and $G$ has the discrete topology, so $\pi_{n}(BG) \cong 0$ for all $n > 1$. Since $EG$ is connected, the tail end of the sequence tells us that $$\pi_1(BG) \cong \pi_0(G) \cong G,$$ since $G$ has the discrete topology. Therefore we see that $BG$ is a $K(G,1)$ when $G$ is a discrete group.
Note that since $EG$ is contractible, $\pi_{n+1}(BG) \cong \pi_n(G)$ for all $n \geq 1$, so if $G$ is equipped with a topology for which $\pi_n(G) \not\cong 0$ for $n \geq 1$, then $BG$ cannot be a $K(G,1)$.
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