Let $p$ and $q$ be distinct prime, and consider the classifying spaces $X=BC_p$ and $Y=BC_q$, where $C_p$ is the cyclic group of order $p$. Prove that the join $X*Y$ is contractible.
I am not sure how to think topologically about the classifying space of a cyclic group. Also, is the correct way to think about the join is as a homotopy pushout?
The Seifert-Van Kampen theorem implies that $\pi_1(BC_p\ast BC_q)$ is the pushout of $$C_p\leftarrow C_p\times C_q\rightarrow C_q,$$ which is trivial if $p$ and $q$ are distinct primes. I claim that $H_\ast(BC_p\ast BC_q)$ vanishes in positive degrees. Then Hurewicz implies that all homotopy groups of $BC_p\ast BC_q$ vanish.
For any two spaces $X$ and $Y$ there is a split short exact sequence $$0\to \tilde{H}_k(X\ast Y)\to \tilde{H}_{k-1}(X\times Y)\to \tilde{H}_{k-1}(X)\oplus \tilde{H}_{k-1}(Y)\to 0,$$ see this question. This is a straightforward application of the Mayer-Vietoris sequence and uses the fact that the projections $X\times Y\to X$ and $X\times Y\to Y$ have sections. It thus suffices to show that $\tilde{H}_k(BC_p\times BC_q)\to \tilde{H}_k(BC_p)\oplus \tilde{H}_k(BC_q)$ is injective. For any $n$ we have $$H_k(BC_n) = \begin{cases} \mathbb{Z} &\text{if } k=0,\\ \mathbb{Z}/n &\text{if } k \text{ is odd},\\ 0 &\text{otherwise}, \end{cases}$$ and since $$\mathrm{Tor}^1_\mathbb{Z}(\mathbb{Z}/n;\mathbb{Z}/m)=0=\mathbb{Z}/n\otimes\mathbb{Z}/m$$ if $n$ and $m$ are coprime, the Künneth formula implies that $$\tilde{H}_k(BC_p\times BC_q)\cong\mathbb{Z}/p\oplus \mathbb{Z}/q\cong\tilde{H}_k(BC_p)\oplus \tilde{H}_k(BC_q)$$ for odd $k$ and is zero otherwise. We see that $$\tilde{H}_k(BC_p\times BC_q)\to \tilde{H}_k(BC_p)\oplus \tilde{H}_k(BC_q)$$ is a surjection between finite groups of the same cardinality, hence it must be also be injective.