Third cohomology group of abelian groups

Question

What is the third cohomology group $H^3(C_n,\mathbb{Q}/\mathbb{Z})$ of the cyclic group $C_n$ of order $n$?

How can I calculate the third cohomology group of finite abelian group $A$ with coefficients in $\mathbb{Q}/\mathbb{Z}$? Is there any formula like Knuth formula for homology?

Answer 1

You can use the Universal coeffcient theorem for cohomology which states that $$0\to\text{Ext}^1_\mathbb{Z}(H_{n-1}(X),M)\to H^n(X,M)\to\text{Hom}_\mathbb{Z}(H_n(X),M)\to 0$$ is exact, with $M$ being an abelian group, the coefficients, and $X$ being our group or space in question.

Let's remember that We have for $C_n\cong\mathbb{Z}_n$ and that $H_n(\mathbb{Z}_n)=\mathbb{Z}_n$ for even $n$, and $0$ otherwise, except when $n=0$ at which it is $\mathbb{Z}$.

Then using this handy guide we get that $$\text{Hom}_\mathbb{Z}(H_n(\mathbb{Z}_m),\mathbb{Q}/\mathbb{Z})=\text{Hom}_\mathbb{Z}(\mathbb{Z}_m,\mathbb{Z}_m)\cong\mathbb{Z}_m$$ using this for help, this is when $n$ is even.

For ext we have that $\text{Ext}^1(\mathbb{Z}_m,\mathbb{Q}/\mathbb{Z})=0$

I think you can take it from here with it being a split sequence.