Why $H^1(E(C), Z/nZ)$ is isomorphic to $H^1(E(C), Z)\bigotimes Z/nZ$?

Question

Let $E(C)$ be an elliptic curve over complex field.

Why $H^1(E(C), Z/nZ)$ is isomorphic to $H^1(E(C), Z)\bigotimes Z/nZ$?

$H^1(E(C), Z/nZ)$＝$H^1(E(C), Z\bigotimes Z/nZ)$

From here, can we take out $\bigotimes$ from homology group?

P.S If I could solve this, I can prove $H^1(E(C), Z/nZ)$ is isom to $T_\ell(E(C))$ as group.

Answer 1

Expanding my comment into an answer: the Universal Coefficient Theorem for cohomology states that, given a topological space $X$ and an abelian group $G$, the sequence \begin{align*} \DeclareMathOperator{\Ext}{Ext} \DeclareMathOperator{\Hom}{Hom} \newcommand{\C}{\mathbb{C}} \newcommand{\Z}{\mathbb{Z}} 0 \to \Ext(H_{i-1}(X), G) \to H^i(X, G) \to \Hom(H_i(X), G) \to 0 \end{align*} is exact for each $i \geq 0$. (Here $\Ext = \Ext_{\Z}^1$.) Since $E(\C)$ is homeomorphic to a torus, then $H_0(E(\C)) \cong \Z$ and $H_1(E(\C)) \cong \Z^2$. Note that $\Hom(\Z^2, \Z/n\Z) \cong (\Z/n\Z)^2$ and since $\Z$ is free, then $\Ext(\Z, \Z/n\Z) = 0$. Thus taking $X = E(\C)$, $G = \Z/n\Z$, and $i = 1$, we obtain the short exact sequence $$ 0 \to 0 \to H^1(E(\C), \Z/n\Z) \to (\Z/n\Z)^2 \to 0 \, , $$ so $H^1(E(\C), \Z/n\Z) \cong (\Z/n\Z)^2$.