Proving $N/N'\cong$ Hom$_\mathbb Z(M'/M,\mathbb C^*)$ I am studying the following proof and need help understanding some of the details-

Theorem : Let $N$ be a free $\mathbb Z$ - module of finite rank and let $N'$ have finite index in $N$ with quotient $G=N/N'$. Then $G\cong \operatorname{Hom}_{\mathbb Z}(M'/N,\mathbb C^*)$ where $M=\operatorname{Hom}_{\mathbb Z}(N,\mathbb Z)$ and $M'=\operatorname{Hom}_{\mathbb Z}(N',\mathbb Z)$
Proof : we have the following exact sequence - $$0\longrightarrow M\longrightarrow M'\longrightarrow M'/M\longrightarrow 0$$
Now, since $\operatorname{Hom}_{\mathbb Z}(\_,\mathbb C^*)$ is left exact and $\mathbb C^*$ is divisible, we have that $\operatorname{Hom}_{\mathbb Z}(\_,\mathbb C^*)$ is exact so that we have the exact sequence - $$0\longrightarrow \operatorname{Hom}_{\mathbb Z}(M'/M,\mathbb C^*)\longrightarrow \operatorname{Hom}_{\mathbb Z}(M',\mathbb C^*)\longrightarrow \operatorname{Hom}_{\mathbb Z}(M,\mathbb C^*)\longrightarrow 0$$
Since $N'$ is of finite index in $N$ we have the inclusions $$N'\subseteq N\subseteq N_{\mathbb Q} \qquad \text{and}\qquad M\subseteq M'\subseteq M_{\mathbb Q}$$ where, $N_\mathbb Q=N\otimes \mathbb Q$ and $M_\mathbb Q=M\otimes \mathbb Q$.
The pairing between $M$ and $N$ induces a pairing $\langle\cdot,\cdot\rangle:M_\mathbb Q\times N_\mathbb Q\to \mathbb Q$. Hence the map $$M'/M\times N/N'\to \mathbb C^* \quad\text{given by}\quad ([m'],[u])\mapsto e^{2\pi i\langle m',u\rangle}$$ is wel defined and induces $G\cong \operatorname{Hom}_{\mathbb Z}(M'/M,\mathbb C^*)$.

My questions :

*

*Why does finite index of $N'$ in $N$ imply the two inclusions $N'\subseteq N\subseteq N_{\mathbb Q}$ and $M\subseteq M'\subseteq M_{\mathbb Q}$?


*Why is the last map well defined?
If $([m'_1],[u_1])=([m'_2],[u_2])\Rightarrow m'_1-m'_2\in M$ and $u_1-u_2\in N'$. But I'm not sure how to show $\langle m'_1,u_1\rangle=\langle m'_2,u_2\rangle$
Thank you.
 A: (1)  The inclusions $N'\subseteq N\subseteq N_\mathbb{Q}$ are trivial.  Writing "$M\subseteq M'$" really means that the canonical map $M\to M'$, given by taking a homomorphism $\varphi:N\to\mathbb{Z}$ and restricting it to the subgroup $N'$, is injective.  This injectivity follows from the fact that $N'$ has finite index in $N$, since for every $x\in N$, there is some nonzero $a\in\mathbb{Z}$ such that $ax\in N'$ (since the image of $x$ in $N/N'$ has finite order).  So if $\varphi|_{N'}=0$, then $0=\varphi(ax)=a\varphi(x)$, which implies $\varphi(x)=0$.  Since $x$ is arbitrary, this means that if $\varphi|_{N'}=0$, then $\varphi=0$, which is exactly the injectivity we wanted.
Furthermore, since $M$ and $M'$ have the same rank (since $N$ and $N'$ have the same rank), $M$ must have finite index in $M'$.  So, tensoring the short exact sequence $0\to M\to M'\to M'/M\to 0$ with $\mathbb{Q}$, we find that the natural map $M_\mathbb{Q}\to M'_\mathbb{Q}$ is an isomorphism.  The inclusion $M'\subseteq M_\mathbb{Q}$ is then just the natural inclusion $M'\subseteq M'_\mathbb{Q}$ combined with this isomorphism.
(2)  Fix $m'\in M$ and $u\in N$ and $m\in M$ and $u'\in N$.  We want to show that $e^{2\pi i\langle m',u\rangle}=e^{2\pi i\langle m'+m,u+u'\rangle}$, or equivalently that $\langle m'+m,u+u'\rangle-\langle m',u\rangle$ is an integer.  Since $\langle\cdot,\cdot\rangle$ is bilinear, it suffices to show that $\langle m',u'\rangle$ and $\langle m,u+u'\rangle$ are integers.  But $\langle m',u'\rangle$ is just the pairing of $m'\in M'$ with $u'\in N'$, which is an integer since $m'$ is a homomorphism $N'\to\mathbb{Z}$.  Similarly, $m$ is a homomorphism $M\to\mathbb{Z}$ and $u+u'\in M$, so $\langle m,u+u'\rangle$ is an integer.
