Show that $G/2G \cong (\mathbb{Z}/2)^n$, where $G$ is an abelian group and $G \cong \mathbb{Z}^n$

Question

I tried to show this by constructing an explicit map as follows :

Since $G \cong \mathbb{Z}^n$, let $(m_1,\dots,m_n) \in \mathbb{Z}^n$, where $m_i \in \mathbb{Z}$, is the image of $m \in G$. Define $\phi : G/2G \to (\mathbb{Z}/2)^n$ by setting $$ \phi (m+2G) = ([m_1],\dots,[m_n]),\quad \text{where } [m_i] \in \mathbb{Z}/2. $$ Is this correct ?. Seems like this is well-defined (not really sure). Since the author state this as "easy", i thought maybe there are any other simpler proof ? (maybe by first isomorphism theorem, etc). The statement above is in Lemma 9.18 of Lee's Topological Manifold 2nd ed. Thanks.

Answer 1

As has been said, we can construct the homomorphism $f:G \to (\Bbb{Z}/2\Bbb{Z})^n$, $(m_1,\dots,m_n)\mapsto ([m_1],\dots,[m_n])$. Then $\ker(f)=2G$, so by the $1^\text{st}$ isomorphism theorem, $G/2G\cong (\Bbb{Z}/2\Bbb{Z})^n$