In the syllabus for my course on group theory, there is the following theorem which is proven: "If $H \subset \mathbb{Z}^n$ is a subgroup, then $H \cong \mathbb{Z}^k$ for $0 \leq k \leq n$".
I understand most of the proof, but there's a simple line that is intuitive, yet doesn't mathematically make sense to me.
In the proof, which is done by induction, the following homomorphism is defined $$\pi: \mathbb{Z}^{n+1} \rightarrow \mathbb{Z}: \pi(m_1,...,m_{n+1}) = m_{n+1}.$$ The kernel of the map is considered, which are all vectors of the form $(m_1,...,m_n,0)$.
Then the syllabus reads "Since $H \subset \mathbb{Z}^{n+1}$ is a subgroup, also $H \cap \text{ Ker}(\pi) \subset \mathbb{Z}^n$ is a subgroup."
That is the part that confuses me. Both $H$ and Ker$(\pi)$ are subsets of $\mathbb{Z}^{n+1}$. I realize that Ker$(\pi) \cong \mathbb{Z}^n$, but I don't see how it follows that $H \cap \text{ Ker}(\pi)$ is a subset of $\mathbb{Z}^n$, rather than $\mathbb{Z}^{n+1}$. As far as I know, even though the last element of each vector in the intersection will always be equal to 0, that does not mean that the vectors are of dimension $n$ rather than $n+1$.
Can someone explain to me why the intersection can be considered a subgroup of $\mathbb{Z}^n$?
You actually already said the answer. $\mathrm{Ker}(\pi)$ is isomorphic to $\mathbb Z^n$ and $H \cap \mathrm{Ker}(\pi)$ is a subgroup of $\mathrm{Ker}(\pi)$, and so we can consider it as a subgrouop of $\mathbb Z^n$.