Frattini subgroup of $(\mathbb{Z}^n, +)$

Question

Compute the Frattini subgroup of $(\mathbb{Z}^n, +)$.

Hint: you may use the fact that $\operatorname{Aut}(\mathbb{Z}^n)$ is $\operatorname{GL}(n,\mathbb{Z})$, and that the $\operatorname{GL}(n,\mathbb{Z})$-$e_1$ orbit is the set of vectors $(k_1, \ldots, k_n)$ in $\mathbb{Z}^n$ such that their $\gcd$ is $1$.

Answer 1

In the comments, the OP stated that she already partially solved the case $n=1$. Indeed, for any prime $p$, there is a maximal subgroup $p\mathbb{Z} \subset \mathbb{Z}$. Now, for any nonzero $m \in \mathbb{Z}$ there is some prime $p$ which does not divide it, namely $m \notin p\mathbb{Z}$, which means that the intersection of all the subgroups $p\mathbb{Z}$ over all primes is the trivial subgroup $\{0\}$.

The subgroups $p\mathbb{Z}$ can be interpreted as the kernel of the reduce-mod-$p$ homomorphism $\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$, where every integer is mapped to its minimal residue modulo $p$. Since the image is a group with no non-trivial subgroups, the kernel is a maximal subgroup.

Switching over to the larger group $\mathbb{Z}^n$ we see that similar homomorphisms exist: pick an index $1 \leq i \leq n$ and map $(m_1,m_2,\ldots,m_n)$ to $\mathbb{Z}/p \mathbb{Z}$ by looking at $m_i \bmod{p}$. Once again, the kernel is a maximal subgroup. What is the intersection of all of these subgroups?

(I wasn't able to find a way to utilize the hint provided).