$\mathbb Z^n/\langle (a,...,a) \rangle \cong \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle$

I am trying to show the isomorphism

$$\mathbb Z^n/\langle (a,...,a) \rangle \cong \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle.$$

I've tried to define $\psi:\mathbb Z^n \to \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle$ an epimorphism with $\ker(\psi)=\langle (a,...,a) \rangle$ so to apply the first isomorphism theorem, but I couldn't come up with an appropiate morphism. Any hints or suggestions would be greatly appreciated.

Solution 1:

The following: $$(1,0,\dots,0), (0,1,0,\dots,0), \dots, (0,\dots,0,1,0), (1,1,\dots, 1)$$ is a basis of $\mathbb{Z}^n$ and $(a,\dots,a)$ corresponds to $(0,\dots,0,a)$ in this new basis. Your group is thus the isomorphic to the quotient: $$\mathbb{Z}^n / \langle (0, \dots, 0, a) \rangle \cong \mathbb{Z}^{n-1} \oplus \mathbb{Z}/(a).$$

Solution 2:

If you want to use the first isomorphism theorem, then define $\psi:\mathbb Z^n \to \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle$ by $\psi(x_1,\dots,x_n)=(x_2-x_1,\dots,x_n-x_1,\bar x_1)$. Check that $\psi$ is a surjective homomorphism and $\ker\psi=\langle (a,\dots,a) \rangle$.