I have run into a statement in literature and cannot figure out its meaning. The statement is:
if $0\to Z^{n}\to G\to K\to 1$ (equivalently $G/Z^{n}=K$), group $G$ is said to be an extension of $Z^{n}$ by $K$ with $\phi:K\to GL(n, Z)$. Here, $Z^{n}$ is a free abelian group.
I cannot figure out what $GL(n,Z)$ means in this statement. It does not seem to be a general linear group because $Z$ is just a group. Also, I don't know what exactly is $\phi$.
Can anyone make sense out of this statement?
$GL(n,\mathbb{Z})$ is the group of self-isomorphisms of $\mathbb{Z}^n$. It may also be thought of as the group of invertible $\mathbb{Z}-$linear maps $\mathbb{Z}^n \to \mathbb{Z}^n$, and you can identify it with the set of invertible $n$ by $n$ matrices with integer coefficients, whose inverse also has integer coefficients.
A homomorphism $\phi:K \to GL(n,\mathbb{Z})$ defines an extension by specifying a semidirect product $\mathbb{Z}^n \rtimes_\phi K$. See wikipedia: https://en.wikipedia.org/wiki/Semidirect_product