Yes, every homomorphism $\Gamma \to \mathbb{Z}$ is trivial.

We may assume that $G$ is simply connected, thus it decomposes as a product of simple factors. Let's consider two cases:

1. $G$ has exactly one non-compact simple factor.

2. $G$ has at least two non-compact simple factors.

In case 1 $G$ has property (T), so also does $\Gamma$ and the result follows. In case 2 the result follows from theorem 0.8 in

Shalom, Yehuda Rigidity of commensurators and irreducible lattices. Invent. Math. 141 (2000), no. 1, 1–54.

Formally, the above theorem applies only for $\Gamma<G$ cocompact, but in fact it works in all cases. For what I say now you need some familiarity with Shalom's proof of the above theorem. The proof needs square integrability for $\Gamma$ in $G$ with respect to some fundamental domain. This property holds when $G$ has trivial center as explained in the paper (then $G$ is algebraic and assumption 0.1 holds). The case where $G$ has finite center follows easily. assume $G$ has an infinite center and denote by $Z$ the center of $\Gamma$, which is of finite index in the center of $G$. Then $G/\Gamma \simeq (G/Z)/(\Gamma/Z)$ and square integrability follows from the finite center case.

Edit: After thinking further, the square integrability of a non-uniform lattice in a group with infinite center doesn't seem to follow immediately by reduction modulo the center. So for anyone who wants to use the above answer: take it on your own risk. Or better: please check properly that square integrability does hold. It should.

Yes, every homomorphism $\Gamma \to \mathbb{Z}$ is trivial.

We may assume that $G$ is simply connected, thus it decomposes as a product of simple factors. Let's consider two cases:

1. $G$ has exactly one non-compact simple factor.

2. $G$ has at least two non-compact simple factors.

In case 1 $G$ has property (T), so also does $\Gamma$ and the result follows. In case 2 the result follows from theorem 0.8 in

Shalom, Yehuda Rigidity of commensurators and irreducible lattices. Invent. Math. 141 (2000), no. 1, 1–54.

Formally, the above theorem applies only for $\Gamma<G$ cocompact, but in fact the proof shows that you need 2-integrability of $\Gamma$ in $G$, which holds by Proposition 7.1 here, see the preceding discussion for the definition.

The above is an edit of an earlier partial answer I gave, based on the answer of Mikael de la Salle. See Mikael's answer and YCor's comments for further details.

Yes, every homomorphism $\Gamma$ to $\mathbb{Z}$ is trivial.

We may assume that $G$ is simply connected, thus it decomposes as a product of simple factors. Let's consider two cases:

1. $G$ has exactly one non-compact simple factor.

2. $G$ has at least two non-compact simple factors.

In case 1 $G$ has property (T), so also does $\Gamma$ and the result follows. In case 2 the result follows from theorem 0.8 in

Shalom, Yehuda Rigidity of commensurators and irreducible lattices. Invent. Math. 141 (2000), no. 1, 1–54.

Formally, the above theorem applies only for $\Gamma<G$ cocompact, but in fact it works in all cases. For what I say now you need some familiarity with Shalom's proof of the above theorem. The proof needs square integrability for $\Gamma$ in $G$ with respect to some fundamental domain. This property holds when $G$ has trivial center as explained in the paper (then $G$ is algebraic and assumption 0.1 holds). The case where $G$ has finite center follows easily. assume $G$ has an infinite center and denote by $Z$ the center of $\Gamma$, which is of finite index in the center of $G$. Then $G/\Gamma \simeq (G/Z)/(\Gamma/Z)$ and square integrability follows from the finite center case.

Yes, every homomorphism $\Gamma \to \mathbb{Z}$ is trivial.

We may assume that $G$ is simply connected, thus it decomposes as a product of simple factors. Let's consider two cases:

1. $G$ has exactly one non-compact simple factor.

2. $G$ has at least two non-compact simple factors.

In case 1 $G$ has property (T), so also does $\Gamma$ and the result follows. In case 2 the result follows from theorem 0.8 in

Shalom, Yehuda Rigidity of commensurators and irreducible lattices. Invent. Math. 141 (2000), no. 1, 1–54.

Formally, the above theorem applies only for $\Gamma<G$ cocompact, but in fact it works in all cases. For what I say now you need some familiarity with Shalom's proof of the above theorem. The proof needs square integrability for $\Gamma$ in $G$ with respect to some fundamental domain. This property holds when $G$ has trivial center as explained in the paper (then $G$ is algebraic and assumption 0.1 holds). The case where $G$ has finite center follows easily. assume $G$ has an infinite center and denote by $Z$ the center of $\Gamma$, which is of finite index in the center of $G$. Then $G/\Gamma \simeq (G/Z)/(\Gamma/Z)$ and square integrability follows from the finite center case.

Edit: After thinking further, the square integrability of a non-uniform lattice in a group with infinite center doesn't seem to follow immediately by reduction modulo the center. So for anyone who wants to use the above answer: take it on your own risk. Or better: please check properly that square integrability does hold. It should.