To count the number of homomorphisms from the group \(\mathbb{Z}\) (the integers under addition) to the group \(\mathbb{Z}_2\) (the integers modulo 2 under addition), we need to consider the properties of homomorphisms.

A homomorphism \(f: \mathbb{Z} \rightarrow \mathbb{Z}_2\) preserves the group operation, which means:

\[

f(a+b)=f(a)+f(b)

\]

Since \(\mathbb{Z}\) is a cyclic group generated by 1 , the homomorphism is completely determined by the image of the generator. Let's denote the image of 1 as \(f(1)\). There are two possibilities for \(f(1)\), either 0 or 1 in \(\mathbb{Z}_2\).

1. If \(f(1)=0\) in \(\mathbb{Z}_2\), then for every integer \(a, f(a)=a \cdot f(1)=a \cdot 0=0\). This defines the trivial homomorphism that maps every element of \(\mathbb{Z}\) to 0 in \(\mathbb{Z}_2\).

2. If \(f(1)=1\) in \(\mathbb{Z}_2\), then for every integer \(a, f(a)=a \cdot f(1)=a \cdot 1=a\) in \(\mathbb{Z}_2\). This defines the homomorphism that maps every element of \(\mathbb{Z}\) to its equivalence class in \(\mathbb{Z}_2\)

So, there are two homomorphisms from \(\mathbb{Z}\) to \(\mathbb{Z}_2\).

\[

\text{Number of homomorphisms } f: \mathbb{Z} \rightarrow \mathbb{Z}_2 \\

f(1) = 0 \\

f(1) = 1

\]

Thus, there are 2 homomorphisms from \(\mathbb{Z}\) into \(\mathbb{Z}_2\).