Certainly! Let's solve this step by step.
Given groups:
- [tex]\( G = \mathbb{Z} \)[/tex]: the group of integers under addition.
- [tex]\( G' = \mathbb{Z}_n \)[/tex]: the group of integers modulo [tex]\( n \)[/tex] under addition.
Define a map [tex]\(\varphi: G \rightarrow \mathbb{Z}_n\)[/tex] by [tex]\(\varphi(m) = [m]\)[/tex], where [tex]\([m]\)[/tex] denotes the equivalence class of [tex]\(m\)[/tex] modulo [tex]\(n\)[/tex].
We need to show that [tex]\(\varphi\)[/tex] is a homomorphism and that [tex]\(\varphi\)[/tex] maps onto [tex]\(\mathbb{Z}_n\)[/tex].
### Homomorphism Verification
To verify that [tex]\(\varphi\)[/tex] is a homomorphism, we need to check if [tex]\(\varphi\)[/tex] respects the addition operation in both groups. Specifically, we need to show that for all [tex]\(m, r \in \mathbb{Z}\)[/tex],
[tex]\[ \varphi(m + r) = \varphi(m) + \varphi(r). \][/tex]
1. Consider [tex]\(m, r \in \mathbb{Z}\)[/tex]:
[tex]\[ \varphi(m + r) = [(m + r)] \][/tex]
2. Apply the definition of addition in [tex]\(\mathbb{Z}_n\)[/tex]:
[tex]\[ \varphi(m) = [m] \quad \text{and} \quad \varphi(r) = [r] \][/tex]
3. Addition in [tex]\(\mathbb{Z}_n\)[/tex] is defined by:
[tex]\[ [m] + [r] = [m + r] \][/tex]
4. Thus:
[tex]\[ \varphi(m + r) = [m + r] = [m] + [r] = \varphi(m) + \varphi(r) \][/tex]
Since [tex]\(\varphi(m + r) = \varphi(m) + \varphi(r)\)[/tex], [tex]\(\varphi\)[/tex] is a homomorphism.
### Surjectivity
Next, to show that [tex]\(\varphi\)[/tex] is onto (surjective), we need to show that every element in [tex]\(\mathbb{Z}_n\)[/tex] is the image of some integer under [tex]\(\varphi\)[/tex].
1. Consider any element [tex]\([k] \in \mathbb{Z}_n\)[/tex], where [tex]\(k \in \{0, 1, 2, \ldots, n-1\}\)[/tex].
2. Let [tex]\(m = k\)[/tex]. Since [tex]\(k \in \mathbb{Z}\)[/tex], [tex]\(\varphi(m) = [m]\)[/tex] is an element in [tex]\(\mathbb{Z}_n\)[/tex].
- Thus, [tex]\(\varphi(k) = [k]\)[/tex].
Because [tex]\([k]\)[/tex] (for any [tex]\(k \in \{0, 1, 2, \ldots, n-1\}\)[/tex]) is the image of [tex]\(k\)[/tex] under [tex]\(\varphi\)[/tex], every element in [tex]\(\mathbb{Z}_n\)[/tex] has a pre-image in [tex]\(\mathbb{Z}\)[/tex]. Therefore, [tex]\(\varphi\)[/tex] is surjective.
### Conclusion
Since we have verified that [tex]\(\varphi\)[/tex] is both a homomorphism and surjective, [tex]\(\varphi\)[/tex] is a homomorphism of [tex]\(\mathbb{Z}\)[/tex] onto [tex]\(\mathbb{Z}_n\)[/tex].
This analysis confirms that [tex]\(\varphi(m) = [m]\)[/tex] defines a valid homomorphism from [tex]\(\mathbb{Z}\)[/tex] to [tex]\(\mathbb{Z}_n\)[/tex].
