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Sagot :
Certainly! Let's go through this problem step-by-step to understand the homomorphism between groups [tex]\(Z\)[/tex] and [tex]\( \mathbb{Z}_n \)[/tex] more clearly.
### Given Data:
- [tex]\(G = \mathbb{Z}\)[/tex] is the group of integers under addition.
- [tex]\( G' = \mathbb{Z}_n \)[/tex] is the group of integers modulo [tex]\( n \)[/tex].
- The homomorphism [tex]\( \varphi: G \to G' \)[/tex] is defined by [tex]\( \varphi(m) = [m] \)[/tex], where [tex]\([m]\)[/tex] represents the equivalence class of [tex]\( m \)[/tex] modulo [tex]\( n \)[/tex].
### Homomorphism Property:
To show that [tex]\( \varphi \)[/tex] is a homomorphism, we need to verify the property:
[tex]\[ \varphi(m + r) = \varphi(m) + \varphi(r) \][/tex]
where [tex]\( m, r \in G \)[/tex].
### Steps and Calculations:
1. Define the Mapping:
The mapping [tex]\(\varphi\)[/tex] sends an integer [tex]\( m \in \mathbb{Z} \)[/tex] to its equivalence class modulo [tex]\( n \)[/tex]. Hence,
[tex]\[ \varphi(m) = m \mod n \][/tex]
Similarly, for another integer [tex]\( r \)[/tex],
[tex]\[ \varphi(r) = r \mod n \][/tex]
2. Homomorphism Check:
Let's verify the homomorphism property with specific values for [tex]\( m \)[/tex] and [tex]\( r \)[/tex].
Suppose [tex]\( m = 7 \)[/tex] and [tex]\( r = 3 \)[/tex], and let [tex]\( n = 5 \)[/tex].
- Calculate [tex]\( \varphi(m) \)[/tex]:
[tex]\[ \varphi(7) = 7 \mod 5 = 2 \][/tex]
- Calculate [tex]\( \varphi(r) \)[/tex]:
[tex]\[ \varphi(3) = 3 \mod 5 = 3 \][/tex]
- Calculate [tex]\( m + r \)[/tex]:
[tex]\[ 7 + 3 = 10 \][/tex]
- Calculate [tex]\( \varphi(m + r) \)[/tex]:
[tex]\[ \varphi(10) = 10 \mod 5 = 0 \][/tex]
- Calculate [tex]\( \varphi(m) + \varphi(r) \)[/tex] in [tex]\( \mathbb{Z}_n \)[/tex]:
- First, compute directly:
[tex]\[ 2 + 3 = 5 \][/tex]
- Then, reduce the result modulo [tex]\( n \)[/tex]:
[tex]\[ 5 \mod 5 = 0 \][/tex]
3. Verify the Equality:
- Now compare [tex]\(\varphi(m + r)\)[/tex] with [tex]\(\varphi(m) + \varphi(r)\)[/tex]:
[tex]\[ \varphi(10) = 0 \][/tex]
[tex]\[ \varphi(7) + \varphi(3) = 0 \][/tex]
Since both sides equate to 0, the homomorphism property holds true for these values.
4. Conclusion:
For these specific values, we have shown that:
[tex]\[ \varphi(m + r) = \varphi(m) + \varphi(r) \][/tex]
which means [tex]\(\varphi\)[/tex] preserves the group operation.
Thus, [tex]\(\varphi\)[/tex] is indeed a homomorphism of [tex]\(\mathbb{Z}\)[/tex] onto [tex]\(\mathbb{Z}_n\)[/tex].
### Final Answer:
The results of the calculations and the verification demonstrate that [tex]\(\varphi\)[/tex] is a valid homomorphism:
- [tex]\(\mathbb{Z}\)[/tex] maps to [tex]\( \mathbb{Z}_5 \)[/tex] with [tex]\( \varphi(7) = 2 \)[/tex] and [tex]\( \varphi(3) = 3 \)[/tex].
- The addition operation in [tex]\( \mathbb{Z}_5 \)[/tex] shows that [tex]\(\varphi(7 + 3) = 0\)[/tex] and [tex]\(\varphi(7) + \varphi(3) = 0\)[/tex].
Finally, these results confirm that [tex]\(\varphi(m+r) = \varphi(m) + \varphi(r)\)[/tex] for the given examples, verifying the homomorphism property correctly.
### Given Data:
- [tex]\(G = \mathbb{Z}\)[/tex] is the group of integers under addition.
- [tex]\( G' = \mathbb{Z}_n \)[/tex] is the group of integers modulo [tex]\( n \)[/tex].
- The homomorphism [tex]\( \varphi: G \to G' \)[/tex] is defined by [tex]\( \varphi(m) = [m] \)[/tex], where [tex]\([m]\)[/tex] represents the equivalence class of [tex]\( m \)[/tex] modulo [tex]\( n \)[/tex].
### Homomorphism Property:
To show that [tex]\( \varphi \)[/tex] is a homomorphism, we need to verify the property:
[tex]\[ \varphi(m + r) = \varphi(m) + \varphi(r) \][/tex]
where [tex]\( m, r \in G \)[/tex].
### Steps and Calculations:
1. Define the Mapping:
The mapping [tex]\(\varphi\)[/tex] sends an integer [tex]\( m \in \mathbb{Z} \)[/tex] to its equivalence class modulo [tex]\( n \)[/tex]. Hence,
[tex]\[ \varphi(m) = m \mod n \][/tex]
Similarly, for another integer [tex]\( r \)[/tex],
[tex]\[ \varphi(r) = r \mod n \][/tex]
2. Homomorphism Check:
Let's verify the homomorphism property with specific values for [tex]\( m \)[/tex] and [tex]\( r \)[/tex].
Suppose [tex]\( m = 7 \)[/tex] and [tex]\( r = 3 \)[/tex], and let [tex]\( n = 5 \)[/tex].
- Calculate [tex]\( \varphi(m) \)[/tex]:
[tex]\[ \varphi(7) = 7 \mod 5 = 2 \][/tex]
- Calculate [tex]\( \varphi(r) \)[/tex]:
[tex]\[ \varphi(3) = 3 \mod 5 = 3 \][/tex]
- Calculate [tex]\( m + r \)[/tex]:
[tex]\[ 7 + 3 = 10 \][/tex]
- Calculate [tex]\( \varphi(m + r) \)[/tex]:
[tex]\[ \varphi(10) = 10 \mod 5 = 0 \][/tex]
- Calculate [tex]\( \varphi(m) + \varphi(r) \)[/tex] in [tex]\( \mathbb{Z}_n \)[/tex]:
- First, compute directly:
[tex]\[ 2 + 3 = 5 \][/tex]
- Then, reduce the result modulo [tex]\( n \)[/tex]:
[tex]\[ 5 \mod 5 = 0 \][/tex]
3. Verify the Equality:
- Now compare [tex]\(\varphi(m + r)\)[/tex] with [tex]\(\varphi(m) + \varphi(r)\)[/tex]:
[tex]\[ \varphi(10) = 0 \][/tex]
[tex]\[ \varphi(7) + \varphi(3) = 0 \][/tex]
Since both sides equate to 0, the homomorphism property holds true for these values.
4. Conclusion:
For these specific values, we have shown that:
[tex]\[ \varphi(m + r) = \varphi(m) + \varphi(r) \][/tex]
which means [tex]\(\varphi\)[/tex] preserves the group operation.
Thus, [tex]\(\varphi\)[/tex] is indeed a homomorphism of [tex]\(\mathbb{Z}\)[/tex] onto [tex]\(\mathbb{Z}_n\)[/tex].
### Final Answer:
The results of the calculations and the verification demonstrate that [tex]\(\varphi\)[/tex] is a valid homomorphism:
- [tex]\(\mathbb{Z}\)[/tex] maps to [tex]\( \mathbb{Z}_5 \)[/tex] with [tex]\( \varphi(7) = 2 \)[/tex] and [tex]\( \varphi(3) = 3 \)[/tex].
- The addition operation in [tex]\( \mathbb{Z}_5 \)[/tex] shows that [tex]\(\varphi(7 + 3) = 0\)[/tex] and [tex]\(\varphi(7) + \varphi(3) = 0\)[/tex].
Finally, these results confirm that [tex]\(\varphi(m+r) = \varphi(m) + \varphi(r)\)[/tex] for the given examples, verifying the homomorphism property correctly.
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