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Sagot :
Sure, let's explore this problem step by step:
1. Understanding the Group [tex]\(G\)[/tex]:
- [tex]\(G\)[/tex] is given as an abelian (commutative) group.
- This means for any elements [tex]\(a, b \in G\)[/tex], we have [tex]\(ab = ba\)[/tex].
2. Defining the Homomorphism [tex]\(\varphi\)[/tex]:
- We have a map [tex]\(\varphi: G \rightarrow G\)[/tex] defined by [tex]\(\varphi(a) = a^2\)[/tex].
- We need to show that this map is a homomorphism.
3. Homomorphism Condition:
- To verify [tex]\(\varphi\)[/tex] is a homomorphism, we need to check whether [tex]\(\varphi(ab) = \varphi(a)\varphi(b)\)[/tex] for all [tex]\(a, b \in G\)[/tex].
4. Checking [tex]\(\varphi(ab)\)[/tex]:
- Consider two elements [tex]\(a\)[/tex] and [tex]\(b\)[/tex] in [tex]\(G\)[/tex].
- Calculate:
[tex]\[ \varphi(ab) = (ab)^2 \][/tex]
- Since [tex]\(G\)[/tex] is abelian, [tex]\(ab = ba\)[/tex]. Thus:
[tex]\[ (ab)^2 = ab \cdot ab = a^2 \cdot b^2 = \varphi(a)\varphi(b) \][/tex]
- Therefore, [tex]\(\varphi(ab) = \varphi(a)\varphi(b)\)[/tex], so [tex]\(\varphi\)[/tex] is indeed a homomorphism.
5. Image of the Homomorphism [tex]\(\varphi\)[/tex]:
- To determine whether [tex]\(\varphi\)[/tex] is onto, let's analyze a specific example with a given group [tex]\(U_8\)[/tex].
6. Group [tex]\(U_8\)[/tex]:
- The group [tex]\(U_8\)[/tex] refers to the group of units modulo 8. These are the integers in [tex]\(\{1, 2, 3, 4, 5, 6, 7\}\)[/tex] that are coprime to 8.
- The elements of [tex]\(U_8\)[/tex] are [tex]\(\{1, 3, 5, 7\}\)[/tex]. Each of these elements has a multiplicative inverse in [tex]\(U_8\)[/tex].
7. Applying [tex]\(\varphi\)[/tex] in [tex]\(U_8\)[/tex]:
- Consider [tex]\( \varphi(a) = a^2 \mod 8\)[/tex].
- We need to compute [tex]\(a^2\)[/tex] for each element in [tex]\(U_8\)[/tex]:
[tex]\[ 1^2 \equiv 1 \pmod{8} \][/tex]
[tex]\[ 3^2 \equiv 9 \equiv 1 \pmod{8} \][/tex]
[tex]\[ 5^2 \equiv 25 \equiv 1 \pmod{8} \][/tex]
[tex]\[ 7^2 \equiv 49 \equiv 1 \pmod{8} \][/tex]
- Thus, for any [tex]\(a \in U_8\)[/tex], [tex]\(\varphi(a) = a^2 \equiv 1 \pmod{8}\)[/tex].
8. Conclusion:
- The image of [tex]\(\varphi\)[/tex] in [tex]\(U_8\)[/tex] is [tex]\(\{1\}\)[/tex].
- Hence, [tex]\(\varphi(U_8) = \{1\} = \{e\}\)[/tex], where [tex]\(e\)[/tex] is the identity element of the group.
We have shown that [tex]\(\varphi(a) = a^2\)[/tex] is a homomorphism, but it is not necessarily onto, as demonstrated with the group [tex]\(U_8\)[/tex] where all elements map to the identity element [tex]\(e\)[/tex].
1. Understanding the Group [tex]\(G\)[/tex]:
- [tex]\(G\)[/tex] is given as an abelian (commutative) group.
- This means for any elements [tex]\(a, b \in G\)[/tex], we have [tex]\(ab = ba\)[/tex].
2. Defining the Homomorphism [tex]\(\varphi\)[/tex]:
- We have a map [tex]\(\varphi: G \rightarrow G\)[/tex] defined by [tex]\(\varphi(a) = a^2\)[/tex].
- We need to show that this map is a homomorphism.
3. Homomorphism Condition:
- To verify [tex]\(\varphi\)[/tex] is a homomorphism, we need to check whether [tex]\(\varphi(ab) = \varphi(a)\varphi(b)\)[/tex] for all [tex]\(a, b \in G\)[/tex].
4. Checking [tex]\(\varphi(ab)\)[/tex]:
- Consider two elements [tex]\(a\)[/tex] and [tex]\(b\)[/tex] in [tex]\(G\)[/tex].
- Calculate:
[tex]\[ \varphi(ab) = (ab)^2 \][/tex]
- Since [tex]\(G\)[/tex] is abelian, [tex]\(ab = ba\)[/tex]. Thus:
[tex]\[ (ab)^2 = ab \cdot ab = a^2 \cdot b^2 = \varphi(a)\varphi(b) \][/tex]
- Therefore, [tex]\(\varphi(ab) = \varphi(a)\varphi(b)\)[/tex], so [tex]\(\varphi\)[/tex] is indeed a homomorphism.
5. Image of the Homomorphism [tex]\(\varphi\)[/tex]:
- To determine whether [tex]\(\varphi\)[/tex] is onto, let's analyze a specific example with a given group [tex]\(U_8\)[/tex].
6. Group [tex]\(U_8\)[/tex]:
- The group [tex]\(U_8\)[/tex] refers to the group of units modulo 8. These are the integers in [tex]\(\{1, 2, 3, 4, 5, 6, 7\}\)[/tex] that are coprime to 8.
- The elements of [tex]\(U_8\)[/tex] are [tex]\(\{1, 3, 5, 7\}\)[/tex]. Each of these elements has a multiplicative inverse in [tex]\(U_8\)[/tex].
7. Applying [tex]\(\varphi\)[/tex] in [tex]\(U_8\)[/tex]:
- Consider [tex]\( \varphi(a) = a^2 \mod 8\)[/tex].
- We need to compute [tex]\(a^2\)[/tex] for each element in [tex]\(U_8\)[/tex]:
[tex]\[ 1^2 \equiv 1 \pmod{8} \][/tex]
[tex]\[ 3^2 \equiv 9 \equiv 1 \pmod{8} \][/tex]
[tex]\[ 5^2 \equiv 25 \equiv 1 \pmod{8} \][/tex]
[tex]\[ 7^2 \equiv 49 \equiv 1 \pmod{8} \][/tex]
- Thus, for any [tex]\(a \in U_8\)[/tex], [tex]\(\varphi(a) = a^2 \equiv 1 \pmod{8}\)[/tex].
8. Conclusion:
- The image of [tex]\(\varphi\)[/tex] in [tex]\(U_8\)[/tex] is [tex]\(\{1\}\)[/tex].
- Hence, [tex]\(\varphi(U_8) = \{1\} = \{e\}\)[/tex], where [tex]\(e\)[/tex] is the identity element of the group.
We have shown that [tex]\(\varphi(a) = a^2\)[/tex] is a homomorphism, but it is not necessarily onto, as demonstrated with the group [tex]\(U_8\)[/tex] where all elements map to the identity element [tex]\(e\)[/tex].
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