IDNLearn.com: Your trusted source for finding accurate and reliable answers. Join our community to receive prompt, thorough responses from knowledgeable experts.
Sagot :
First, let's review the fundamental concepts in the problem:
1. Abelian Group: A group [tex]\( G \)[/tex] is abelian if the group operation is commutative, i.e., [tex]\( ab = ba \)[/tex] for all [tex]\( a, b \in G \)[/tex].
2. Homomorphism: A function [tex]\( \varphi: G \rightarrow G \)[/tex] is a homomorphism if for all [tex]\( a, b \in G \)[/tex], [tex]\( \varphi(ab) = \varphi(a)\varphi(b) \)[/tex].
3. Function [tex]\( \varphi \)[/tex]: In this problem, the function [tex]\( \varphi \)[/tex] is defined by [tex]\( \varphi(a) = a^2 \)[/tex] for [tex]\( a \in G \)[/tex].
### Check If [tex]\(\varphi\)[/tex] Is a Homomorphism:
Given [tex]\( G \)[/tex] is abelian and [tex]\( \varphi(a) = a^2 \)[/tex], we need to check if [tex]\( \varphi \)[/tex] is a homomorphism.
For any [tex]\( a, b \in G \)[/tex]:
[tex]\[ \varphi(ab) = (ab)^2 \][/tex]
Since [tex]\( G \)[/tex] is abelian, we have:
[tex]\[ (ab)^2 = ab \cdot ab = a \cdot a \cdot b \cdot b = a^2 b^2 \][/tex]
So:
[tex]\[ \varphi(ab) = a^2 b^2 \][/tex]
Also, since [tex]\( \varphi(a) = a^2 \)[/tex] and [tex]\( \varphi(b) = b^2 \)[/tex], we have:
[tex]\[ \varphi(a) \varphi(b) = a^2 b^2 \][/tex]
Thus:
[tex]\[ \varphi(ab) = \varphi(a) \varphi(b) \][/tex]
This confirms that [tex]\( \varphi \)[/tex] is indeed a homomorphism.
### Check If [tex]\(\varphi\)[/tex] Is Onto:
Next, we must determine if [tex]\( \varphi \)[/tex] is onto (surjective). It is surjective if every element in [tex]\( G \)[/tex] is the image of some element under [tex]\( \varphi \)[/tex].
To evaluate this, consider the group [tex]\( U_8 \)[/tex], the multiplicative group of units modulo 8. The elements of [tex]\( U_8 \)[/tex] are [tex]\( \{1, 3, 5, 7\} \)[/tex]. Let's calculate [tex]\( a^2 \)[/tex] for each element:
- [tex]\( 1^2 \mod 8 = 1 \)[/tex]
- [tex]\( 3^2 \mod 8 = 9 \equiv 1 \mod 8 \)[/tex]
- [tex]\( 5^2 \mod 8 = 25 \equiv 1 \mod 8 \)[/tex]
- [tex]\( 7^2 \mod 8 = 49 \equiv 1 \mod 8 \)[/tex]
We observe that for every element [tex]\( a \in U_8 \)[/tex], we have [tex]\( a^2 \equiv 1 \mod 8 \)[/tex], implying that [tex]\( a^2 = e \)[/tex] where [tex]\( e \)[/tex] represents the identity element in [tex]\( U_8 \)[/tex].
Therefore, for every [tex]\( a \in U_8 \)[/tex], [tex]\( \varphi(a) = 1 \)[/tex]. This means [tex]\( \varphi: U_8 \rightarrow U_8 \)[/tex] maps every element to the identity element [tex]\( 1 \)[/tex].
Hence, the image of [tex]\( \varphi \)[/tex] is just [tex]\( \{1\} \)[/tex].
Thus, [tex]\( \varphi \)[/tex] is not onto in [tex]\( U_8 \)[/tex] since the only element in the image is the identity element and not the entire group [tex]\( U_8 \)[/tex].
### Conclusion:
In summary, the function [tex]\( \varphi(a) = a^2 \)[/tex] is a homomorphism for the abelian group [tex]\( G \)[/tex]. However, in the specific case of the group [tex]\( U_8 \)[/tex], [tex]\( \varphi \)[/tex] is not surjective, as it maps every element to the identity element [tex]\( 1 \)[/tex], thus [tex]\( \varphi(G) = \{1\} \)[/tex].
1. Abelian Group: A group [tex]\( G \)[/tex] is abelian if the group operation is commutative, i.e., [tex]\( ab = ba \)[/tex] for all [tex]\( a, b \in G \)[/tex].
2. Homomorphism: A function [tex]\( \varphi: G \rightarrow G \)[/tex] is a homomorphism if for all [tex]\( a, b \in G \)[/tex], [tex]\( \varphi(ab) = \varphi(a)\varphi(b) \)[/tex].
3. Function [tex]\( \varphi \)[/tex]: In this problem, the function [tex]\( \varphi \)[/tex] is defined by [tex]\( \varphi(a) = a^2 \)[/tex] for [tex]\( a \in G \)[/tex].
### Check If [tex]\(\varphi\)[/tex] Is a Homomorphism:
Given [tex]\( G \)[/tex] is abelian and [tex]\( \varphi(a) = a^2 \)[/tex], we need to check if [tex]\( \varphi \)[/tex] is a homomorphism.
For any [tex]\( a, b \in G \)[/tex]:
[tex]\[ \varphi(ab) = (ab)^2 \][/tex]
Since [tex]\( G \)[/tex] is abelian, we have:
[tex]\[ (ab)^2 = ab \cdot ab = a \cdot a \cdot b \cdot b = a^2 b^2 \][/tex]
So:
[tex]\[ \varphi(ab) = a^2 b^2 \][/tex]
Also, since [tex]\( \varphi(a) = a^2 \)[/tex] and [tex]\( \varphi(b) = b^2 \)[/tex], we have:
[tex]\[ \varphi(a) \varphi(b) = a^2 b^2 \][/tex]
Thus:
[tex]\[ \varphi(ab) = \varphi(a) \varphi(b) \][/tex]
This confirms that [tex]\( \varphi \)[/tex] is indeed a homomorphism.
### Check If [tex]\(\varphi\)[/tex] Is Onto:
Next, we must determine if [tex]\( \varphi \)[/tex] is onto (surjective). It is surjective if every element in [tex]\( G \)[/tex] is the image of some element under [tex]\( \varphi \)[/tex].
To evaluate this, consider the group [tex]\( U_8 \)[/tex], the multiplicative group of units modulo 8. The elements of [tex]\( U_8 \)[/tex] are [tex]\( \{1, 3, 5, 7\} \)[/tex]. Let's calculate [tex]\( a^2 \)[/tex] for each element:
- [tex]\( 1^2 \mod 8 = 1 \)[/tex]
- [tex]\( 3^2 \mod 8 = 9 \equiv 1 \mod 8 \)[/tex]
- [tex]\( 5^2 \mod 8 = 25 \equiv 1 \mod 8 \)[/tex]
- [tex]\( 7^2 \mod 8 = 49 \equiv 1 \mod 8 \)[/tex]
We observe that for every element [tex]\( a \in U_8 \)[/tex], we have [tex]\( a^2 \equiv 1 \mod 8 \)[/tex], implying that [tex]\( a^2 = e \)[/tex] where [tex]\( e \)[/tex] represents the identity element in [tex]\( U_8 \)[/tex].
Therefore, for every [tex]\( a \in U_8 \)[/tex], [tex]\( \varphi(a) = 1 \)[/tex]. This means [tex]\( \varphi: U_8 \rightarrow U_8 \)[/tex] maps every element to the identity element [tex]\( 1 \)[/tex].
Hence, the image of [tex]\( \varphi \)[/tex] is just [tex]\( \{1\} \)[/tex].
Thus, [tex]\( \varphi \)[/tex] is not onto in [tex]\( U_8 \)[/tex] since the only element in the image is the identity element and not the entire group [tex]\( U_8 \)[/tex].
### Conclusion:
In summary, the function [tex]\( \varphi(a) = a^2 \)[/tex] is a homomorphism for the abelian group [tex]\( G \)[/tex]. However, in the specific case of the group [tex]\( U_8 \)[/tex], [tex]\( \varphi \)[/tex] is not surjective, as it maps every element to the identity element [tex]\( 1 \)[/tex], thus [tex]\( \varphi(G) = \{1\} \)[/tex].
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Your search for solutions ends at IDNLearn.com. Thank you for visiting, and we look forward to helping you again.
