Answer :
Certainly! Let's dive into the details of the question about the given function [tex]\(\varphi\)[/tex] in the context of the abelian group [tex]\(G\)[/tex].
1. Understanding the Group and Homomorphism Definition:
- Let [tex]\(G\)[/tex] be an abelian group, meaning that [tex]\(G\)[/tex] is a set with a binary operation (let's denote it by [tex]\(\)[/tex]) that satisfies the group axioms: closure, associativity, identity element, inverse element, and commutativity.
- There is also a mapping [tex]\(\varphi: G \rightarrow G\)[/tex] defined by [tex]\(\varphi(a) = a^2\)[/tex] for any [tex]\(a \in G\)[/tex].
2. Homomorphism Verification:
- To verify that [tex]\(\varphi\)[/tex] is a homomorphism, we need to check that [tex]\(\varphi(ab) = \varphi(a)\varphi(b)\)[/tex] for all [tex]\(a, b \in G\)[/tex].
- Given [tex]\(G\)[/tex] is abelian, the operation [tex]\(\)[/tex] is commutative. Let [tex]\(a, b \in G\)[/tex],
[tex]\[ \varphi(ab) = (ab)^2. \][/tex]
- Using the commutativity property of the abelian group,
[tex]\[ (ab)^2 = ab ab = a a b b = a^2 b^2 = \varphi(a) \varphi(b). \][/tex]
- Hence, [tex]\(\varphi\)[/tex] is indeed a homomorphism, because for any [tex]\(a, b \in G\)[/tex], [tex]\(\varphi(ab) = \varphi(a)\varphi(b)\)[/tex].
3. Considering the Group [tex]\(U_8\)[/tex]:
- Take [tex]\(U_8\)[/tex], the group of units modulo 8. [tex]\(U_8 = \{1, 3, 5, 7\}\)[/tex] with the group operation being multiplication modulo 8.
- We need to verify if [tex]\(\varphi(a) = a^2\)[/tex] for all [tex]\(a \in U_8\)[/tex] and look at its behavior.
- Calculate [tex]\(a^2\)[/tex] for all elements of [tex]\(U_8\)[/tex]:
[tex]\[ \begin{aligned} &1^2 = 1 \equiv 1 \pmod{8}, \\ &3^2 = 9 \equiv 1 \pmod{8}, \\ &5^2 = 25 \equiv 1 \pmod{8}, \\ &7^2 = 49 \equiv 1 \pmod{8}. \end{aligned} \][/tex]
- From this, we see that for every [tex]\(a \in U_8\)[/tex], [tex]\(a^2 \equiv 1 \pmod{8}\)[/tex], meaning [tex]\(\varphi(a) = e\)[/tex], where [tex]\(e\)[/tex] is the identity element in [tex]\(U_8\)[/tex].
4. Conclusion:
- The image under [tex]\(\varphi\)[/tex], denoted [tex]\(\varphi(G)\)[/tex], must be carefully considered. For [tex]\(U_8\)[/tex], every element squares to the identity element [tex]\(e\)[/tex]. Hence, [tex]\(\varphi(G) = \{e\}\)[/tex].
- Therefore, the homomorphism [tex]\(\varphi\)[/tex] maps every element in [tex]\(U_8\)[/tex] to the identity, making [tex]\(\varphi\)[/tex] not surjective (onto).
Given these explanations, the mapping [tex]\(\varphi(a) = a^2\)[/tex] in [tex]\(G = U_8\)[/tex] results in the image of [tex]\(\varphi\)[/tex] being the set containing only the identity element. This illustrates a peculiar property where despite being a homomorphism, [tex]\(\varphi\)[/tex] is not an onto function.
I hope these steps clarify the verification and implications of the homomorphism [tex]\(\varphi\)[/tex] within the specific group [tex]\(U_8\)[/tex].
1. Understanding the Group and Homomorphism Definition:
- Let [tex]\(G\)[/tex] be an abelian group, meaning that [tex]\(G\)[/tex] is a set with a binary operation (let's denote it by [tex]\(\)[/tex]) that satisfies the group axioms: closure, associativity, identity element, inverse element, and commutativity.
- There is also a mapping [tex]\(\varphi: G \rightarrow G\)[/tex] defined by [tex]\(\varphi(a) = a^2\)[/tex] for any [tex]\(a \in G\)[/tex].
2. Homomorphism Verification:
- To verify that [tex]\(\varphi\)[/tex] is a homomorphism, we need to check that [tex]\(\varphi(ab) = \varphi(a)\varphi(b)\)[/tex] for all [tex]\(a, b \in G\)[/tex].
- Given [tex]\(G\)[/tex] is abelian, the operation [tex]\(\)[/tex] is commutative. Let [tex]\(a, b \in G\)[/tex],
[tex]\[ \varphi(ab) = (ab)^2. \][/tex]
- Using the commutativity property of the abelian group,
[tex]\[ (ab)^2 = ab ab = a a b b = a^2 b^2 = \varphi(a) \varphi(b). \][/tex]
- Hence, [tex]\(\varphi\)[/tex] is indeed a homomorphism, because for any [tex]\(a, b \in G\)[/tex], [tex]\(\varphi(ab) = \varphi(a)\varphi(b)\)[/tex].
3. Considering the Group [tex]\(U_8\)[/tex]:
- Take [tex]\(U_8\)[/tex], the group of units modulo 8. [tex]\(U_8 = \{1, 3, 5, 7\}\)[/tex] with the group operation being multiplication modulo 8.
- We need to verify if [tex]\(\varphi(a) = a^2\)[/tex] for all [tex]\(a \in U_8\)[/tex] and look at its behavior.
- Calculate [tex]\(a^2\)[/tex] for all elements of [tex]\(U_8\)[/tex]:
[tex]\[ \begin{aligned} &1^2 = 1 \equiv 1 \pmod{8}, \\ &3^2 = 9 \equiv 1 \pmod{8}, \\ &5^2 = 25 \equiv 1 \pmod{8}, \\ &7^2 = 49 \equiv 1 \pmod{8}. \end{aligned} \][/tex]
- From this, we see that for every [tex]\(a \in U_8\)[/tex], [tex]\(a^2 \equiv 1 \pmod{8}\)[/tex], meaning [tex]\(\varphi(a) = e\)[/tex], where [tex]\(e\)[/tex] is the identity element in [tex]\(U_8\)[/tex].
4. Conclusion:
- The image under [tex]\(\varphi\)[/tex], denoted [tex]\(\varphi(G)\)[/tex], must be carefully considered. For [tex]\(U_8\)[/tex], every element squares to the identity element [tex]\(e\)[/tex]. Hence, [tex]\(\varphi(G) = \{e\}\)[/tex].
- Therefore, the homomorphism [tex]\(\varphi\)[/tex] maps every element in [tex]\(U_8\)[/tex] to the identity, making [tex]\(\varphi\)[/tex] not surjective (onto).
Given these explanations, the mapping [tex]\(\varphi(a) = a^2\)[/tex] in [tex]\(G = U_8\)[/tex] results in the image of [tex]\(\varphi\)[/tex] being the set containing only the identity element. This illustrates a peculiar property where despite being a homomorphism, [tex]\(\varphi\)[/tex] is not an onto function.
I hope these steps clarify the verification and implications of the homomorphism [tex]\(\varphi\)[/tex] within the specific group [tex]\(U_8\)[/tex].
