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].
