2. Let's break down the problem step-by-step and highlight the provided information as well as draw the necessary conclusions.
Given:
- [tex]\( G \)[/tex] is an abelian group.
- A function [tex]\(\varphi: G \rightarrow G\)[/tex] is defined by [tex]\(\varphi(a) = a^2\)[/tex].
First, we need to verify that [tex]\(\varphi\)[/tex] is a homomorphism. To do this, take any two elements [tex]\(a, b \in G\)[/tex] and consider [tex]\(\varphi(ab)\)[/tex]:
[tex]\[
\varphi(ab) = (ab)^2
\][/tex]
Since [tex]\(G\)[/tex] is abelian, the elements [tex]\(a\)[/tex] and [tex]\(b\)[/tex] commute:
[tex]\[
(ab)^2 = (ab)(ab) = a(ba)b = a(ab)b = a^2 b^2
\][/tex]
So,
[tex]\[
\varphi(ab) = a^2 b^2 = \varphi(a) \varphi(b)
\][/tex]
Thus, [tex]\(\varphi\)[/tex] is indeed a homomorphism.
Next, we need to consider whether [tex]\(\varphi\)[/tex] is onto. To assess this, we look at the group [tex]\(U_8\)[/tex], which is the set of units modulo 8, or [tex]\(\{1, 3, 5, 7\}\)[/tex]:
For each element [tex]\(a \in U_8\)[/tex]:
[tex]\[
1^2 \equiv 1 \mod 8
\][/tex]
[tex]\[
3^2 \equiv 9 \equiv 1 \mod 8
\][/tex]
[tex]\[
5^2 \equiv 25 \equiv 1 \mod 8
\][/tex]
[tex]\[
7^2 \equiv 49 \equiv 1 \mod 8
\][/tex]
In this specific group (i.e. [tex]\( U_8 \)[/tex]), squaring any element results in the identity element [tex]\( e \)[/tex] (which is 1 in this case). So,
[tex]\(\varphi(G) = \{e\} = \{1\}\)[/tex].
This means that [tex]\(\varphi\)[/tex] maps every element of [tex]\(U_8\)[/tex] to 1, indicating that [tex]\(\varphi\)[/tex] is not onto, since it does not cover every element of the set [tex]\(U_8\)[/tex] but only maps to the identity element.
Finally, we conclude:
- [tex]\(\varphi\)[/tex] is a homomorphism because [tex]\(\varphi(ab) = \varphi(a) \varphi(b)\)[/tex] for all [tex]\(a, b \in G\)[/tex].
- In the example where [tex]\(G = U_8\)[/tex], the image [tex]\(\varphi(G)\)[/tex] is just the identity element, showing that [tex]\(\varphi\)[/tex] is not necessarily onto.
