Given that [tex]\( G \)[/tex] is an abelian group and [tex]\(\varphi: G \rightarrow G\)[/tex] is defined by [tex]\(\varphi(a) = a^2\)[/tex], we want to check if [tex]\(\varphi\)[/tex] is a homomorphism and consider its properties for [tex]\( G = U_8 \)[/tex].
1. Checking if [tex]\( \varphi \)[/tex] is a Homomorphism:
To verify that [tex]\(\varphi\)[/tex] is a homomorphism, we need to show that for all [tex]\( a, b \in G \)[/tex], [tex]\(\varphi(ab) = \varphi(a) \varphi(b)\)[/tex].
Given that [tex]\(\varphi(a) = a^2\)[/tex] and [tex]\( \varphi(b) = b^2 \)[/tex]:
[tex]\[
\varphi(ab) = (ab)^2
\][/tex]
Since [tex]\( G \)[/tex] is abelian, the group operation is commutative, i.e., [tex]\( ab = ba \)[/tex]. Therefore,
[tex]\[
(ab)^2 = ab \cdot ab = a \cdot b \cdot a \cdot b = a \cdot a \cdot b \cdot b = a^2 b^2 = \varphi(a) \varphi(b)
\][/tex]
Thus, [tex]\(\varphi(ab) = \varphi(a) \varphi(b)\)[/tex], confirming that [tex]\(\varphi\)[/tex] is a homomorphism.
2. Considering [tex]\( \varphi \)[/tex] in Context of [tex]\( U_8 \)[/tex]:
The group [tex]\( U_8 \)[/tex] is the group of units modulo 8, consisting of elements that are relatively prime to 8. The elements of [tex]\( U_8 \)[/tex] are:
[tex]\[ U_8 = \{ 1, 3, 5, 7 \} \][/tex]
Let's apply [tex]\(\varphi\)[/tex] to each element in [tex]\( U_8 \)[/tex]:
[tex]\[
\varphi(1) = 1^2 \equiv 1 \pmod{8}
\][/tex]
[tex]\[
\varphi(3) = 3^2 \equiv 9 \equiv 1 \pmod{8}
\][/tex]
[tex]\[
\varphi(5) = 5^2 \equiv 25 \equiv 1 \pmod{8}
\][/tex]
[tex]\[
\varphi(7) = 7^2 \equiv 49 \equiv 1 \pmod{8}
\][/tex]
In each case, we find that [tex]\(\varphi(a) = a^2 \equiv 1 \pmod{8}\)[/tex]. Therefore, for every [tex]\(a \in U_8\)[/tex], [tex]\(\varphi(a) = 1\)[/tex], which means the image of [tex]\( \varphi \)[/tex]:
[tex]\[
\varphi(U_8) = \{ 1 \}
\][/tex]
Hence, [tex]\(\varphi(G) \)[/tex] is the trivial subgroup [tex]\(\{e\}\)[/tex], where [tex]\( e \)[/tex] is the identity element of the group, which in this case corresponds to 1.
In summary, [tex]\(\varphi\)[/tex] is indeed a homomorphism, but it is not onto because the image of [tex]\(\varphi\)[/tex] for [tex]\( G = U_8 \)[/tex] is the trivial subgroup [tex]\(\{1\}\)[/tex].
