Let's review the proof for Lemma 2.5.2:
(a) Proof that [tex]\(\varphi(e) = e^{\prime}\)[/tex]:
Given that [tex]\(\varphi\)[/tex] is a homomorphism from group [tex]\(G\)[/tex] to group [tex]\(G'\)[/tex].
1. Let [tex]\(e\)[/tex] be the identity element in group [tex]\(G\)[/tex].
2. For any element [tex]\(a \in G\)[/tex], by the definition of the identity element,
[tex]\[ a \cdot e = a. \][/tex]
3. Apply the homomorphism [tex]\(\varphi\)[/tex]:
[tex]\[ \varphi(a \cdot e) = \varphi(a). \][/tex]
4. By the property of homomorphisms, [tex]\(\varphi(a \cdot e) = \varphi(a) \cdot \varphi(e)\)[/tex]. Therefore, we have:
[tex]\[ \varphi(a) = \varphi(a) \cdot \varphi(e). \][/tex]
5. To isolate [tex]\(\varphi(e)\)[/tex], we multiply both sides of the equation by the inverse of [tex]\(\varphi(a)\)[/tex]:
[tex]\[ \varphi(a)^{-1} \cdot \varphi(a) = \varphi(a)^{-1} \cdot (\varphi(a) \cdot \varphi(e)). \][/tex]
6. Simplifying,
[tex]\[ e' = e' \cdot \varphi(e), \][/tex]
where [tex]\(e'\)[/tex] denotes the identity element in [tex]\(G'\)[/tex].
7. By the property of the identity element in [tex]\(G'\)[/tex], we have:
[tex]\[ e' \cdot \varphi(e) = \varphi(e). \][/tex]
8. Therefore,
[tex]\[ e' = \varphi(e). \][/tex]
So, we have shown that [tex]\(\varphi(e) = e'\)[/tex], the identity element of [tex]\(G'\)[/tex].
(b) Proof that [tex]\(\varphi(a^{-1}) = \varphi(a)^{-1}\)[/tex] for all [tex]\(a \in G\)[/tex]:
1. For any element [tex]\(a \in G\)[/tex], we know the following by the definition of the inverse element:
[tex]\[ a \cdot a^{-1} = e \][/tex]
where [tex]\(e\)[/tex] is the identity element in [tex]\(G\)[/tex].
2. Apply the homomorphism [tex]\(\varphi\)[/tex] to both sides of the equation:
[tex]\[ \varphi(a \cdot a^{-1}) = \varphi(e). \][/tex]
3. By the property of homomorphisms, we have:
[tex]\[ \varphi(a) \cdot \varphi(a^{-1}) = \varphi(e). \][/tex]
4. From part (a) of the lemma, we know [tex]\(\varphi(e) = e'\)[/tex]. Therefore:
[tex]\[ \varphi(a) \cdot \varphi(a^{-1}) = e'. \][/tex]
5. This shows that [tex]\(\varphi(a^{-1})\)[/tex] is indeed the inverse of [tex]\(\varphi(a)\)[/tex] in [tex]\(G'\)[/tex]. Hence, we can write:
[tex]\[ \varphi(a^{-1}) = \varphi(a)^{-1}. \][/tex]
Thus, we have proved both parts of Lemma 2.5.2.
