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Chapter 16: Problem 6

Let \(f: G \rightarrow H\) be a group homomorphism onto \(H\). If \(G\) is abelian, prove that \(H\) is abelian.

Short Answer

Expert verified
If \( G \) is abelian and \( f: G \rightarrow H \) is a group homomorphism onto \( H \), then \( H \) is also abelian.

Step by step solution

01

Define Arbitrary Elements

Let's start by choosing arbitrary elements \( h_1, h_2 \) in \( H \). Since \( f \) is a surjective homomorphism, there exist elements in \( G \), lets denote them by \( g_1, g_2 \) respectively, such that \( f(g_1) = h_1 \) and \( f(g_2) = h_2 \).
02

Use Surjective Nature

Taking into account the surjective nature of the mapping function, apply the homomorphism property. That is, \( f(g_1g_2) = h_1h_2 = f(g_1)f(g_2) \) and \( f(g_2g_1) = h_2h_1 = f(g_2)f(g_1) \).
03

Apply Abelian Property

Because \( G \) is an abelian group, the order of multiplication does not matter. Hence, \( g_1g_2 = g_2g_1 \). This implies \( f(g_1g_2) = f(g_2g_1) \). But we know from step 2 that \( f(g_1g_2) = h_1h_2 \) and \( f(g_2g_1) = h_2h_1 \). Therefore, \( h_1h_2 = h_2h_1 \).
04

Conclude Proof

Thus, as per the definition of an abelian group, it can be concluded that \( H \) is abelian as for all \( h_1, h_2 \in H \), \( h_1h_2 = h_2h_1 \).

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