Product with Inverse on Homomorphic Image is Group Homomorphism/Mistake
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Source Work
1996: John F. Humphreys: A Course in Group Theory:
- Chapter $8$: The Homomorphism Theorem:
- Exercise $3$
Mistake
- Suppose that $H$ is an abelian group and let $\vartheta: G \to H$ be a homomorphism. Define a map $\phi: G \times G \to H$ by
- $\map \phi {g_1, g_2} = \map \phi {g_1} \map \phi {g_2}^{-1}$.
- Prove that $\phi$ is a homomorphism.
Correction
The condition is wrong.
It should read:
- $\map \phi {g_1, g_2} = \map \vartheta {g_1} \map \vartheta {g_2}^{-1}$
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $8$: The Homomorphism Theorem: Exercise $3$