Images of elements in a group homomorphism

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SUMMARY

The image of elements in a group homomorphism is fundamentally dependent on the image of the identity element, denoted as 1, due to its simplicity in calculations. The homomorphism property allows for the expression of any element a as a product of the identity element, leading to the conclusion that the image can be computed as a times the image of 1, specifically illustrated with the calculation of φ(7). This demonstrates that while other generators can be utilized, the identity element provides a more straightforward approach to understanding homomorphic images.

PREREQUISITES
  • Understanding of group theory concepts, particularly group homomorphisms.
  • Familiarity with the properties of identity elements in algebraic structures.
  • Basic knowledge of mathematical notation and operations involving functions.
  • Experience with calculations involving specific examples in group theory.
NEXT STEPS
  • Study the properties of group homomorphisms in greater depth.
  • Explore examples of homomorphisms using different generators beyond the identity element.
  • Learn about the implications of homomorphic images in various algebraic structures.
  • Investigate the role of identity elements in other mathematical contexts, such as rings and fields.
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Mathematicians, students of abstract algebra, and anyone interested in the properties and applications of group homomorphisms will benefit from this discussion.

Terrell
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Why does the image of elements in a homomorphism depend on the image of 1? Why not the other generators?
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Other generators would work as well, but ##1## is especially easy to handle:
$$\varphi(a)=\varphi(a\cdot 1)=\varphi (\underbrace{1+\ldots +1}_{a\text{ times }})=\underbrace{\varphi(1)+\ldots +\varphi(1)}_{a\text{ times }}=a\cdot \varphi(1)$$
Now convince me with such a calculation by the use of ##\varphi(7)##.
 
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fresh_42 said:
Other generators would work as well, but ##1## is especially easy to handle:
$$\varphi(a)=\varphi(a\cdot 1)=\varphi (\underbrace{1+\ldots +1}_{a\text{ times }})=\underbrace{\varphi(1)+\ldots +\varphi(1)}_{a\text{ times }}=a\cdot \varphi(1)$$
Now convince me with such a calculation by the use of ##\varphi(7)##.
I was not sure it was out of pure convenience. Thanks!
 

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