I Images of elements in a group homomorphism

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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