Undergrad Images of elements in a group homomorphism

[Terrell]

Why does the image of elements in a homomorphism depend on the image of 1? Why not the other generators?

 

[fresh_42]

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 $\varphi(7)$.

 

[Terrell]

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!