TITLE: Tricks - Prove Homomorphism Maps Identity to Identity - Fraleigh p. 128 Theorem 13.12(1.)
QUESTION [10 upvotes]: Let $\phi$ be a homomorphism of a group G into a group G'.
If $e =$ the identity element in G, then $\phi(e) =$ the identity element in G'.
Is this what Sharkos is trying to answer:
About homomorphisms, you only know $\phi(a) \phi(b) = \phi(ab)$.
ab is more complicated than anything we want to think about, hence just presuppose $b = e$.
Then $ \begin{align} \phi(a)\phi(e) & = \phi(a\color{magenta}{e}) \\ & = \phi(a) \end{align} $
Left multiply the last equation  by $\color{green}{\phi(a)^{-1}}$: $\quad \phi(e) =  id_{G'}.$     

(1.) How do you predestine to rewrite $a$ as   $ a = a\color{magenta}{e}$?
  Or to 'presuppose $b = e$' ? I understand neither tricks.
(2.) What's the intuition?

REPLY [2 votes]: Without digging into any thing deep or complicated, you only have three pieces of knowledge:

$e'$ is the only element that ever satisfies an equation in $G'$ of the form $e'x = x$ or $ye' = y$
$e$ satisfies equations in $G$ of the form $ex = x$ and $ye = y$
$\phi(xy) = \phi(x) \phi(y)$  (and similarly for products of more than two things)

By the first point, if you want to prove $\phi(e) = e'$, the only option we have (if we want to use these simple pieces of knowledge) is to find an equation $\phi(e)x = x$ or $y \phi(e) = y$. Trying to figure out how to use the other pieces of information to do this leads to a proof.
The proof could also have been derived from a simple approach of "I have facts, I'll do something with them and see if the result is useful". e.g. looking at the second and third points and seeing that you can combine them... so you do so, and see if the result is useful. (and it turns out to be exactly what you need to use the first point to complete the proof)