Alternate Proof to $f(e_G)=e_H$

Question

Is this proof correct?

Proof: Let for all $a$ and $e_G \in G$ we know that if $f$ is a homomorphism from ${(G,*)}$ to $(H,o)$ then,

$f(a*e_G)=f(a)=f(a)$ o $f(e_G)$. Similairly $f(e_G*a)=f(a)=f(e_G)$ o $f(a)$. So we now know the following:

$f(e_G)$ o $f(a)=f(a)$ o $f(e_G)=f(a)$

At this point can one state that $f(e_G)=e_H$? Or will this proof be sufficient if $f$ is a bijection?

Note that I have gone through all the proofs in this question but no proof given is similar to the one I have written above.

Answer 1

Your idea is fine, however it would be better to write $f(a) = f(a\ast e_G) = f(a)\circ f(e_G)$ instead of $f(a\ast e_G) = f(a) = f(a)\circ f(e_G)$ as this seems you are not using the axioms of a group directly. But, if $f(a) = f(a)\circ f(e_G)$, then multiply on the left by the inverse of $f(a)$ in $H$ to obtain $$e_H= f(a)^{-1}\circ f(a)= f(a)^{-1}\circ f(a)\circ f(e_G) = e_H \circ f(e_G) = f(e_G) $$

Answer 2

It is possible to deduce from $f(e)f(a)=f(a)f(e)=f(a)$ for all $a\in G$ that $f(e)$ is identity in $H$.

Alternate way I would say: under homomorphism, image of a subgroup is again a subgroup [can be proved by subgroup-test], then what is possibility of $f(\{e_G\})$?