Two groups are defined as $(G_{1}, \circ)$ and $(G_{2}, \bullet)$, where $e_{1}$ is the identity element of the first group and $e_{2}$ of the second group. I will prove that if $$f: (G_{1}, \circ) \rightarrow (G_{2}, \bullet)$$ is group homomorphism, then $$f(e_{1})=e_{2}.$$ So there's my proof that i'm not really certain of:
If f is a homomorphism of those two groups it means that $$f(x\circ y)=f(x)\bullet f(y), so\\f(x)=f(x\circ e_{1})=f(x)\bullet f(e_{1}), \text{and as we know }e_{2} \text{ is an identity element of the second group so we can write:}\\f(x)\bullet e_{2}=f(x)\bullet f(e_{1}),\\\text{and now, (using the mathematical law, that I have no idea how to call in English) I have my thesis}\\e_{2}=f(e_{1}).$$Am I even close to thinking right?
Your proof is correct. From
$(*)$ $f(x) \bullet e_2=f(x) \bullet f(e_1)$
we get $e_2=f(e_1)$ by multiplying (*) from the left with $f(x)^{-1}$