An isomorphism $f:G_1 \to G_2$ maps the identity of $G_1$ to the identity of $G_2$

Question

Here is my attempted proof:

Suppose $e_1$ is the identity of $G_1$ and likewise $e_2$ is the identity of $G_2$. Then we have to show $f(e_1)=e_2$.

Since $f$ is an isomorphism, $f$ is bijective and hence surjective. Therefore, $f(x)=e_2$ for some $x\in G_1$.

Also, $f(x\cdot e_1)=e_2$. This implies $f(x)f(e_1)=e_2$. Also the product of $f(x)$ and $f(e_1)$ gives the identity, hence they must be inverses of each other, i.e. $f(e_1)=f(x)^{-1}$. But $f(x)=e_2$ and hence $f(e_1)=e_{2}^{-1}=e_2$.

Is my proof correct? Are there any alternative ways?

Answer 1

Your proof is correct, but you are explicitly using surjectivity. You don't actually need this, but instead all you need is for the map $f$ to be a homomorphism (so not necessarily a surjective or injective):

If $f: G_1\rightarrow G_2$ is a group homomorphism then $f(e_1)=e_2$.

Proof. Write $g:=f(e_1)$. Then $g^2=f(e_1)^2=f(e_1^2)=f(e_1)=g$. That is, $g^2=g$. Then: $$\begin{align*} g^2&=g\\ \Rightarrow g^2\cdot g^{-1}&=g\cdot g^{-1}\\ \Rightarrow g&=e_2 \end{align*}$$ Therefore, $f(e_1)=e_2$ as required.