Prove that if $G_1, G_2$ are groups and $\phi: G_1 \to G_2$ is a multiplicative map $(\phi (ab)=\phi (a) \phi(b))$, then $\phi(1_{G_1})=1_{G_2}$ and for every $x \in G_1, \phi(x^{-1})=\phi(x)^{-1}$.
My attempt:
Consider an arbitrary $x \in G_1$ and take the identity from $G_1$, which is $1_{G_1}$. Then we have
$$\phi(x) \phi (1_{G_1})=\phi(x 1_{G_1})=\phi(x)$$
$$ \phi (1_{G_1}) \phi(x)=\phi(1_{G_1} x)=\phi(x)$$
This proofs that $\phi (1_{G_1})=1_{G_2}$.
For the second part, I think I should take the following:
$$1_{G_2}=\phi(1_{G_1})=\phi(x x^{-1})=\phi(x) \phi(x^{-1}) \Rightarrow \phi(x^{-1})=\phi(x)^{-1}$$
Here we used, that since $G_2$ is a group, $\phi(x)$ must have an inverse there.
Is my solution correct? Any help appreciated for either part.
The first part is not correct. You proved (correctly) that $\phi(1_{G_1})$ is such that, when muliplied by any element $h$ of the image of $\phi$, it gives $h$ again. But being $1_{G_2}$ means the when we multiply it by any element of $G_2$, we get that element again.
You can prove that the identity element $1_G$ of any group $G$ is the only element $g\in G$ such that $g.g=g$. Since $\phi(1_{G_1}).\phi(1_{G_1})=\phi(1_{G_1})$, $\phi(1_{G_1})=1_{G_2}$.