Prove that $g∶ H \to f(H),$ given by $g(x)=f(x)$ is an isomorphism, given $f$ is an isomorphism. 
Prove that $g∶ H \to f(H),$ given by $g(x) = f(x)$ is an isomorphism, given $f$ is an isomorphism.

Let $f∶ K \to G$ be an isomorphism of groups.
I also proved before this question that $f(H) = \{y ∈ G ∶ y = f(x) \text{ for some } x \in H \}$ is a subgroup of $G$.
So I have to show $g$ is a homomorphism, injective, and surjective. Let $x,y \in H$. Then $g(xy)=f(xy)=f(x)f(y)=g(x)g(y).$ (Since $f$ is a homomorphism). Let $y=f(x) \in f(H),$ and $x \in H$. Then $g(x)=f(x)=y.$ Hence it is surjective.
I know that I could prove an injective homomorphism by using $g(a)=e_{f(H)}$ implies $a=e_H$. I have trouble with injectivity sometimes. Also is what I wrote for surjective ok?

Let $h,k\in H$. We have 
$$\begin{align}
g(hk)&=f(hk)\\
&=f(h)f(k)\\
&=g(h)g(k),
\end{align}$$
so $g$ is an homomorpism.
Let $y\in f(H)$. Then there is some $h\in H$ with $y=f(h)$. But $g(h)=f(h)$. Hence $g$ is surjective.
Suppose $g(h)=g(k)$ for $h, k\in H$. Then 
$$\begin{align}
f(h)&=g(h)\\
&=g(k)\\
&=f(k).
\end{align}$$
But $f$ is injective since it is an isomorphism. Hence $h=k$. Hence $g$ is injective.
Hence $g$ is an isomorphism.
