Let $G$ and $H$ be nontrivial groups such that $G$ is simple and let $f : G \to H$ be a surjective homomorphism. Show that $f$ is an isomorphism. Having some discomfort in my solution and was wondering if there was an easier way to do this, Thanks.
If $G$ is simple, then the kernel of $f$ is either $\{1\}$ or $G$ itself.
Since the kernel is a normal subgroup, by the first isomorphism theorem we have that $G/\{1\} = G \cong {\rm Im}(f)$. Now, if $f$ is a surjective homomorphism, the image of $f$ is the group $H$. Therefore we have that $G \cong H$.
 A: This is my response to a question that basically asked "is my proof of the following statement OK?", but has now been mostly erased.
The statement, paraphrased, was this: Suppose $f:G \to H$ is a surjective homomorphism between nontrivial groups, and $G$ is simple; then $f$ is an isomorphism.
I have one small suggestion about proof-writing.
Use the word 'if' only if something is uncertain, or you're introducing one of two cases, as in

If $n$ is odd, then...; on the other hand, if $n$ is even, then ...

So your proof would be better written

Because $G$ is simple, the kernel of $f$ is either $\{1\}$ or $G$.
The kernel is a normal subgroup, so by the first
isomorphism theorem we have that $G/\{1\} = G \cong Im(f)$.** Now, because
$f$ is a surjective homomorphism, the image of $f$ is the group $H$.
Therefore $G \cong H$, as required.

** There's actually a flaw in your proof at this point. You've established that $ker(f)$ is either $\{1\}$ or $G$, but now you're assuming that it's $\{1\}$. The other possibility is that it's all of $G$, which leads to $G/G \cong Im(f)$, so that $im(f)$ is a one-element group. But it's also (as your next sentence observes) isomorphic to $H$, which is assumed nontrivial, so there's a contradiction, and this case cannot happen. Let's try a complete rewrite:
$$ 
\newcommand{\ker}{\operatorname{ker}}
\newcommand{\Im}{\operatorname{Im}}
$$

Let $K = \ker(f)$. $K$ is normal, so $G/K \cong \Im(f)$ by the first isomorphism theorem. And $f$ is surjective, so $\Im(f) = H$, hence
$$
G/K \cong H \tag{1}
$$
But $G$ is simple, so $K = \{1\}$ or $K = G$.


Case 1: Suppose that $K = G$. Then $G/K = \{1\}$, so $H$ is a trivial group, which contradicts the hypotheses.


Case 2: $K = \{1\}$. Then $G/K \cong G$, and combining with equation 1, we get $G \cong H$.

Notice that nowhere in this proof did we use the hypothesis that $G$ was nontrivial.
A: Hint: $ker(f)$ is a normal subgroup of $G$. And use the fact that in general $G/ker(f) \cong im(f)$.
