For any finite group, there is a homomorphism whose image is simple This is for homework.  The question asks
"Show that, for any finite group $G$, there is a homomorphism $f$ such that $f(G)$ is simple."
My thought was this.  Since $G$ is finite, there are only a finite number of normal subgroups of $G$, call them $N_1, \dotsc, N_k$.  Now consider the canonical homomorphism $\pi : G \to G / N_1 \times \dotsb \times N_k$.  Call $G / N_1 \times \dotsb \times N_k = Q$.  I have shown in a previous exercise that $H$ is a normal subgroup of $G$ if and only if $\pi(H)$ is a normal subgroup of $Q$.  So, say there is some normal subgroup $N$ of $Q$.  By that exercise, this implies $\pi^{-1}(N)$ is a normal subgroup of $G$, and thus $\pi^{-1}(N) = N_i$ for some $i$.  Now $N = \pi(N_i)$. But isn't $\pi(N_i)$ a trivial subgroup of $Q$?  Am I done at this point?

If $G$ is already simple then any isomorphism will do, so assume $G$ is not simple. Because $G$ is finite, there must exist a maximal nontrivial proper normal subgroup $N$, meaning $N \neq 1$ and whenever 
$N\leq H \leq G$ with $H$ normal in $G$ then  $H=G$ or $H=N$. Now show that $G/N$ must be simple.
The problem with your argument is that there is nothing stopping $Q$ being trivial.
