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?

Instead of the (not entirely clear to me) idea you proposed, it is easiest to do this by induction on the size of the group. The base case is problematic, as the trivial group does not admit such a homomorphism. So you must first repair the question by adding "nontrivial" before "finite". Then the base case has been taken care of.
Assuming now $|G|>1$, if $G$ is itself a simple group the identity homomorphism $G\to G$ will do.
Otherwise ($G$ is neither trivial nor simple), $G$ has a nontrivial proper normal subgroup, say $N$, and $1<|G/N|<|G|$. Then the induction hypothesis applies to $G/N$, and gives a surjective homomorphism $f':G/N\to H$ to a simple group $H$. Now the composition $f=f'\circ\pi:G\to H$, where $\pi:G\to G/N$ is the canonical projection, is surjective, so it does the job.
