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 $G\overset{\phi}{\longrightarrow}H\overset{\psi}{\longrightarrow}K$ are homomorphisms then $G\overset{\phi\psi}{\longrightarrow}K$ is a homomorphism. If the theorem is false, then there is a finite group for which the image of no homomorphism is simple.  So for every chain of homomorphisms $G=H_0\overset{\phi_0}{\longrightarrow}H_1\overset{\phi_1}{\longrightarrow}H_2\cdots \overset{\phi_n}{\longrightarrow}H_n$, where each $H_k=\phi_k(H_{k-1})$, we can cut $\phi:=\prod_{i=0}^n\phi_i$ into $\varphi_j:=\prod_{i=0}^j\phi_i$ and $\psi_j:=\prod_{i=j}^n\phi_i$.  Assuming the $H$s are finite, we know that if $\left| H_k/\operatorname{ker}(\phi_{k}) \right|=\left| H_k \right|$, $\phi_{k}$ is an isomorphism.  If we reach an $n\in\mathbb{N}$ for which the only homomorphisms available (into any group) for $\phi_k$ are isomorphisms and the trivial homomorphism, then by definition, $H_k$ is simple, so $\varphi_j$ is a homomorphism from $G$ to a simple image.  On the other hand, $|G|$ is finite, so since $\left| H_k \right|$ is divisible by $\left| H_k/\operatorname{ker}(\phi_{k}) \right|$, so we must eventually end up with a simple group or the trivial group.  
