# 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?

• What do you mean by $N_1\times\cdots\times N_k$? Probably $G$ will not contain a subgroup that is (isomorphic to) the direct product of these normal subgroups. – Marc van Leeuwen Sep 28 '13 at 4:04

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.

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.

• I'm OK if $Q$ is trivial. I want $Q$ to have no nontrivial normal subgroups. – tylerc0816 Sep 28 '13 at 2:24
• Clearly the question wants you to show that $f(G)$ is not trivial, otherwise there really is nothing to prove, because every group can be mapped into a trivial group. Check the definition of "simple group". – TBrendle Sep 28 '13 at 2:28

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.