Warning: contains maths, algebra even.
This is probably one of the most satisfying proofs that I've ever come up with (or at least remember coming up with). This is partly due to the fact it is non trivial, and partly due to the amount of time it took to come up with it.
Anyway, here's how it goes:
Suppose there is an onto homomorphism \(\varphi\) from \(S_5\) to \(S_4\).
By the first isomorphism theorem, \(S_4 \cong \, ^{S_5}\!/_{\mathrm{ker} \varphi}\), so \(\left| \mathrm{ker} \varphi \right| = \frac{\left| S_5 \right|}{\left| S_4 \right|} = 5\).
As this is prime, \(\mathrm{ker} \varphi\) is cyclic.
Let \(x\) be an element that generates \(\mathrm{ker} \varphi\). \(x\) is an even permutation as \(x^5 = 1_{\mathrm{ker} \varphi}\), so \(\mathrm{sgn}(x)^5 = 1\) and \(\mathrm{sgn}(x) = 1\).
Therefore all elements of \(\mathrm{ker} \varphi\) are even, so \(\mathrm{ker} \varphi \subseteq A_5\), the alternating group.
But \(\mathrm{ker} \varphi \lhd S_5\), so \(\mathrm{ker} \varphi \lhd A_5\), which is a contradiction as \(A_5\) is simple.
So there does not exist such a homomorphism.
Sure, I assumed that \(A_5\) was simple, but we all know that, right?
