A_n is the only normal subgroup of S_n

Lemma Suppose $1 \not = N \lhd G$ has trivial intersection with $[G,G]$, then it lies in the center.
proof: Let $n \in N$ then $$n^g [g,n] = g^{-1} n g [g,n] = g^{-1} g n = n$$ and we know $n^g \in N$ so $[g,n] \in N$ so it equals $1$ so $n$ commutes with $g$.

Lemma $S_n$ for $n \ge 3$ has trivial center.
proof: If $z$ lies in the center then $zg = gz$ for all $\pi$. We show that $g^z = g$ for all $g$ implies $z=1$: Take any three symbols from the group $a,b,c$ then consider:

• $(a\;b)^z = (az\;bz)$ so $az=a,bz=b$ or $az=b,bz=a$.
• $(a\;b\;c)^z = (az\;bz\;cz)$ so (using the previous) $cz=c$.

Corollary For $n \ge 5$ the only nontrivial normal subgroup of $S_n$ is $A_n$ (since it must meet the simple group $A_n$).

Note: $S_4$ has just one normal Klein-4 subgroup (even though it has other non-normal Klein-4 subgroups).

• http://math.stackexchange.com/questions/159994/proving-that-a-n-is-the-only-proper-nontrivial-normal-subgroup-of-s-n-n-ge 
• http://groupprops.subwiki.org/wiki/S4