TITLE: No normal subgroup of a subgroup of $S_n$ imply the subgroup is the one of even permutations or consists of two elements
QUESTION [3 upvotes]: The following is an old exam question from a n introduction to group
theory course:

Let $G$ be a proper subgroup of $S_{n}$, $n\geq3$.
Prove that if $G$ does not have any non-trivial normal subgroup (i.e
  $\neq\{e\},G$) then $|G|=2$ or $G\leq A_{n}$

I don't know where to start, can someone please hint me ?

REPLY [3 votes]: Consider the sign map $sgn :S_n\to \mathbb{Z}/2\mathbb{Z}$, taking a permutation to its sign. Then $A_n = \ker(sgn)$.
Restrict this homomorphism to $G$ to consider the $G\subset A_n$ case. 
Now if $sgn|_G \equiv 1$, then all nontrivial elements  $x,\:y\in G$ have odd sign except the identity. What happens when you consider $x^2$? $xy?$
Can you conclude $|G| =2$?
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