If a group doesn't have subgroups of index 2 and 3, then any subgroup of index 4 is normal. Let $G$ be this group and $H$ be any subgroup of index 4. 
$G$ acts on the set of left cosets of $H$ in $G$, which is a homomorphism $\varphi: G\to Aut(G/H) = S_4$. It is easy to see that $\ker \varphi\subset H$. We obtain that $$|G/\ker\varphi| = [G:\ker\varphi] = [G:H][H:\ker\varphi] = 4[H:\ker\varphi].$$
But $G/\ker\varphi \hookrightarrow S_4$, thus $|G/\ker\varphi|$ divides $4! = 24$. Therefore $G/\ker \varphi$ can have order $4,8, 12, 24$. 
$G/\ker \varphi$ cannot have order $8$, since by Sylow first theorem we have a normal subgroup of order 4, which is index 2. This subgroup lifted to $G$ will be a subgroup of index 2. $G/\ker \varphi$ cannot have order 12, since the Sylow subgroup of 2 is index 3 and can be lifted to a subgroup of index 3 in $G$. $G/\ker \varphi$ cannot have order 24 for similar reason, or one can argument $S_4$ has $A_4$ sitting in it. 
Therefore $G/\ker \varphi$ has order 4, meaning $H = \ker \varphi\lhd G$. Please correct me if I am wrong.

This, of course, never happens: suppose $\;H\le G\;,\;\;[G:H]=4\;$ . As you proved, we have that $\;H\lhd G\;$, but then no matter what of the two groups of order four is $\;G/H\;$ , it will have a subgroup $\;K/H\;$ of order two, with $\;H\le K\le G\;$ ,  and by the correspondence theorem we then have that
$$[G/H:K:H]=2\iff [G:K]=2$$
and we thus have a subgroup of index two...
