Converse of Lagranges theorem on the group other than symmetric group $S_{n}$ and Alternating group $A_{n}$ I know converse of Lagranges theorem is not true in general.  I find example on $A_{n}$ and $S_{n}$.
My question is : can we find an example of a group other than $S_{n}$ and $A_{n}$ where converse of Lagranges theorem is not true?  
 A: Consider $G=H/Z(H)$ where $H=SL(2,\Bbb Z_7)$.This group is a simple group of order $168$. So this group has no subgroup of order $84$, even though $84$ divides $168$. 
A: One counterexample is given by $G = \text{SL}_2(\mathbb{F}_3)$ as $G$ has no subgroup $H \subset G$ of order $12$. That is because the quotient would have order $2$ and therefore is an abelian group such that we get a surjective homomorphism $G^{\text{ab}} \rightarrow H$. Since the order of $G^{\text{ab}}$ is $3$, that cannot be the case. Note that this depends very much on the cardinality of the field $\mathbb{F}_3$, as the group $\text{SL}_n(K)$ is perfect except for $n = 2$ and $K$ of cardinality $2$ or $3$.
Oops, did not see that you excluded $S_n$ as well. Just ignore this:
Another counterexample is the group $S_n$ for $n \geq 5$ as $A_n$ is simple then. Therefore $S_n$ cannot have a subgroup of order $\frac{n!}{4}$.
Let me state a way to find some counterexamples:
Let $G$ be a finite group, $N \subset G$ a normal subgroup and $g \in G$. Then $\text{ord}(\overline{g})$ is a divisor of $\text{ord}(g)$, where $\overline{g}$ is the image of $g$ under the quotient map $G \rightarrow G/N$. If we now assume that $[G:N] = 2$ and that $\text{ord}(g)$ has odd order, then we get $\text{ord}(\overline{g}) = 1$ by Lagrange's theorem, such that $g \in N$. We just showed that every element of odd order needs to be in $N$. Therefore, if more than half of the elements of a group $G$ have odd order, the group cannot have a subgroup of index $2$. That for example holds for $A_4$ since $A_4$ has $9$ elements of odd order. So this is one way of proving that $A_4$ has no subgroup of order $6$.
There is even a characterization for the case of index $2$ subgroups:
Proposition:
A group $G$ has no index $2$ subgroup iff $G = \langle \lbrace g^2 \mid g \in G\rbrace \rangle$.
