Disproving the converse of Lagrange's theorem In this page of wikipedia there is a disproving of the converse of Lagrange's theorem. I would like to see a more simple (or short) disproving of Lagrange's theorem.
 A: Hint: I agree that the Wikipedia proof that $A_4$ has no subgroup of order $6$ is a bit convoluted. An alternative proof is to note that there are only two groups of order 6: the cyclic group $C_6$ and the dihedral group $D_3$. $A_4$ has no subgroup isomorphic to $C_6$ because it contains no element of order $6$. It has no subgroup isomorphic to $D_3$ because it has no elements $\sigma$ and $\tau$ such that $\sigma^3 = \sigma$, $\tau^2 = \tau$ and $\tau\sigma\tau = \sigma^{-1}$. The calculations for the last part aren't too tedious, because you can assume w.l.o.g, that $\sigma = (1\,2\,3)$ and there are only three values of $\tau$ to try.
A: I prefer the following proof:
Let $G$ be a finite group, $N \subset G$ a normal subgroup and $g \in G$. Then we obviously have that $\text{ord}(\overline{g})$ is a divisor of $\text{ord}(g)$, where $\overline{g}$ is the image of $g$ under the quotient map $G \rightarrow G/N$. 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 no group having more than $50$% elements of odd order can have a subgroup of index $2$. Since $A_4$ has $9$ elements of odd order it cannot have a subgroup of order $6$.
