'Non-example' to the Lagrange theorem.

Question

Lagrange's theorem: Let G be a finite group and let H be a subgroup of G. Then, $|H| | |G|$ The converse does not hold in general.

'Non-example':

$G=A_{4}$ where $A_{4}$ is the alternating group of degree n

Then, |G|=12.

Conclude that G does not have a subgroup of order 6. Where did this conclusion follows from?

Answer 1

There are only two subgroups of order $6$.One is isomorphic to $Z_6$ and the other is isomorphic to $S_3$. So if $A_4$ were to have a subgroup of order $6$.it should be one of them. It can't be $Z_6$ because of the simple fact that $S_4$ itself does not contain an element of order $6$. Next if $A_4$ has a subgroup isomorphic to $S_3$ then that subgroup must contain $2$ elements of order $3$ and $3$ elements of order $2$ that DO NOT COMMUTE.but unfortunately or fortunately all the order $2$ elements that $A_4$ has they all commute with each other because they themselves form the Klein $4$ group.. So in all you can't have any subgroup of order $6$

Answer 2

Suppose such a subgroup exists and call it $H$. Then since it has index 2 it must be normal. If K were another subgroup, then since $H$ is normal, $HK$ is a subgroup. Find $K$ such that $H \cap K = \{1\}$. Then $|HK| = |H||K|$. Pick $K$ such that $|H||K|$ does not divide $|G|$ and this will contradict the LaGrange's theorem.

If $|H| = 6$. Then you can easily count that there are 8 3-cycles in $A_{4}$. So that H cannot contain all of them. Hence let $K$ be the cyclic subgroup generated by a three-cycle not contain in $H$. Then you have the contradiction you need because 24 would have to divide 12 according to Lagrange's theorem.