If $h$ divides $|G|$, not necessary that $G$ has a subgroup of order $h$

Question

if $h$ divides $\# G$, then it is not necessarily that $G$ has a subgroup of size $h$

which can be related to Lagrange Theorem $[G:H]=\frac{\#G}{\# H}$ in the case when the subgroup exists, however it is not generally true such as in the case of $\mathbb A_4$ with order of 12 and having no subgroup of 6. So

How can you demonstrate the proposition of the highlight in GAP?

Answer 1

If you want to check whether a particular group would be a counterexample to the (wrong) converse of Lagrange's theorem you probably cannot do better than to calculate the subgroups. In GAP, the command

ord:=Set(List(ConjugacyClassesSubgroups(G),x->Size(Representative(x))));
Difference(DivisorsInt(Size(G)),ord);

will return a list with the orders of subgroups of $G$ and in the second command determine the orders for which there are no subgroups.

Answer 2

Let $G =A_{4}$. Then 6 divides the order of G but G does not have a subgroup of order 6.

Proof: Note that a 2-cycle is always an odd permutation and any m-cycle can be written as a product of (m-1) 2-cycles and so an m-cycle is odd if and only if m is even, this tells us the shapes of cycles contained in $A_{4}: id, (ab)(cd), (abc)$. The number of p-cycles in $S_{n}$ is given by $(p-1)!\binom{n}{p}$ and in particular there are 8 cycles of length 3 in $A_{4}$.

Now suppose G has a subgroup H of order 6. Then since $[G: H] = \frac{12}{6} = 2$, H is normal in G. Since G has 8 elements of order 3 not all of them are contained in H so we pick $\sigma$ which is not in H. $K = \langle \sigma \rangle$ is a subgroup of G.

Since H is normal, HK is a subgroup of G and we have the important formula: $|HK| = \frac{|H||K|}{|H \cap K|}$ giving us that HK is a subgroup of G of order 18 but this contradicts the LaGrange's Theorem and so such an H cannot exist.