Converse of Lagranges theorem question:

Question

For a finite group $G$ and a subgroup $H,$ Lagrange's theorem says that $|G|=|G:H||H|,$ where $|G:H|$ is the number of cosets of $H$ in $G.$
My question is for any subgroup $H$ can we find another subgroup with order $|G:H|$?

Answer 1

Nope. The symmetric group on five letters $S_5$ (order $120$) has a subgroup $H$ of order $4$ so this has index $120/4=30$ but $S_5$ has no subgroups of order $30$.

Look under the column "Order of Subgroups". You will find $4$ but not $30$.

Answer 2

No. $A_4$, the group of even permutations on $4$ symbols, is a counterexample. It has order $12$, and there are elements of order $2$ (e.g. $(1,2)(3,4)$) which generate subgroups of order $2$, but no subgroup of order $6$ (see here).