Prove: If a group $G$, of order 36 has a subgroup of order 18 ,$H$, then $G$ either has a normal subgroup of order 9, or a normal subgroup of order 4.
This came about while reading the same article asked about in Group of order 36 stackexchange. What I just repeated is the first line in the proof.
$H$ must have a subgroup of order 9$,K$ characteristic in H. If #$Syl_3(G)$=4 is possible, then $H$ is not characteristic in $G$. Thus I can't really see how an argument that uses transitivity of characteristic subgroups would work without working out the structures of order 18.
There is probably a simple way that is a one liner because it was the first step in the proof and the person moves on without comment. Can I have a hint as to what the short way might be?
Some other things I deduced that might go in the right direction: Suppose there isn't a normal subgroup of order 9 and thus #$Syl_3(G)$=4. Then let $\pi$ be the permutation representation of G in $S_4$ through conjugation. By assumption $Ker \pi \neq G$, and also $n=|ker \pi|\mid36$ and $36/n=|ran \pi|\mid|S_4|=24$,so $|ker \pi|=$ is one of $18,12,9,6,3$. Maybe there is some way I can use this to show that there are not 3 or 9 sylow-4 subgroups?