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?


It seems to me that there must be a normal subgroup of order $9$.

The number of $3$-Sylow subgroups of $H$ must divide $2$ but also be $1$ mod $3$. Thus there is exactly one $3$-Sylow subgroup $K$ of $H$. Then $K$ must also be a $3$-Sylow subgroup of $G$.

Since $H$ has index $2$ in $G$, it must be normal, hence invariant under conjugation. Therefore all conjugates of $K$ in $G$ are contained in $H$. But since $K$ is the unique $3$-Sylow in $H$, these conjugates must in fact be equal to $K$. Therefore $K$ is normal in $G$, and $|K|=9$.


If $H$ is order $18$, then $H$ is normal because it is index $2$, and just by the lack of flexibility for cosets, when $g\not\in H$, $Hg=gH$.

Groups of order $18$ are easy to classify. The Sylow theorems tell you that $H$ has only one Sylow-3 subgroup, and it is of order $9$. Since there is only $1$, it is normal in $H$. But it's also normal in $G$, because under conjugation, where is it going to go? $H$ is preserved under general conjugation by an element from $G$, so this group of order $9$ we have identified has nowhere to go under conjugation by $g$ but to an order-9 subgroup of $H$...of which there is only one.

So it seems to me you have a lock on the normal subgroup of order $9$ (given that $H$ is order $18$), and don't need to worry about order $4$.


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