# sylow theory and simple groups

i have to prove group of order $36$ is not simple by using Sylow's theorems. please read whole solution as i have difficulties and i indicate them by " ? " now as $|G|= 36 = 3^2 . 2^2$ and $3^2 | 36$ then by Sylow I theorem $G$ must have Sylow $3$- subgroups of order $3^2 = 9$. now let number of Sylow $3$- subgroup be $n_3$ then according to Sylow 2nd and 3rd theorem $n_3 = 1 + 3k$ and $n_3 | O(G)$ this imply $1+3k | 36$ which is hold if $k= 0$ or $k =1$. if $k= 0$ then $n_3 = 1$ and hence we get unique Sylow $3$-subgroup of order " ? " and hence it is normal and hence $G$ can not be simple and hence we are done.

But if $k= 1$ then $n_3 = 4$ hence we get $4$ Sylow $3$- subgroups each of order $9$ and hence there are $8$ elements of order $9$, in each of these sylow-$3$-subgroups, is it true? hence there $4 . 8 = 32$ elrments of order $9$ is it true? and as $|G| = 36$ and hence remaining $4$ elements forms unique sylow $2$-subgroups of order $4$ is it true? i have confusion because number of sylow $2$- subgroup i.e. $n_2 = 1+ 2k$ must divides $|G|$ i.e $1+ 2k | 36$ it holds if $k = 0$ or $k= 1$ or $k=4$ and hence we get $n_2 = 1$ or $n_2 = 3$ or $n_2 = 9$ hence i have doubt .

• Why should the $3$-Sylow-subgroups be necessarily cyclic? – Sebastian Schoennenbeck Nov 10 '14 at 7:32

First of all $G$ is not simple, but not by the logic which you are applying. You are saying that in the Sylow $3$ subgroup tere will be $8$ elements of order $9$- that's not true. Consider for example $\mathbb{Z}_3\times \mathbb{Z}_3$, which is of order $9$ and it does not have $8$ elements of order $9$.
Instead, let $H$ be the Sylow-$3$ subgroup of $G$. Then there is a homomorphism of $G$ to $A(S)$ where $A(S)$ is the set of all permutations on $S$= set of all right cosets of $H$ in $G$ and kernel of this homomorphism is the largest normal subgroup of $G$ contained in $H$. Now in our case $|S|=4$ and hence $|A(S)|=24$. So if $\phi$ is the above mentionaed homomorphism then clearly $|ker(\phi)|>1$ , which is a non-trivial normala subgroup of $G$ contained in $H$. Hence $G$ is not simple.