I have to solve the following exercise of group theory:
Let $G$ be a non-abelian group of order $36$ for which there exists $\varphi:G \rightarrow H$ a group morphism such that $|Im\varphi|=4$. The exercise ask to find all possible value of $n_2$, the number of the $2$-Sylow subgroups.
Since $|Im\varphi|=4$, in $G$ we have a normal subgroup of order $9$, the kernel $K$. For the Sylow's theorem the number of $2$-Sylow's subgroups $n_2$ must divide $9$ and it is equal to $1$ in $\mathbb{Z}_2$. So we have as a possibility for $n_2$ the following values: $1,3,9$. If $n_2=1$, there is a unique $N$ subgroup of order $4$ so it is normal. Since $K$ is normal and $K \cap N=\{e\}$, $KN$ is a subgroup of $G$ and must coincide with $G$. Moreover $K$ and $N$ ar abelian subgroup since their order is the square of a prime number, and since $K \cap N=\{e\}$, each element of $N$ commutes with each element of $K$. From this reasoning I can conclude that $G$ is an abelian group and this goes against the hypothesis. Thus $n_2 \neq 1$, is there an error in my reasoning?
Best Answer
There are $12$ groups of order $36$ having a normal subgroup of order $9$, namely the $4$ abelian groups, and $8$ non-abelian ones, e.g., a nontrivial extension of the dihedral group $D_9$ by $C_2$, $C_9\ltimes (C_2\times C_2)$, $(C_3\times C_3)\ltimes C_4$, $(C_3\times C_3)\ltimes (C_2\times C_2)$, and $D_{18}$.
Reference: Notes by Wodzicki.