Normal subgroups of a group of order 12 Let $G$ be a group of order 12 and suppose $G$ does not have normal subgroups of order $4$. How to show that $G$ has a subgroup of order $6$?
 A: By Sylow's Theorem, the number of Sylow 2-subgroups $n_2 = |G:N_G(P)| \equiv 1 \pmod 2$ where $P$ is any Sylow 2-subgroup of G and NG denotes the normalizer. Since $P$ cannot be normal in $G$, $|G:N_G(P)|\neq 1$ and must be $3$ (the only other odd divisor of $12$). Assuming the Sylow 3-subgroup is not normal either, results to a number of $4$ subgroups of order $4$ by the same argument. This would add up to one neutral element, 8 elements of order $3$, $3$ elements of one Sylow 2-subgroup and at least another element for another Sylow 2-subgroup,totaling at least 13 group elements of $G$.
Hence $G$ has a normal subgroup $N$ of order $3$. $G$ also has a subgroup $H$ of order $2$ contributed to Cauchy (also part of Sylow's Theorem). Now since $N$ is normal in $G$ and $N\cap H = 1$, we have a semidirect product $NH \le G$ of order $6$.
Added after Steve D's comment:
Let $N\triangleleft G$ with $|N|=3$ and $N = \langle\sigma\rangle$. Let $H\le G$ be a subgroup of order $4$. Assume first that $H$ is cyclic and $\tau\in H$ of order $4$. Then $\sigma^\tau \in \{\sigma, \sigma^2\}$, hence $\sigma^{\tau^2}= \sigma$ for the involution $\tau^2$. If $H$ on the other hand is isomorphic to Klein's four-group, and two element $\tau,\tau'\in H$ of order $2$ both map $\sigma^\tau = \sigma^{\tau'} = \sigma^2$ its product, the third involution $\tau\tau'$ maps $\sigma^{\tau\tau'} = \sigma$.
In both cases there is an element $\rho \in H\le G$ of order $2$ that commutes with $\sigma$, hence the semidirect product is direct and cyclic of order $6$.
A: An alternative way to finish Peter Patzt's argument is as follows: Once you know that $G$ has a normal subgroup $N$ of order $3$, note that $G/N$ is of order $4$ and therefore abelian. Since normal subgroups of the quotient correspond to normal subgroups of $G$ that contain $N$, and every group of order $4$ contains a normal subgroup of order $2$, this normal subgroup of $G/N$ of order $2$ lifts to a normal subgroup $K$ of $G$ that contains $N$ with $[K:N]=2$. Since $|N|=3$, that means that $K=6$, so you get a normal subgroup of order $6$.
