I am trying to show that every non-abelian group $G$ of order $6$ has a non-normal subgroup of order $2$ using Sylow theory.
First, Sylow's Theorem says the number of Sylow $2$-subgroups $n_2$ is either $1$ or $3$. Assume that $n_2=1$. Then $G$ has a normal subgroup $P$ of order $2$. By index considerations, any subgroup $N$ of order $3$ will be normal. We know $G=PN$, and does this somehow derive a contradiction? I'd like to contradict the non-abelianness of $G$ to deduce that $n_2=3$, and hence $G$ has $3$ non-normal Sylow $2$-subgroups.
Best Answer
Let the non-trivial element of the normal subgroup of order $2$ be $g$. Then all conjugates of $g$ must be $g$ (it cannot be identity as identity is only conjugate to itself). Hence $g$ commutes with everything. We know an element of order $3$ exists (by Cauchy), so call this $h$. Then $g$ and $h$ generate the group and commute, and it is now clear that $G$ must be $C_6$, which is abelian.