Is there a nonabelian group of order 28 whose 2-Sylow subgroup is isomorphic to $\mathbb{Z}/4\mathbb{Z}$?
My reasoning is that by Sylow's Theorem, there is a 2-Sylow subgroup of order $4$. Since it is of order $p^2=2^2$, then this subgroup is abelian. By the fundamental theorem for finite abelian groups, this group is isomorphic to either $\mathbb{Z}/4\mathbb{Z}$ or $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$.
Is this reasoning enough?
Best Answer
There are exactly two nonabelian groups of order $28$, namely the dihedral group $D_{14}$ and another group, namely the so-called dicyclic group of order $28$, which is a semidirect product of $C_7$ and $C_4$.
Now the Sylow $2$-subgroups of $D_{14}$ are not cyclic. They are isomorphic to $C_2\times C_2$. What about the other group?
Reference: These notes, which contain the solution to the (popular) question on the Sylow $2$-subgroups of a group of order $28$.