I have been asked to say how many non-abelian groups of order $28$ are there so that there is at least an element of order 4.
Using Sylow's Theorem, it must be that $n_2 = 1 (\mod \ 2)$ and $n_2 | 7$, and therefore there are two options for $n_2$: $n_2 \in \{1,7\}$. For a group of order $28$, a Sylow $2$-subgroup has order $4$. Therefore, the question is up to finding the value of $n_2$ and how many of these Sylow $2$-subgroups have elements of order $4$.
The useful fact is that if a group has order $4$, then or it is cyclic or it is isomorphic to the Klein group. If a Sylow $2$-subgroup is isomporphic to the Klein group, then there is no element of order $4$. Therefore, the only thing to see is how many of those Sylow $2$-subgroups are cyclic.
Any hints? By the way, using again Sylow it is easy to obtain that $n_7 = 1$, and therefore the Sylow $7$-subgroup is normal and isomorphic to $\mathbb{Z}_7$.