A group of order 28 with a normal Sylow 2-subgroup is abelian How to prove that the group of order 28 which have normal subgroup of order 4 is Abelian group?


This was an answer to a merged question.

No, any group of order $28$ which contains a normal subgroup of order $4$ is necessarily abelian. 
The number of $7$-Sylow subgroups must be a divisor of $4$, and also congruent to $1$ modulo $7$, hence there is a unique $7$-Sylow subgroup which is necessarily normal. If $G$ contains a normal subgroup of order $4$, then the $2$-Sylow subgroup is normal. 
So since all the Sylow subgroups are normal, $G$ is isomorphic to the direct product of its Sylow subgroups,
$$
G\simeq P_7\times P_2
$$
where $P_7$ is the Sylow $7$-subgroup, and $P_2$ is the $2$-Sylow subgroup. Thus $G$ is abelian. 
