Prove a group of order $28$ with a normal subgroup of order $4$ is abelian without Sylow Theorems I have made some headway with the proof, but I can't quite finish it off. Please could I have some help? Please note that at no point are Sylow Theorems to be used during this proof.
Let $G$ be a group such that $|G| =28$. We are given $H$, such that $|H|=4$, and $H$ is a normal subgroup of $G$. Previously, I have proven that $G$ must also contain a normal subgroup, $K$, where $|K|=7$. This was done without Sylow. I noticed that $H$ must be isomorphic to $C_4$ or $C_2\times C_2$ because these are the only groups of order 4, up to isomorphism. Since I'm trying to show $G$ is abelian, I guessed that $G$ will be $C_{28}$ or $C_2 \times C_{14}$. 
To try and show this, I started using the Direct Product Theorem. In either case of the identity of $H$, $H \cap K = e$, because $H$ will not contain any elements of order 7, and all the elements of $K$ are order 7 apart from the identity. Also, $H$ and $K$ are normal, so their elements commute with each other: For $h \in H$ and $k \in K$, $(khk^{-1})h^{-1} \in H$ and $k(hk^{-1}h^{-1}) \in K$ means $khk^{-1}h^{-1} = e$. 
But I cannot work out how to show any element of $G$ is the product of elements in $H$ and $K$. Should I perhaps consider the order of elements in $G$? When $H$ is $C_4$, $G$ will be $C_{28}$ and so must contain an element of order 28. 
Any help is very much appreciated. 
 A: By Cauchy there is an element of order $7$, and hence a subgroup $K\cong C_7$. Since $H$ is normal, $HK$ is a subgroup of $G$ order $|HK|=\frac{|H||K|}{|H\cap K|}=\frac{4\cdot 7}{1}=28$, so $HK=G$. Therefore, $G\cong K\ltimes H$. But $K$ acts trivially on $H$, regardless of the structure of $H$ and even less of that of $\operatorname{Aut}(H)$ (see e.g. here, in particular the last paragraph). So, $G$ is isomorphic to the direct product of abelian groups, and hence it is abelian in turn.
A: $G$ is the semidirect product of $H$ and $K$, now, up to isomorphisms there are two groups of order 4,$Z/4Z$ and $Z/2Z\times Z/2Z$,the group of Automorphisms of $Z/4Z$ is isomorphic to $Z/2Z$,and the automorphism group of $Z/2Z\times Z/2Z$ is isomorphic to $S_{3}$, in both cases the only homomorphism from $K=Z/7Z$ to $Aut(H) $ is the trivial homomorphism, So in fact $G$ is the direct product of $H$ and $K$, since both are abelian, we deduce that $G$ is Abelian.
Of course semidirect product can be avoided, since u proved that both subgroups are normal, indeed $G=HK$ by the formula for product, and the intersection is trivial. Let $x\in H, y\in K$, then $xyx^{-1}y^{-1}\in H \cap K$ because $xyx^{-1} \in K$ since K is normal, so $xyx^{-1}y^{-1}\in K$ the same reasoning shows that it belongs also to $H$, since the intersection is trivial $xyx^{-1}y^{-1}=1$,i.e $xy=yx$, this shows that $G$ is isomorphic to the direct product of $H$ and $K$, hence it is Abelian.
A: If $o$ denotes order, $o(HK)=(o(H)o(K))/o(H\cap K) $.
Here, $o(HK)=7*4=28$ ,so every element $g$ of the group $G$ is in $HK$. So it can be written as $g=hk$.
