Group of order 28 with normal subgroup of order 4 is abelian Herstein ch2.11 q19

Prove that if $G$ of order 28 has normal subgroup of order 4, then $G$ is abelian.

My attempt: The 7-sylow subgroup lies in center. So $\circ(Z)=7, 14$ or $28$. 
For $\circ(Z)=14$, $G/Z$ is cyclic. But this argument fails for $\circ(Z)=7$.
I have not utilized the fact $G$ has normal subgroup of order 4.
Please give a hint. Please do not give solution. Thanks!
(This looks problematic. Also in one of the comments, meaning of $[\operatorname{Aut} H :1]$ is unclear)
Edit: Thanks to @DietrichBurde 's comment, this answers this question. So my post is a duplicate.
 A: Let $K$ and $N$ be normal subgroups of $G$ with $N\cap K=\{e\}$
I shall prove $nk=kn, \forall n\in N,k\in K$
Proof: Let $k\in K$ and $n\in N$
$nkn^- \in K$ since $K\triangleleft G$
and  $kn^-k^- \in N$ since $N \triangleleft G$.
Hence $(nkn^-)k^-=n(kn^-k^-) \in N\cap K =\langle e \rangle$
This gives $nk=kn,\forall n\in N, k\in K$
QED
If $H$ and $K$ be the normal subgroup of order $7$ and $4$ respectively, then $G=HK$ is easy to see.
So every element of $G$ is of the form $hk$ for some $h\in H, k\in K$
Let $g_1=h_1k_1$ and $g_2=h_2k_2$
Then $g_1g_2=h_1k_1h_2k_2=h_2k_2h_1k_1$
=$g_2g_1$
Note that I have used here the fact that groups of order $4$ and $7$ are abelian to commute between  $h_1,h_2$ and $k_1,k_2$ and the above theorem to commute $h_i,k_j,i,j=1,2$
A: Let $H$  be the subgroup of order 4. It is well known that $H$ must be abelian.
Now, the map that send $g \in G$ to the conjugation by $g$ define an omomorphism from $G$ to $Aut H$.
What can you say about the image and the kernel of this omomorphism?
Can you see how to conclude?
