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.
Best Answer
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?