Let $H$ be a normal subgroup of a finite group $G$ such that $|H|=4$. Then $G$ contains a normal subgroup of order $2$ or a subgroup of index $3$.
My attempt: By Lagrange's theorem, $4\mid |G|$. In particular, $G$ has a subgroup of order $2$. If $G$ is Abelian, this subgroup is normal. So, assume that $G$ is not Abelian. Thus $|G|\geq 8$.
My idea was to embed $G$ into a larger group $K$ such that $G$ is a maximal subgroup of $K$ and $|K:G|=4$. Then $G$ would have a subgroup of index $3$ using: Let $H$ be a maximal subgroup of a finite group $G$ such that $|G:H|=4$. Then there exists $K\leq H$ such that $|H:K|=3$.
Suppose that $|G|=4n\ (n\geq 2)$. Then we need $|K|= 16n$. I tried using induction, but couldn't succeed.
Best Answer
The subgroup $H$ must be either cyclic, or isomorphic to the Klein $4$-group.
If $H$ is cyclic, then its subgroup of order $2$ must be normal in $G$, since $H$ Is mapped to itself, and this is the only subgroup of order $2$ of $H$. (The subgroup of order $2$ is characteristic in $H$, and characteristic subgroups of normal subgroups are normal.)
If $H$ is isomorphic to the Klein $4$-group, then the action of $G$ by conjugation on $H$ permutes the three subgroups of $H$ of order $2$, yielding a homomorphism $G\to S_3$.
If the image contains $A_3$, then the image has a subgroup of index $3$ which can be pulled back to $G$. If the image does not contain $A_3$, then you either get the trivial morphism (which means the action is trivial, so all three subgroups of order $2$ of $H$ are normal in $G$ and you are done); or the image is cyclic of order $2$, which means that the action of $G$ exchanges two of the subgroups of order $2$ and fixes the other one. The subgroup fixed by the action is normal in $G$.