$H$ is finite abelian group $\implies$ $H$ has no element of order $4$ iff every subgroup of $Q_8\times H$ is normal

Question

Let $H$ be a finite abelian group. The claim is that

1. If every subgroup of $Q_8\times H$ is normal, then $H$ has no element of order 4.
2. If $H$ has no element of order $4$, then every subgroup of $Q_8\times H$ is normal.

To prove (1) I could take an element of order 4 in $H$ and show that there is a non-normal subgroup of $Q_8\times H$. I found this question which I believe solves (1).

To prove (2), I could take a non-normal subgroup of $Q_8\times H$ and show that there is an element of order 4 in $H$.

I have no idea/intuition of what is going on with this question. I'm guessing that it is a special case of some kind of advanced group theory, but this is an introductory textbook so I'm looking for elementary solutions. Any hints or suggestions?

Answer 1

Intuition: The only thing that can stop a subgroup of $Q_8\times H$ from being normal is if some element $(\ell, h)$ is in the subgroup, but $(-\ell, h)$ isn't, where $\ell$ is one of $\pm i$, $\pm j$ or $\pm k$ and $h\in H$. That's because $\pm \ell$ are all the conjugates of $\ell$, and $h$ is the only conjugate of $h$. And a subgroup is normal iff it contains all the conjugates of all its elements.

So if there is an element $h\in H$ of order $4$, then e.g. $\langle(i,h)\rangle$ is not normal, as $$(j,1)(i,h)(j,1)^{-1}=(-i,h)$$ is not in that group.

On the other hand, if no element of $H$ is of order $4$, then for any subgroup, and any $(\ell, h)$ in that subgroup, there is some power of $h$ of the form $h^{3+4n}$ which is equal to $h$. Thus $$(\ell, h)^{3+4n}=(-\ell,h)$$ is also in that subgroup. Making the subgroup normal.