Here is a theorem:
I could go inside the theorem and know some few points of it. It's told that
If all elements of $H_e$ are of finite order so the group $H_e$ is Hamiltonian.
My question is how is it possible? In fact, if $H_e$ has an element of infinite order it will be ableian. I am asking for any hints for understanding what is happening inside the theorem. Thanks
I think, this has not too much to do with the cited theorem --as far as I see-- it only uses that $H_e$ is already a group.
Now, if we have a quasicommutative group $G$ and an arbitrary subgroup, say $U\le G$, then for any $g\in G,\ u\in U$, by quasicommutativity we have an $r\in\Bbb N$ such that $gu=u^rg$, so $gug^{-1}=u^r\in U$, proving that all subgroups are normal.
The given statement without more conditions is not true, because there exist finite Abelian groups. I'm also skeptic about your second statement... what about the group $\Bbb Z\times Q_8$?
Update: This latter one, $\Bbb Z\times Q_8$ is not quasicommutative, and direct product is not that nice among them.. By the basic results of Hamiltonian groups (which I was not aware before), the second statement is indeed true -- but for simply group theoretic reasons:
If there is an element of infinite order in a quasicommutative group, then it is indeed Abelian, see for example this for a proof (first, if $a,b$ doesn't commute then it shows that both must have finite order, then for any other element $x$, if $x$ happened to commute with both $a$ and $b$, then $ax$ is not going to commute with $b$, so $ax$, thus $x$ also has finite order).
So, it seems that, the highlighted statement would correctly conclude that '$H_e$ is either Hamiltonian or Abelian'. And then all is in its right place.
Of course, if $S$ is a finite Abelian group with unit $1$ (and multiplication), then the largest subgroup $H_1$ containing $1$ is of course $S$ itself, and $S$ is quasicommutative, and, since is a group, regular. And all its elements have finite order.