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Here is a theorem:

enter image description here

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

Mikasa
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  • Where is quasicommutativity used in the proof of the cited theorem? Does your question include also the mentioned case where $H_e$ has an element of ininite order? If not, please share it with us.. – Berci Feb 18 '13 at 12:32
  • Nice question! +1 – amWhy Feb 18 '13 at 14:05
  • By 'Hamiltonian', also noncommutativity is meant? Because in that form it is just not true: let $S$ be any finite Abelian group. – Berci Feb 18 '13 at 17:36
  • @Berci:See this http://dml.cz/dmlcz/101114. Yes noncommutativity is considered. I think, I am missing something clear and simple in this theorem, but cannot see that. :-( – Mikasa Feb 18 '13 at 18:10
  • Aha, you should have started with that link.. Also, wikipedia says a different definition for quasi-commutative semigroup (http://en.wikipedia.org/wiki/Commutative_semigroup), but probably this is the same as in the paper. I still don't get what your exact question is and what do you mean by 'go inside the theorem'? – Berci Feb 18 '13 at 18:24
  • @Berci: I focus to know more about this theorem, and so I am asking why that structure should be Hamiltonian. Honestly, my mind is distracted when I change my area from Group to semigroup. – Mikasa Feb 18 '13 at 18:30
  • ...but $H_e$ is already a group, isn't it? Isn't the semigroup $S_3$ a counterexample for this claim? Ah, no, because of quasicommutativity. But isn't it equivalent to saying: any (noncommutative) quasicommutative group with only finite orders is Hamiltonian? – Berci Feb 18 '13 at 18:34
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    Your question seems to really be asking for an explanation of the first remark following Theorem 2 in the document you linked to in the comment. I think it would be a lot clearer if you said that. – Tara B Feb 19 '13 at 16:00
  • @TaraB: Yest it was. By the way, Berci made me a bit sharp. However, he focused on quasicommutative ones below. Thank you for your consideration. – Mikasa Feb 20 '13 at 16:32

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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.

Berci
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