I was thinking in light of this question, where the groups G in discussion is of order 21 and question is to find number of non-trivial normal sub-groups of G.
In the answers it is mentioned that
The 3-Sylow subgroup is not normal, since otherwise G would be cyclic (and therefore abelian).
But on further search I came across "Dedekind Groups" and "Hamiltonian Groups", where non-abelian groups may also have all subgroups normal, confirmed by this question in SE.
Now my questions are:
How is the line in the above answer justified? Does it have anything to do with all subgroups being Sylow and normal?
In general, are there any special conditions under which a (finite) Dedekind group should be abelian? (Asked here, but it is not answered: the link to page in Google books is not accessible).
Edit: 3. While thinking, these are the fact that I felt may be used to justify the above, but I was not able to proceed further. Can the above line be justified using (any of) the below facts?
(i) G has a non trivial center since it is non-abelian,
(ii) G can accommodate 7 3-Sylow subgroups (if there are 7 3-Sylow subgroups, then 14(+1 identity) elements in all of them, and then there is 6(+1 for identity) elements left for the one 7-Sylow subgroup of G)
Thanks in advance.