# Case when all subgroups of a group are normal

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:

1. How is the line in the above answer justified? Does it have anything to do with all subgroups being Sylow and normal?

2. 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)

I think the key word here is "cyclic".

Since the Sylow subgroups of the same order are conjugate to each other, if the Sylow subgroups of orders $3$ and $7$ are normal then there is a unique subgroup of order $3$ and order $7$. But these do not exhaust the elements of the group, so there must be elements with order other than $1, 3, 7$ - and the only possible order for those is $21$.

• That's neat and simple! Thank you! – Jesse P Francis Apr 20 '15 at 9:24

A finite group with all subgroups normal is the direct product of its (unique) Sylow subgroups. Hence it is abelian if and only if all Sylow subgroups are abelian.

In your case the Sylow Subgroups are necessarily abelian, since their order is $3$ and $7$ respectively. So a non-abelian group of order $21$ must have non-normal Sylow subgroups.

• Can I restate "A finite group with all subgroups normal is the direct product of its (unique) Sylow subgroups. Hence it is abelian if and only if all Sylow subgroups are abelian." as " A finite Dedekind group is abelian if and only if all Sylow subgroups are abelian"? – Jesse P Francis Apr 20 '15 at 9:17
• Also, regarding second part of your answer: I think you might have to rewrite it a bit differently: Say it has 7 3-Sylow subgroups, it still has only subgroups of order 3 and 7 alone, and being of prime order, won't they all be abelian? – Jesse P Francis Apr 20 '15 at 9:40

In fact, non-abelian Dedekindian groups can be completely classified.

They have the form $Q_8\times A\times B$ where $Q_8$ is the quaternion group of order $8$, $A$ is an elementary abelian $2$-group and $B$ is an abelian group of odd order (alternatively they have the form $Q_8\times C$ where $C$ is abelian with no element of order $4$).

A proof can be found in for example Berkovich's Groups of Prime Power Order I, Theorem 1.20.

So in the case at hand, it could be seen simply from the order of the group that it could not be non-abelian and Dedekindian.

• Thank you. I had noticed it in the Wikipedia Article about Dedekind groups! – Jesse P Francis Apr 20 '15 at 9:41