Hi

**NonCommAlg**.

We have either

or else

for some

In either case,

has finite order, say

.

Indeed

must be prime. For suppose

for some

Then

for some

contrary to the minimality of

Now let

be any nontrivial element of

Then

for some

Hence

has finite order too.

Since

divides the order of

by Lagrange’s theorem, so the order of

is of the form

for some positive integers

with

coprime with

If

then

would be divisible by a prime

and then

would have a Sylow

-subgroup which cannot possibly contain

, an element of order

Hence

and so every nontrivial element

of

has order a power of

Note that if

is any finite subgroup of

of order

then, since

we must have

Hence

is the unique finite subgroup of

of order

So let

have order

Assume that

Then

has a totally ordered chain of subgroups

containing every one of its subgroups. (Indeed this is a composition series in Jordan–Hölder theory.) (Note that

Of course if

then

and so all subgroups of

(namely

and {1}) are comparable as well. Hence every finite cylic subgroup of

has subgroups that are all comparable.

The next thing is to show that every finite subgroup

of

is cyclic. It would then follow by the above that all subgroups of any finite subgroup of

are comparable. Since

is Abelian, so is

If

is not cyclic, then, by the theory of finite Abelian groups,

would be the internal direct product of cyclic subgroups

This would imply that

if

meaning that

can’t be in all the

’s. As this is not what we want, we must take it that

is cyclic.

(NB: For more on direct products and the classification of finite Abelian groups, see Chapters 13 and 14 of John F. Humphreys,

*A Course in Group Theory*, Oxford University Press, 1996, which contains an excellent and readable account of the subject.

)

Now we are ready to prove the main result. Let

be subgroups of

and suppose that

i.e.

with

Let

As

is Abelian, it is easily checked that

is a subgroup of

containing both

and

Since it is also finite, all its subgroups are comparable by the results established above; in particular,

and

are comparable. Since

we must have

Thus

is in

for any

Hence

proving that

and

are comparable.