Thread: True or false questions about groups

1. If every proper subgroup of a group G is cyclic, then G itself must be cyclic.

I know that if a group G is cyclic, then every subgroup of G is also cyclic. However does the converse hold? And how would you prove it?

2. A group with only a finite number of subgroups must be finite.

I think this statement is true, but I can't seem to gather a concrete proof for it.

Any help would be greatly appreciated!

Cheers

1. consider the group Q8 = {1,-1,i,-i,j,-j,k,-k} where i^2 = j^2 = k^2 = -1, ij = k, jk = i, and ki = j. what can you say about the proper subgroups of Q8?

2. this is a first-rate question with which to test your knowledge of groups. here is an idea: consider any non-identity element g1, you can always form <g1>.

now if <g1> was infinite, it would be isomorphic to Z, which has an infinite number of proper subgroups, which would then be an infinite number of proper subgroups of G.

so we may safely conclude that <g1> is finite. if G = <g1>, we are done. if not, then there is some g2 not in <g1>. consider <g2>.

now, prove that the number of subgroups we find in this way must be finite. why does this suffice to show that G is finite?

Thanks for the help Deveno, however I still have some queries. When you say that if <g1> = G, then we are done, is that because <g1> is isomorphic to where n is the order of <g1> and since has a finite number of subgroups, then so does G? But how can we actually prove that has a finite number of subgroups?

So if we consider a g2 not in <g1> then <g2> will be another DIFFERENT subgroup, by 'different' i mean <g1> and <g2> are NOT the same set.

We know that both <g1> and <g2> are finite sets. However I am still a bit lost on how to complete the proof.

First why don't we also consider elements IN <g1>? Since these elements could also produce a different subgroup, ie, consider the subgroup <2> of . If we consider the element 6 in <2>, then <6> is a different subgroup as well...
Second, I don't see how picking g1, g2, g3... and then forming cyclic groups out of these elements suffices to show G is finite.

Cheers

Zn is finite. any finite set has a finite number of subsets. for example, a set with n elements has at most 2^n subsets. not of all of these subsets will be subgroups, but a smaller number than some finite number is still finite.

we are trying to prove that G is finite. if G is isomorphic to a finite group (such as Zn) then it is finite. every element of G must be of finite order (since otherwise G has a subgroup isomorphic to Z, and thus G would have an infinite number of subgroups...the subgroups of the cyclic group <g1> of infinite order). so the only way G could fail to be finite, is if it had an infinite number of elements of finite order.

now <g1> may not be all of G (this is usually the case). but if it is not, then we can take g2 not in <g1> and form <g2>, which although it may have some overlap with <g1>, will still contain additional elements of G not in <g1>. if every element of G is in <g1> or <g2>, then G = <g1> U <g2>, which is a finite union of finite sets, and so must be finite.

but if not, then we have some g3 which is not in <g1> or <g2>. we can then form <g3>, which contains additional elements of G not in <g1> or <g2>. again, if G = <g1> U <g2> U <g3>, then G is finite, otherwise we find g4 not in <g1> U <g2> U <g3>, and form <g4>.

we thus arrive with a sequence g1,g2,g3,g4..... each of which generates a subgroup that contains elements that are not in the union of all the previously generated cyclic groups. can this sequence be infinite? no, because G only has a finite number of subgroups. so the process must terminate with some gn, whence G = <g1> U <g2> U....U <gn>, and being a finite union of finite sets, G must be finite.

let's see how this works with Z18 (which we know is finite). now if we picked g1 = 1, we would get all of Z18, and boom! we're done. but let's say we weren't so lucky. let's say our first choice was, say 6. so we form <6> = {0,6,12}. well that's not all of G, so let's pick something not in <6>, say 4. now we form <4> = {0,4,8,12,16,2,6,10,14}. but <6> U <4> still isn't all of G, we only have {0,2,4,6,8,10,12,14,16} so far. but 3 is not in this set, so we form <3> = {0,3,6,9,12,15}. this gets us 3 new members of G, so now we've accounted for:

{0,2,3,4,6,8,9,10,12,14,15,16}. 5 isn't in this set, so next we form <5> = {0,5,10,15,2,7,12,17,4,9,14,1,6,11,16,3,8,13}, finally we have accounted for all of Z18.

so we have two different finite sequences, here: 1, and 6,4,3,5. we're really not interested in the proof in what the internal structure of G really is, we just want to show that:

finite # of subgroups --> finite # of elements. so the only thing that is pertinent is that both sequences are FINITE.

Ahhh that was great, so we basically created a surjection from elements of G to cyclic subgroups, nice~

That was a very comprehensive explanation!! Thanks so much Deveno!

Originally Posted by usagi_killer

Ahhh that was great, so we basically created a surjection from elements of G to cyclic subgroups, nice~

That was a very comprehensive explanation!! Thanks so much Deveno!

No...I mean, there exist infinite simple groups, so not every group has a homomorphic image. What you are doing is looking at the cyclic subgroups of your group - so showing that these embed into your group. That is, taking injections from cyclic subgroups into your group...

right. it is important to realize that we are not "decomposing G" into cyclic subgroups, but rather "covering G" with finite cyclic subgroups. this is only possible if G has a finite number of subgroups to begin with.

does anyone else here see a parallel between this and the heine-borel theorem?

Originally Posted by Deveno

does anyone else here see a parallel between this and the heine-borel theorem?

I'm not sure...

So Heine-borel says, ( subset of ),

closed and bounded every open cover admits a finite subcover ( compact).

What we have here is,

finite every cover by subgroups admits a finite subcover.

Now, let have the discrete topology (open sets are just elements of your group, etc.). What we have shown is that is finite it is a compact topological space under the discrete topology. So basically you just need to get rid of the `subgroups' in the statement above - you need to prove that,

finite every open cover admits a finite subcover ( compact).

So, if every open cover admits a finite subcover then is finite - it is an easy exercise in playing with definitions to see that this is what we have done.

Clearly if is finite every open cover admits a finite subcover, and so we are done!

But I can't see where the `closed and bounded' would come in...

well, obviously compact groups is a larger class than "discrete" groups. but they seem to me to be an important subclass. furthermore, discrete groups are still closed and bounded in other topologies besides the discrete one, for example, finite cyclic groups can be seen as subgroups of the circle group , and so are compact under the usual topology on .

Originally Posted by Deveno

well, obviously compact groups is a larger class than "discrete" groups. but they seem to me to be an important subclass. furthermore, discrete groups are still closed and bounded in other topologies besides the discrete one, for example, finite cyclic groups can be seen as subgroups of the circle group , and so are compact under the usual topology on .

Yes - but you are wanting a class of groups which is compact if and only if it is finite!

Groups which look like a (set-theoretic) direct product of their generators might work - think abelian groups (but, more generally, semidirect products of cyclic groups, or ever Zappa-Szep products of cyclic groups...) - as these kinda look like Euclidean space...ish...