Abstract algebra question (math olympiad)

[AdrianZ]

Let G be a non-cyclic group of order pⁿ where p is a prime number. Prove that G has at least p+3 subgroups.

Could anyone offer a solution to this problem?

 

[micromass]

How many cyclic subgroups does such a group have?? (i.e. generated by one element).

 

[Deveno]

have you tried induction, using n = 2 as your base case:

G = Z_p x Z_p

subgroups:

G
Z_p x {0}
{0} x Z_p
<(1,k)>, for p-1 choices of k (k ≠ 0)
{(0,0)}

total: 1+1+1+(p-1)+1 = p+3

for the induction step, you have two cases to prove:

G has a non-cyclic subgroup of order p^k, k > 1 (easy case)
G has only cyclic subgroups (hard case)

for the second case (i haven't worked out all the details), i think you can use the fact that G has a non-trivial center, to reduce it to the case that G is abelian.

now if G is abelian and has only one subgroup of order p^k, for every k < n, then G would be cyclic, and by supposition G is non-cyclic (appeal to sylow for existence of a subgroup of order p^k)

so we can assume that for some 1< k < n, G has two distinct subgroups of order p^k. but then we have two distinct subgroups of order p (by Cauchy), say H and K, and since G is abelian HK is a subgroup of G.

then: since G is supposed to be non-cyclic with only cyclic subgroups, HK must be cyclic, contradicting the fact that it has 2 distinct subgroups (H and K) of order p.

so G must not be abelian, and that should take care of it (unless I've done something stupid...could be).

 

[morphism]

Here's a quick proof. Let \Phi(G) be the Frattini subgroup of G (= intersection of maximal subgps of G). Then G/\Phi(G) = (\mathbb Z/p\mathbb Z)^k for some k\geq 1 and moreover k=1 iff G is cyclic. Since G isn't cyclic, we see that G/\Phi(G) is an elementary abelian p-group of rank at least 2, hence has at least (p^2-1)/(p-1)=p+1 subgroups of index p. Consequently G has at least p+1 subgroups of index p. Now throw in the trivial subgroup and G itself to see that G has at least p+3 subgroups.

 

[Deveno]

Here's a quick proof. Let \Phi(G) be the Frattini subgroup of G (= intersection of maximal subgps of G). Then G/\Phi(G) = (\mathbb Z/p\mathbb Z)^k for some k\geq 1 and moreover k=1 iff G is cyclic. Since G isn't cyclic, we see that G/\Phi(G) is an elementary abelian p-group of rank at least 2, hence has at least (p^2-1)/(p-1)=p+1 subgroups of index p. Consequently G has at least p+1 subgroups of index p. Now throw in the trivial subgroup and G itself to see that G has at least p+3 subgroups.

by gosh, that's slick...let me ask you this, at what level do you think students get their first exposure to the Frattini subgroup?

 

[micromass]

Sigh... I was hoping that the OP would discover the answer by our hints...

The different proofs are all nice though.

 

[Deveno]

Sigh... I was hoping that the OP would discover the answer by our hints...

The different proofs are all nice though.

in a perfect world, which this is not, the OP will examine each reply and turn, and think deeply on them. if that happy event comes to pass, perhaps they will gain more than just "the answer".

 

[morphism]

by gosh, that's slick...let me ask you this, at what level do you think students get their first exposure to the Frattini subgroup?

I'd say graduate-level, if at all.

re: micromass: in principle, I generally agree with you. That said, I think this problem is somewhat unlike the others that get posted here, and I genuinely think that it hasn't been ruined. Like Deveno says, each of the first three posts above contains ideas (and mini-exercises) for the OP to ponder.

 

[AdrianZ]

I read the answers, I got the idea but didn't fully understand them. I haven't studied p-sylow groups in details yet, I guess I need to work on the class equation and sylow theorem and reply later.
morphism's answer using the Frattini group sounds interesting because it's short and quick, so I hope one day I would have the opportunity to study the Frattini group as well.

Thanks for the help guys.