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Does someone have an idea about how to prove the following claim?

If $G_1 $ , and $G_2 $ are two p-groups, of the same order, but $G_2 $ has bigger rank, then $G_2 $ has more normal subgroups than $G_1$ [ rank of a group = minimal number of generators ]

Thanks in advance

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This can't possibly be true, any simple group gives a counterexample. Are you missing an assumption, e.g. that they're p-groups? –  Noah Snyder Oct 16 '12 at 11:23
    
@NoahSnyder : Yes, you're right... Sorry... I indeed forgot to mention that I'm interested in p-groups Thanks! –  joshua Oct 16 '12 at 12:01
    
"Rank" has many meanings, so what does it mean here? –  Derek Holt Oct 16 '12 at 12:03
    
@DerekHolt: I think I mean here the usual meaning of "rank"- The minimal number of generators(i.e. - the size of a basis) –  joshua Oct 16 '12 at 12:07
    
For reference, although that meaning of rank is "standard" (e.g. it has a wikipedia page) I don't think it's widely known. I certainly didn't know about it. –  Noah Snyder Oct 16 '12 at 12:16

2 Answers 2

The claim is false.

For example: $Q_{16} \times C_2$ has 22 normal subgroups and rank 3, but there is a rank 2 group called SmallGroup(32,2) in GAP with 26 normal subgroups and presentation $$\langle a,b | a^4 = b^4 = [a,b]^2 = [[a,b],a] = [[a,b],b] = 1 \rangle$$

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So it won't be true to say that elementary abelian p- group of order $p^n$ have the maximal number of normal subgroups that a p-group of order $p^n$ can have? Thanks ! –  joshua Oct 16 '12 at 14:50
    
I think the elementary abelian group probably does have the most normal subgroups. Maybe the most subgroups period. It certainly has a lot. The cyclic group has the least number of subgroups. However, the middle ranks may not be strictly ordered. –  Jack Schmidt Oct 16 '12 at 15:11
    
Thanks ! But what might be a possible argument for this fact? Got an idea? –  joshua Oct 16 '12 at 15:47

I'm still somewhat skeptical that this claim is true, do you have a good reason for believing it? If it is true here's a sketch of how the argument probably begins.

For p-groups, by the Burnside basis theorem a set generates G if and only if it generates $G/\Phi(G)$ where $\Phi(G)$ is the Frattini subgroup generated by commutators and pth powers (so $G/\Phi(G)$ is the universal elementary abelianization). This means that the rank is k if and only if the index of the Frattini subgroup is $p^k$.

Since $G/\Phi(G)$ is abelian, any subgroup containing $\Phi(G)$ is normal. This gives a large number of normal subgroups of $G_2$. So the idea would be to get an effective bound on the number of normal subgroups in $\Phi(G_1)$. But I don't see how to do this.

If such a theorem is true it's almost certainly somewhere in Huppert's Endliche Gruppen.

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So it won't be true to say that elementary abelian p- group of order $p^n$ have the maximal number of normal subgroups that a p-group of order $p^n$ can have? Thanks ! –  joshua Oct 16 '12 at 14:51
    
@joshua:I haven't worked the whole argument out, but it sure looks to me like elementary abelian p-groups will have the maximal number of subgroups of a p-group of that size. –  Noah Snyder Oct 16 '12 at 15:12
    
Thanks ! Got any idea for any possible direction one can use in order to prove this fact? –  joshua Oct 16 '12 at 15:48
    
I'd ignore normality and just try to show that the total number of subgroups of the elementary p-group (which are all necessarily normal) is larger than the total number of subgroups of the other group (Jack suggested something similar). –  Noah Snyder Oct 16 '12 at 15:53
    
I've also tried this... But got no idea why it should be true about elementary abelian groups, but not for any of the "middle ranks " ... Thanks ! –  joshua Oct 16 '12 at 16:10

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