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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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    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.2012-10-16

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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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    Thanks ! But what might be a possible argument for this fact? Got an idea?2012-10-16
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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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    Good idea...I'll try it ... Thanks !2012-10-16