Let $G$ be a finite $p$-group. Since we can embed $Z_2(G)/Z(G)$ in $Hom(G,Z(G))$, we have $d_2 \leq d(G)d(Z(G))$; where $d_2(G)=d(Z_2(G)/Z(G))$ and $d(G)$ denotes the minimal number of generators of $G$. The question is, does the equality $d_2 = d(G)d(Z(G))$ imply that $Z(G)$ is cyclic?
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$\begingroup$ I think that the answer is affirmative. If you have not an answer, can you suggest a way to tackle this question? $\endgroup$– Yassine GuerboussaCommented Mar 2, 2013 at 18:11
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$\begingroup$ But why do you think the answer is yes? $\endgroup$– Derek HoltCommented Mar 2, 2013 at 22:00
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$\begingroup$ For a cyclic group, isn't $d_2(G)$ zero? $\endgroup$– Will SawinCommented Mar 2, 2013 at 23:23
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$\begingroup$ Ok, Prof. Derek Holt; The order of G is greater than the products of the p^{d_i], where d_i is the number of generators of the ith term of the upper central series. Still n is greater than the sum of d_1 and d_2 and the others replaced by 1 ( the last replaced by 2, since the last factor can not be cyclic). Now under our condition, it follows easily that $d_1(d(G)+1) \leq n-c+1$( this inequality is due th A. Abdollahi). Now if G is a counter example, then the coclass of G is at least 5. Also, the class of G is greater than 2(otherwise $d_2 \leq d(G)$). therefore the order of G is at least p^8 $\endgroup$– Yassine GuerboussaCommented Mar 3, 2013 at 7:32
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$\begingroup$ It is not hard to see that our counter example can not be powerful nor p-central. and with some work it can not be a direct product of two p-groups. $\endgroup$– Yassine GuerboussaCommented Mar 3, 2013 at 7:35
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1 Answer
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OK, here is an example of a group $Q$ of class 3 and order $p^{17}$, which will work for $p \ge 5$. We have $d(Q)=5$, $d(Z(Q)) = 2$, $d(Z_2(Q))/Z(Q)) = 10$, with $Z(Q) = \langle Q.16, Q.17 \rangle$ and $Z_2(Q) = \langle Q.6,\ldots,Q.17 \rangle$. All generators have order $p$ - in fact $Q$ has exponent $p$. All pairs of generators commute except for those in the list below. (This is Magma output.)
Q.2^Q.1 = Q.2 * Q.6,
Q.3^Q.1 = Q.3 * Q.7,
Q.3^Q.2 = Q.3 * Q.8,
Q.4^Q.1 = Q.4 * Q.9,
Q.4^Q.2 = Q.4 * Q.10,
Q.4^Q.3 = Q.4 * Q.11,
Q.5^Q.1 = Q.5 * Q.12,
Q.5^Q.2 = Q.5 * Q.13,
Q.5^Q.3 = Q.5 * Q.14,
Q.5^Q.4 = Q.5 * Q.15,
Q.6^Q.1 = Q.6 * Q.16,
Q.7^Q.1 = Q.7 * Q.17,
Q.8^Q.2 = Q.8 * Q.16,
Q.9^Q.2 = Q.9 * Q.17,
Q.10^Q.1 = Q.10 * Q.17,
Q.10^Q.3 = Q.10 * Q.16,
Q.11^Q.2 = Q.11 * Q.16,
Q.11^Q.3 = Q.11 * Q.17,
Q.12^Q.4 = Q.12 * Q.16,
Q.13^Q.4 = Q.13 * Q.17,
Q.14^Q.5 = Q.14 * Q.16,
Q.15^Q.1 = Q.15 * Q.16,
Q.15^Q.2 = Q.15 * Q.17,
Q.15^Q.5 = Q.15 * Q.17.
The conditions that you listed on such an example just mean that examples are moderately large, and so are more difficult to construct. They do not provide any genuine evidence that there are no such examples. There was a conjecture about $p$-groups called the class-breadth conjecture that was open for a long time, but as soon as it became possible to use computers to study larger groups, it became relatively easy to find counterexamples.
answered Mar 3, 2013 at 11:31
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$\begingroup$ @Derek: Is it possible to have a 2-group of class at least 4 satisfying the required condition? Do you expect that such a 2-group exists? Only I would like to know your opinion? Thanks in advance. $\endgroup$ Commented Mar 3, 2013 at 15:43
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$\begingroup$ Wow, I have thinked about this problem for more than a year. I am really grateful Prof. Derek. unfortunately, I know almost any thing about Magma ( also Gap), I try to check your example without computer. $\endgroup$ Commented Mar 3, 2013 at 17:38
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$\begingroup$ Dear Prof. Abdollahi, Now I think that there is a great possibility to find a counter example to the noninner conjecture. And why not among those in small libraries of groups. I don't think that are all checked. $\endgroup$ Commented Mar 3, 2013 at 17:43
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$\begingroup$ @Yassine: Derek's groups are of exponent $p$ and so they do not provide any counterexample to the noninner conjecture: Every non-abelian finite $p$-group has a noninner automorphism of order $p$. Every possible counterexample to this conjecture satisfies the property mentioned by Yassine. $\endgroup$ Commented Mar 3, 2013 at 19:43
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$\begingroup$ @Yassine: Another point is that as you mentioned above a possible counterexample must have order at least $p^8$, in this case GAP cannot help as the small group library is limited to groups of order at most $p^6$. But for the case of 2-groups of orders 512 or 1024 I do not know as some experts have access to the classification of these groups. $\endgroup$ Commented Mar 3, 2013 at 19:47