For some work in equivariant stable homotopy, I am trying to understand the family of finite $p$-groups $P$ with derived subgroup $P'$ of order $p$. There is a 1999 J. Algebra paper by Simon Blackburn (Groups of prime power order with derived subgroup of prime order) that gives a very detailed classification, but I would like to understand these a bit more conceptually, and don't really care about uniqueness of description.

So I am wondering if my group theory friends can tell me (perhaps with a reference) if it is correct that all such groups can be constructed as follows:

(a) Start with an extra special $p$ group $\widetilde V$, so it sits in a nonsplit short exact sequence $$ C_p \rightarrow \widetilde V \rightarrow V,$$ where $V$ is an elementary abelian group of even dimension, and $C_p = \widetilde V^{\prime}$.

(b) Then pullback via a surjective map $\pi: A \rightarrow V$, where $A$ is an abelian $p$ group, yielding a nonsplit short exact sequence $$ C_p \rightarrow \widetilde A \rightarrow A,$$ with $C_p = \widetilde A^{\prime}$.

(c) [See Derek Holt's example, and ensuing comments.] Note that $Z(\widetilde A) = C_p \times \ker \pi$. Let $\alpha: C_p \rightarrow \ker \pi$ be a homomorphism, and let $C < Z(\widetilde A)$ be its graph. Now pushout via an inclusion $C \hookrightarrow C_{p^k}$, yielding a group $P$.

Then $P$ is a $p$-group of the sort I am interested in: $P' = C_p$. Furthermore $Z(P) = C_{p^k} \times \ker \pi$, and $P/Z(P) = V$, which looks rather like the ingredients of Blackburn's classification.

So now my question again: does every finite $p$-group with derived subgroup of order $p$ arise in this way?

  • $\begingroup$ Yes I believe that this is correct, but I found (c) slightly unclear. I was unsure whether the subgroup $C_p$ of $\tilde{A}$ used to define the pushout with $C_p \to C_{p^k}$ was intended to be equal to the derived subgroup $\tilde{A}'$ (which becomes $P'$). I think it could be any central subgroup of $\tilde{A}$ order $p$. $\endgroup$
    – Derek Holt
    Jul 18 '20 at 7:55
  • $\begingroup$ @DerekHolt Yes, I meant for that cyclic subgroup to be the derived subgroup. I am guessing that pushing out by another central subgroup of order p could have been incorporated in the earlier step with a different A. $\endgroup$ Jul 18 '20 at 14:07

As I said in my comment, I am not completely sure whether I understand your construction in (c), but the following example is an interesting test case.

Start with an extraspecial group $\langle a,b,c \rangle$ of order $p^3$ and exponent $p$ (with $p$ odd), with $[a,b]=c$ and $c$ central of order $p$.

Now let $A = C_p \times C_{p^2}$ surjecting onto $V$, and let $\tilde A$ be the pullback as in (b). So now we still have $a^p=1$, have $b^p=d$ with $d$ central of order $p$ and $\langle d \rangle = \ker \pi$.

Finally take a pushout with $C_{p^2} = \langle e \rangle$, but using the subgroup $\langle cd \rangle$ of $\tilde A$, so $e^p=cd$.

Now $P = \langle a,b,c,d,e \rangle$ has order $p^5$ with $P' = \langle c \rangle$, and $Z(P)= \langle d,e \rangle$. So we do have $Z(P) = C_{p^2} \times \ker \pi$, but the element $c \in P'$ is not a $p$-th power in $Z(P)$ (although it is a $p$-th power in $P$).

  • $\begingroup$ Very sneaky. And perhaps I am seeing why Blackburn's classification scheme is quite involved. I'll modify step (c) to take this into account. $\endgroup$ Jul 18 '20 at 18:24
  • $\begingroup$ Now I think my original construction includes your example. Let $A = C_p \times C_{p^2} \times C_p$ map to $V$ by sending both the 2nd and 3rd factors to the second factor of $V$, and pull back $\widetilde V$. I think the resulting group is your $P$. (No third step needed.) $\endgroup$ Jul 18 '20 at 19:42
  • $\begingroup$ It's late at night, but I don't think this group is isomorphic to my $P$. In $P$, a generator of $P'$ is a $p$-th power, but in your group it is not. $\endgroup$
    – Derek Holt
    Jul 18 '20 at 21:02
  • $\begingroup$ Good point. You are right. $\endgroup$ Jul 18 '20 at 21:30

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