# Construction of finite $p$-groups with derived subgroup of order $p$?

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?

• 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 $p$. Jul 18 '20 at 7:55
• @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. 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 $$P$$).

• 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. Jul 18 '20 at 18:24
• 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.) Jul 18 '20 at 19:42
• 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. Jul 18 '20 at 21:02
• Good point. You are right. Jul 18 '20 at 21:30