11
$\begingroup$

$\newcommand{\Z}{\mathbb Z}\newcommand{\F}{\mathbb F}$I would like to define a procedure which turns a finite $p$-group $G$ together with a field $K$ of characteristic $p$ (or even, in fact, any ring of characteristic $p$) into a group $G_K$ such that $G_K$ is obtained by "replacing all the $\Z/p\Z$-factors of $G$ by $K$". That is to say: unscrew the $p$-group until only cyclic factors are left, then base-change those cyclic factors, and screw them again following the original assembly process.

Ideally, I would like this procedure to satisfy the following properties:

  • When $K = \mathbb F_p$, I simply get $G_K = G$, and more generally for $K =\mathbb F_q$ with $q=p^d$ I get a group $G_K$ of order $|G|^d$
  • When $G = \mathbb Z/p\Z$, I get $G_K = K$ (see below for the abelian case)
  • This is functorial in $K$, so any embedding of fields $K \hookrightarrow L$ gives an embedding $G_K \subseteq G_L$; in particular, this would apply to automorphisms $\sigma$ of $K$ and then I would expect the fixed subgroup of $G_K$ under the group automorphism induced by $\sigma$ to be $G_{K^\sigma}$
  • For a fixed field $K$, the functor $G \to G_K$ is "exact" in the sense that for any exact sequence $1 \to A \to B \to C \to 1$ of $p$-groups I recover an exact sequence $1 \to A_K \to B_K \to C_K \to 1$ of $p$-groups

When $G$ is abelian (so a finite $\Z_p$-module), I know how to do this: I can simply consider $G \otimes_{\Z_p} W(K)$, or maybe I can write down an isomorphism $G \simeq \bigoplus \Z/p^{n_i}\Z$ and then have $G_K = \bigoplus W_{n_i}(K)$ (when $K$ is perfect these two options are equivalent).

Another (more general) case where I know how to deal with this is when $G$ has nilpotency class $<p$. In that case, the Lazard correspondence tells us that $G$ corresponds to some finite Lie $\Z_p$-algebra $\mathfrak g$, namely that $G$ can be recovered by equipping $\mathfrak g$ with the "Baker-Campbell-Hausdorff group law" $\circ$ such that $x \circ y = x + y + \frac12 [x,y] + \ldots$ where one truncates the Baker-Campbell-Hausdorff formula as soon as there are $p$-th commutators. But then one can bilinearly extend $\mathfrak g$ and its Lie bracket into a Lie $W(K)$-algebra, and equip that Lie algebra with its group law $\circ$ to obtain the group $G_K = (\mathfrak g \otimes_{\Z_p} W(K), \circ)$.

I do not know how this would go in more generality, but there are certainly instances where I know what I would expect, for example if $G$ is the group of unipotent upper-triangular matrices of size $n$ over $\F_p$, one would expect $G_K$ to be the group of unipotent upper-triangular matrices of size $n$ over $K$, etc. (typically for the group of $\mathbb F_p$-points of an unipotent algebraic group over $\mathbb F_p$) An approach that sounds promising, focusing just on the case of groups of nilpotency class $2$, given as $$ 1 \to Z \to G \to Q \to 1$$ with both $Z = Z(G)$ and $Q=G/Z(G)$ abelian. Then the extension $G$ is characterized by an element of $H^2(Q,Z)$, I know how I want to base-change both $Z$ and $Q$, so I would actually want a map $H^2(Q,Z) \to H^2(Q_K, Z_K)$. Shapiro's lemma comes close to giving me such a map, namely I get an isomorphism $H^2(Q,Z) \simeq H^2(Q_K, \mathrm{Coind}_Q^{Q_K} Z)$. However we do not have an isomorphism $\mathrm{Coind}_Q^{Q_K} Z \simeq Z_K$ unless $K$ is a finite field (in which case this does work!). This is all the more mysterious that I know the map $H^2(Q,Z) \to H^2(Q_K, Z_K)$ exists whenever $p>2$: take the corresponding extension $G$, turn it into a Lie algebra, base-change it, turn it back into a group, and take the corresponding element of $H^2$. It seems very mysterious to me that this kind of "Shapiro-like lemma" with the coinduced replaced by the tensor product would hold unless perhaps when $p=2$, and I do not understand the cohomological significance of the fact that it does hold when the Lazard correspondence applies.

Does anyone have any knowledge on previous work on this question? Do these constructions of "extensions of scalars of $p$-groups" have a common name? Are there obstructions, for certain finite $p$-groups, for an extension of scalars to exist?


EDIT: I'm mentioning another case I know how to deal with: assume that G has nilpotency class 2 and exponent p. Write $1 \to Z \to G \to Q \to 1$ with $Z=Z(G)$ and $Q=G/Z(G)$, and pick an $\F_p$-basis $b_1, \ldots, b_r$ of $Q$. For each basis element pick a lift $\tilde b_i \in G$. Then, choose the "almost morphic" (the single arbitrary choice for which we have to pay a price is some ordering of the basis elements) set-theoretic section $s \colon Q \to G$ defined by $$ \sum \lambda_i b_i \mapsto \tilde b_1^{\lambda_1} \cdots \tilde b_r^{\lambda_r} $$ this is well-defined because $\lambda_i$ is unique as an element of $\F_p$, and each $\tilde b_i$ has order $p$ (if we do not assume $G$ has exponent $p$ we have to deal with carrying in the exponents). Let $z_{i,j} = [\tilde b_i, \tilde b_j] \in Z$. Consider two elements $x = \sum \lambda_i b_i$ and $y = \sum \mu_i b_i$ of $Q$. We have $$ s(x)s(y) = \tilde b_1^{\lambda_1} \cdots \tilde b_r^{\lambda_r} \cdots \tilde b_1^{\mu_1} \cdots \tilde b_r^{\mu_r} = c(x,y) s(x+y) $$ where $c(x,y) = \sum_{i<j} \lambda_i \mu_j z_{i,j}$, so in this particular case we actually do find a bilinear $2$-cochain representing the class in $H^2(Q,Z)$, and then we can simply bilinearly extend this class by tensoring with $K$.

If we do not assume that $G$ has exponent $p$, it may happen that the order of $\tilde b_i$ is larger than that of $b_i$, and then we have to pick an even more uncanonical set-theoretic section, for example by picking the smallest integer as the exponent, and then we have to deal with carrying. I think base-changing should still be possible, by replacing the carrying cocycle by Witt polynomials.

I have to say I would like a cleaner way of constructing this extension, especially a way that makes functoriality clear, and ideally a way that generalizes to any $p$-group of any nilpotency class and any exponent. My general impression, right now, is that any $p$-group, say of order $p^d$, can be represented by having $d$ coordinates in $\F_p$, and the group law is addition "corrected" by some "strictly triangular mapping" (the correction at a given coordinate depends only of the preceding coordinates), where that correction is a mixture of multilinear maps and Witt polynomials. If that could somehow be made correct, then we could simply take the same description but allowing coordinates in a larger ring to obtain extensions of scalars.

$\endgroup$
8
  • 2
    $\begingroup$ This is very similar to a question of mine mathoverflow.net/questions/262745 , which is also a duplicate of mathoverflow.net/questions/69397 . $\endgroup$ Commented Jul 24 at 10:41
  • 1
    $\begingroup$ As I noted in that question, I suspect that there is no way to "enlarge" the dihedral group of 16 over $\mathbb{F}_2^{\text{alg}}$. $\endgroup$ Commented Jul 24 at 10:43
  • $\begingroup$ It seems like maybe a shorter way of formulating this question is "is there a bifunctor from (p-groups) x (characteristic p fields) to (groups) such that $\mathbb{Z}/p \mathbb{Z} \times k$ goes to $k^+$? $\endgroup$ Commented Jul 24 at 11:03
  • 1
    $\begingroup$ The dihedral group of order 16 has nilpotency class 3. The derived subgroup is the cyclic group of order 4, whereas the center is the 2 element subgroup of that cyclic group. $\endgroup$ Commented Jul 24 at 11:28
  • 2
    $\begingroup$ This question brings to mind Serre's concept of "saturation", which he used in his theorem about the tensor product of two low dimensional representations. $\endgroup$ Commented Jul 24 at 16:46

2 Answers 2

10
$\begingroup$

$\def\ZZ{\mathbb{Z}}\def\FF{\mathbb{F}}\def\cG{\mathcal{G}}\def\Ker{\operatorname{Ker}}\def\Image{\operatorname{Image}}\def\Coker{\operatorname{Coker}}$This answer will show that, for a certain interpretation of the question, the answer is "no", and explain why the dihedral group of order $16$ is an important obstacle.

Let $F$ be an infinite field of characteristic $2$.

I'll prove that it is impossible to find an exact functor $\cG$ from $2$-groups to groups such that

(a) $\cG(C_2)$ is $F$ with the ordinary addition operation

(b) if $A$ is a group of order $2^n$, then $\cG(A)$ is $F^n$ as a set, and the group operation on $\cG(A)$ is a polynomial map $F^n \times F^n \to F^n$.

(c) If $A$ and $B$ are groups of orders $2^m$ and $2^n$, and $\alpha : A \to B$ is a group homomorphism, then $\cG(\alpha) : F^m \to F^n$ is a polynomial map.

All groups that appear will be subgroups of products of cyclic of order $\leq 8$ and dihedral groups of order $\leq 16$.

I'll write $C_n$ for the cyclic group of order $n$ and $D_{2n}$ for the cyclic group of order $2n$, namely $C_2 \ltimes C_n$ where the nontrivial element of $C_2$ acts by $-1$.


At first, I'll just study exact functors $\cG$ from $2$-groups to groups such that $\cG(C_2)$ is nontrivial. I'll put a horizontal line where we start using the polynomiality. Sorry for being so lengthy in this portion, but I went through a lot of work convincing myself that I really wasn't using anything except exactness.

Lemma 1: Given a group homomorphism $\alpha : A \to B$, we have $\Ker(\cG(\alpha)) = \cG(\Ker(\alpha))$.

Proof: Kernel is a limit (namely, the equalizer of $\alpha : A \to B$ and of the trivial map $A \to B$); exact maps preserve limits. $\square$

Lemma 2: If $\alpha : A \to B$ is injective, then $\cG(\alpha) : \cG(A) \to \cG(B)$ is injective.

Proof Injective is the same as "kernel is $0$". $\square$

So, if $A$ is a subgroup of $B$, it makes sense to think of $\cG(A)$ as a subgroup of $\cG(B)$, and we will do so from now on.

Lemma 3: If $A \trianglelefteq B$ then $\cG(A) \trianglelefteq \cG(B)$ and $\cG(B/A) \cong \cG(B)/\cG(A)$.

Proof: If $A \trianglelefteq B$ then we may speak about the map $B \to B/A$, and we can say that its kernel is $A$. So $\Ker(\cG(B) \to \cG(B/A))$ is $\cG(A)$. In particular, $\cG(A)$ is normal in $\cG(B)$. Moreover, since $\cG(A) = \Ker(\cG(B) \to \cG(B/A))$, we have $\cG(B/A) \cong \cG(B) / \cG(A)$. $\square$

Lemma 4: Given a group homomorphism $\alpha : A \to B$, $\cG(\Image(\alpha)) \cong \Image(\cG(\alpha))$.

Proof: $\Image(\alpha) = B/\Ker(\alpha)$. $\square$

Lemma 5: For any two groups $A$ and $B$, we have $\cG(A \times B) \cong \cG(A) \times \cG(B)$.

Proof: Product is a limit. $\square$

Lemma 6: Let $H$ be a group with subgroups $A$ and $B$ which commute with each other. Then $\cG(A)$ and $\cG(B)$ commute with each other inside $\cG(H)$.

Proof: The condition that $A$ and $B$ commute (inside $H$) is equivalent to saying that there is a group homomorphism $\alpha : A \times B \to H$ such that $A \times \{ 1 \} \to A \times B \to H$ is the inclusion of $A$ and $\{ 1 \} \times B \to A \times B \to H$ is the inclusion of $B$. Now apply the functor $\cG$ to these maps. $\square$

Lemma 7: If $A$ is abelian, then $\cG(A)$ is abelian.

Proof: Apply Lemma 6 with $A=B=H$. $\square$.

If $A$ is an abelian group, we write $[2]$ for the doubling map $a \mapsto a^2$.

Lemma 8: If $A$ is abelian, then $\cG([2])$ is the $[2]$ map of $\cG(A)$.

Proof: We can write $[2]$ as the composite $A \to A \times A \to A$, where the first map is the diagonal and the second map is the $\alpha$ map from the proof of Lemma 6. $\square$

Lemma 9: $\cG(C_2)$ is a $2$-torsion abelian group.

Proof: $C_2$ is abelian and $[2]=0$ in $C_2$, so these properties are preserved by $\cG$ by Lemmas 7 and $9$. $\square$

Lemma 10: For any cyclic $2$-group $C_{2^k}$, and $j < k$, we have $\Ker {\big(}[2^j] : \cG(C_{2^k}) \to \cG(C_{2^k}){\big)} = 2^{k-j} \cG(C_{2^k})$.

Proof: Use Lemmas 1 and 8. $\square$

Lemma 11: $\cG(D_{2^{k+1}})$ is a semidirect product $\cG(C_2) \ltimes \cG(C_{2^k})$.

Proof: Apply the functor $\cG$ to the right split exact sequence expressing $D_{2^{k+1}}$ as a semidirect product. $\square$.

In particular, we get an action of $\cG(C_2)$ on $\cG(C_{2^k})$. Since $\cG(C_{2^k})$ is an abelian group (Lemma 7); its endomorphism monoid naturally has the structure of a ring; call it $E_k$. So we can write the action of $\cG(C_2)$ on $\cG(C_{2^k})$ as a group homomorphism $\rho_k : \cG(C_2) \to E_k$.

Lemma 12: For every $u \in \cG(C_2)$, we have $\rho_k(u) \equiv 1 \bmod{2 E_k}$.

Proof: Reducing $\rho_k(u)$ modulo $2$ gives $\rho_1$, the action map for $\cG(D_4)$. But $\cG(D_4)$ is abelian, so this action is trivial. $\square$

Thus, we can write $\rho_k(u) = 1 - [2] \sigma_k(u)$.

Lemma 12: Assuming $k \geq 3$ (so our dihedral group has order $\geq 16$), we have $\sigma(u)^2 \equiv \sigma(u) \bmod{2^{k-2} E_k}$.

Proof: Since $[2]$ is $0$ on $\cG(C_2)$ and $\rho_k$ is a group homomorphism, we must have $\rho_k(u)^2=1$ for every $u \in \cG(C_2)$. So $(1-2 \sigma_k(u))^2 = 1$ or, in other words, $4 \sigma_k(u)^2 = 4 \sigma_k(u)$ in $E_k$. Dividing by $4$, we have $\sigma_k(u)^2 \equiv \sigma_k(u) \bmod{\Ker [4]}$. By Lemma 10, this equation descends to $\text{End}(\cG(C_2))$. $\square$

So $\sigma_k$ descends to a map $\sigma : \cG(C_2) \to \operatorname{End}(\cG(C_2))$ sending every element of $\cG(C_2)$ to an idempotent $\sigma(u)$ in $\operatorname{End}(\cG(C_2))$.


Now, finally, we use the polynomiality hypotheses. Then $\sigma(u)$ must be a polynomial endomorphism of $\cG(C_2) = F$. The polynomial endomorphisms of $\cG(C_2)$ are the additive polynomials — those polynomials of the form $\sum_{i=0}^{\infty} a_i x^{2^i}$. I'll write $\mathrm{Add}$ for the ring of additive polynomials (mulitiplication is composition). Moreover, $\sigma : F \to \mathrm{Add}$ must be a polynomial map.

But, moreover, every polynomial in the image of $\sigma$ must be idempotent, with respect to the operation of composition. The only polynomials with $g \circ g = g$ are constants and the identity polynomial $x$; the only additive ones are $0$ and $x$. (We are using that $F$ is infinite so that we can identify the equation $g \circ g = g$ as an equation of functions $F \to F$ with an equality of formal polynomials.) So $\sigma : F \to \mathrm{Add}$ is a polynomial map, from an infinite field, which only takes $2$ values. So it must be constant.

But, returning to the original dihedral group, we want $\rho(0) = 1$ and $\rho(1) = -1$, so $\sigma(0) = 0$ and $\sigma(1) = 1$. So $\sigma$ should not be constant, and we have reached our contradiction.

$\endgroup$
5
  • $\begingroup$ This looks nice. Few small comments: in Lemma 3, you use right exactness without stating it (this is fine); the proof of Lemma 4 has a typo; the statement of Lemma 10 should probably include $\mathcal G$ in a few places. I didn't quite follow what you mean by 'descend to $\operatorname{End}(\mathcal G(C_2))$', though. $\endgroup$ Commented Jul 25 at 22:19
  • $\begingroup$ Thanks for your answer! it seems sound to me but I have to study it a little more to really understand the obstruction. (btw, you use the phrase "the ring of additive polynomials", but I do not see an obvious multiplicative law?) $\endgroup$ Commented Jul 26 at 12:17
  • 2
    $\begingroup$ @Moinsdeuxcat: the multiplication is composition. $\endgroup$ Commented Jul 26 at 18:07
  • $\begingroup$ Does this have any bearing on whether $D_8$ (Dihedral group of order $16$) is the $\mathbb{F}_2$-points of a connected unipotent group? $\endgroup$ Commented Aug 9 at 14:09
  • 1
    $\begingroup$ If there were a functor $\cG$ as I interpreted the question to ask for, then it would be a functor from $2$-groups to algebraic groups such that $\cG(A)(\mathbb{F}_2) \cong A$. The reverse direction is much less clear -- just because there isn't a functor doesn't mean that there isn't some unipotent group whose $\mathbb{F}_2$ points are $D_{16}$, it just means it wouldn't fit into all the commutative diagrams you might hope for. But I hope this argument gestures at why I think $D_{16}$ might not occur. $\endgroup$ Commented Aug 10 at 0:09
1
$\begingroup$

Not a complete argument, but I don't expect this to be possible in general. A vague idea of one thing that what might go wrong is that we might be able to write some $G$ as an extension or iterated extension of smaller groups in two different ways which does not generalize after "tensoring with $K$." One possible example, that I don't know how to complete, is that $D_4$ (the dihedral group of order $8$) is both a semidirect product $C_4 \rtimes C_2$ and a semidirect product $C_2^2 \rtimes C_2$, and this doesn't appear to me to generalize.

More generally, exactness together with functoriality implies that $G \mapsto G_K$ sends split exact sequences to split exact sequences, so sends semidirect products to semidirect products, and this seems to me to produce troublesome examples. For example, if $H$ is any $p$-group we can construct the wreath product $G = H \wr C_p$; this sits in a split exact sequence

$$1 \to H^p \to G \to C_p \to 1$$

which is therefore sent to a split exact sequence

$$1 \to H_K^p \to G_K \to K \to 1.$$

(I'm assuming that $G \mapsto G_K$ preserves finite products, which I think ought to follow from functoriality and exactness, and in any case is a clearly desirable / natural property we'd want.) So we have to somehow turn the permutation action of $C_p$ on $H^p$ into an action of $K$ on $H_K^p$, in a way that is functorial in $H$ and $K$; I don't see any candidate for how to do this.

$\endgroup$
4
  • $\begingroup$ $D_4$ has nilpotency class two, so I have explained how to deal with it using the results of Hughes (bilinearly extend the commutator bracket, and this is the commutator bracket of some group) $\endgroup$ Commented Jul 26 at 12:16
  • $\begingroup$ (In fact, $D_4$ is simply the Heisenberg group over $\mathbb F_2$, so the extension of scalars to a ring $R$ of characteristic 2 is just the Heisenberg group over $R$) $\endgroup$ Commented Jul 26 at 12:22
  • $\begingroup$ @Moinsdeuxcat: that doesn't address my point, the issue is uniqueness, not existence. Using the alternate description as $C_4 \rtimes C_2$ gives something like $W_2(K) \rtimes K$ and I am not convinced either that there is a meaningful action of $K$ on $W_2(K)$ which functorially extends the action of $C_2$ on $C_4$ or that the resulting semidirect product is isomorphic to $H_3(K)$. $\endgroup$ Commented Jul 26 at 18:07
  • $\begingroup$ Well I think that action is clear. The subgroup of matrices of the form $\begin{pmatrix}1&a&b\\&1&a\\&&1\end{pmatrix}$ is your $W_2(K)$, and it is acted upon by $\lambda \in K$ via $a \mapsto a, b \mapsto b-\lambda a$. $\endgroup$ Commented Jul 26 at 18:29

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.