Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Due to the first (and very helpful) answer I received, I've reformulated the question a little: $G$ and $H$ are now assumed to be $p$-groups.

Let $p$ be a prime, and let $\mathbb{F}_p$ be the field of $p$ elements. Let $G,H$ be finite $p$-groups, and let $\mathbb{k}[G]$ denote the group algebra.

Does $\mathbb{K}[G]\simeq\mathbb{K}[H]$ for some field $\mathbb{K}$ of characteristic $p$, imply $\mathbb{F}_p[G]\simeq\mathbb{F}_p[H]$?

share|improve this question

2 Answers 2

No, take $G= Z/4Z$ and $H=(Z/2Z)^2$. If the field k contains a 4th root of 1 then both group algebras are isomorphic to k+k+k+k (I am assuming the the charachterisitc is not 2). But these group algebras are not isomorphic if k does not contian $\sqrt{-1}$, which is the case for F_p if p = 3,5 (mod 8)

My guess is that for p-groups everything is OK

share|improve this answer
$\mathbb{F}_p$ contains $\sqrt{-1}$ precisely when $p \equiv 1 \mod 4$, right? –  Maurizio Monge Apr 5 '11 at 18:11
Sorry, somehow I was thinking about $\sqrt{2}$ –  kassabov Apr 7 '11 at 9:08

I believe I have a counter-example in $p$-groups. I'm hoping someone who really knows non-commutative deformation theory can fill in the last half.

$\def\FF\mathbb{F}$Write $\FF_q$ for the field with $q$ elements. Let $p$ be an odd prime.

I first present my two groups, $G_1$ and $G_2$. For $i=1$, $2$, fix a short exact sequence $0 \to Z_i \to W_i \to V_i \to 0$ of $\FF_p$ vector spaces, with $\dim Z_i = 2$, $\dim V_i = 4$. We write $w \mapsto \overline{w}$ for the quotient map $W_i \to V_i$. Let $\phi_i : \bigwedge^2 V_i \to Z_i$ be a linear map. We extend $\phi_i$ to a map $W_i \times W_i \to Z_i$ by $(w,w') \mapsto \phi_i(\overline{w} \wedge \overline{w})$, and denote this map by $\phi_i$ as well.

The underlying set of $G_i$ will be $W$. Given an element $w$ in $W_i$, we write $a^w$ for the corresponding element of $G_i$. The relations in $G_i$ are

  • $a^z$ is central, for all $z \in Z_i$

  • $a^{w} \cdot a^{w'} = a^{w+w'+\phi_i(w,w')}$.

Now, I must tell you how to choose the $\phi_i$. Dualizing $\phi_i$, we get a linear map $Z_i^{\ast} \to \bigwedge^2 V_i^{\ast}$. In other words, we get a two dimensional subspace of $\bigwedge^2 V_i^{\ast}$ or, in other words, a projective line in $\mathbb{P}\left( \bigwedge^2 V_i^{\ast} \right)$. Now, $\mathbb{P}\left( \bigwedge^2 V_i^{\ast} \right)$ contains the Grasmannian $G(2,4)$ of skew-symettric tensors of rank $1$. We will choose $\phi_1$ to meet $G(2,4)$ at two distinct points defined over $\FF_p$, and choose $\phi_2$ to meet $G(2,4)$ at two Galois conjugate points defined over $\FF_{p^2}$.

Explicit coordinates are as follows: Let $e_1$, $e_2$, $e_3$, $e_4$ be a basis for $V_i$. Take $\phi_1$ to be $(e_1^{\ast} \wedge e_2^{\ast}, e_3^{\ast} \wedge e_4^{\ast})$. Letting $\FF_{p^2} = \FF_p[\sqrt{D}]$, take $\phi_2 = (e_1^{\ast} \wedge e_3^{\ast} + D e_2^{\ast} \wedge e_4^{\ast}, e_1^{\ast} \wedge e_4^{\ast} + e_2^{\ast} \wedge e_4^{\ast})$; note that $(e_1^{\ast} \pm \sqrt{D} e_2^{\ast}) \wedge (e_3^{\ast} \pm \sqrt{D} e_4^{\ast})$ is a linear combination of the components of $\phi_2$.

The point of these choices is that there does NOT exist an $\FF_p$-linear map $W_1 \to W_2$ carrying $\phi_1$ to $\phi_2$, but there DOES exist such a map once we tensor with $\FF_{p^2}$.

Now, we need to show that our construction is reflected in properties of the group rings. Let $R_i = \FF_p[G_i]$. Our first goal is to show that $R_1 \not \cong R_2$. We must show how to canonically recover $V_i$, $Z_i$ and $\phi_i$ from $R_i$.

The only $1$-dimensional representation of $G_i$ over $\FF_p$ is the trivial rep, since $G_i$ is a $p$-group. Therefore, $\FF_p$ is an $R_i$ module in only one way. Let $I_i$ be the kernel of the unique map $R_i \to \mathrm{End}(\FF_p)$. Explicitly, $I_i$ has $\FF_p$ basis given by the elements $g-1$, for $g \in G_i$.

The center of $R_i$ is $\FF_p[Z_i]$. For $z$ and $z' \in Z_i$, we have $(a^{z+z'}-1) - (a^z-1) - (a^{z'}-1) = (a^z-1)(a^{z'}-1)$, so $(a^{z+z'}-1) \equiv (a^z-1)+(a^{z'}-1) \bmod (Z(R_i) \cap I_i)^2$. We see that the $\FF_p$ vector space $Z_i$ is canonically isomorphic to $(Z(R_i) \cap I_i)/(Z_i \cap I_i)^2$, by $z \mapsto a^z-1$. (Expressions like $(Z_i \cap I_i)^2$ mean the square as a two-sided ideal.)

We claim similarly that $V_i \cong I_i /I_i^2$ by the map $v \mapsto a^w-1$, where $w$ is an arbitrary lift of $v\in V_i$ to $W_i$. To this end, we first must prove the formula is well defined. If $w$ and $w+z$ are two different lifts, then we must show that $a^{w+z} - a^w = (a^z-1) a^w \in I_i^2$. In other words, we must show that $a^z-1 \in I_i^2$ for $z \in Z_i$. Now, if $z = \phi_i(w, w')$, then $(a^w-1)(a^{w'}-1) - (a^{w'}-1)(a^w-1) = a^{w+w'+\phi_i(w,w')} - a^{w+w'+\phi_i(w',w)} = (a^{2\phi_i(w,w')}-1) a^{w+w'-\phi_i(w,w')} \in I_i^2$. Therefore, $a^{2 \phi_i(w,w')}-1 \in I_i^2$. Letting $w$ and $w'$ vary, we can obtain $a^z-1 \in I_i^2$ for any $z \in Z_i$.

We have now checked that $v \mapsto a^w-1$ is a well defined map of sets $V_i \to I_i/I_i^2$. To see that it is a map of $\FF_p$ vector spaces, note that $$(a^{w+w'}-1) - (a^w-1)- (a^{w'}-1) = (a^w-1)(a^{w'}-1) + a^{w+w'} (1-a^{\phi_i(w,w')}).$$ Both summands on the right hand side are in $I_i^2$. A bit more work shows that this map of $\FF_p$ vector spaces is an isomorphism.

Now, we must show that we can recover $\phi_i$. Let $\langle Z(R_i) \cap I_i \rangle$ be the two sided ideal of $R_i$ generated by $Z(R_i) \cap I_i$. Fix a section $V_i \to W_i$. Using this section, we can write any element $c$ of $R_i$ uniquely as $\sum_{V \in V_i} c_v a^v$. We have $c \in \langle Z(R_i) \cap I_i \rangle$ if and only if all the $c_v$ are in $Z(R_i) \cap I_i$. For $c \in \langle Z(R_i) \cap I_i \rangle$, define $\sigma(c) = \sum_{v \in V} c_v$. We check that $\sigma(c) \bmod (Z(R_i) \cap I_i)^2$ is independent of the choice of section. So $\sigma$ gives a well defined map $\langle Z(R_i) \cap I_i \rangle \longrightarrow (Z(R_i) \cap I_i)/(Z(R_i) \cap I_i)^2 \cong Z_i$, which we will also denote $\sigma$.

For any $w$ and $w' \in W_i$, we have $a^w a^{w'} - a^{w'} a^w = (a^{\phi_i(w,w')} - a^{\phi_i(w',w)}) a^{w+w'} \in \langle Z(R_i) \cap I_i \rangle$. Thus, it makes sense to talk about $\sigma(x x' - x' x)$ for any $x$ and $x' \in R_i$. Suppose furthermore that $x \in I_i$ and $x' \in I_i^2$. Then I claim that $\sigma(x x' - x' x)=0$. By linearity, we may assume that $x=a^w-1$ and $x' = (a^{w'}-1) (a^{w''}-1)$. So $$x x' - x' x = (a^{\phi_i(w,w'+w'')} - a^{-\phi_i(w,w'+w'')}) a^{w+w'+w''+\phi_i(w', w'')} - (a^{\phi_i(w,w')} - a^{-\phi_i(w,w')}) a^{w+w'} - (a^{\phi_i(w,w'')} - a^{-\phi_i(w,w'')}) a^{w+w''}$$ and $\sigma(x x'-x' x) = \phi_i(w,w'+w'') - \phi_i(w,w') - \phi_i(w,w'')=0$. This proves the claim.

Therefore, $(x,x') \mapsto \sigma(x x' - x' x)$ gives a well defined map $I_i/I_i^2 \times I_i/I_i^2 \to (Z(R_i) \cap I_i)/(Z(R_i) \cap I_i)^2$. Tracing through definitions, this map is $2 \phi_i$.

Whew! That's only half the work. We now want to show that $\FF_{p^2}[G_1] \cong \FF_{p^2}[G_2]$. The idea is to write each side as a noncommutative $\FF_{p^2}$ algebra generated by $V_i$, where we choose some basis $v_1$, $v_2$, $v_3$, $v_4$ for $V_i$ and send these basis elements to $a^{w_1}-1$ etcetera, where $w_s$ is a lift of $v_s$ to $W_i$. We have an $\mathbb{F}_{p^2}$ linear map taking $V_1 \otimes \FF_{p^2}$ to $V_2 \otimes \FF_{p^2}$ in a way that respects $\phi$. The low order terms match up. The intuition is that, with enough knowledge of deformation theory, we should be able to lift this low order match to an isomorphism $\FF_{p^2}[G_1] \cong \FF_{p^2}[G_2]$.

Anyone want to finish the computation?

ADDED I've been thinking more about this, and I am no longer so optimistic, although I'd still love to know what an expert thinks. Quillen associates a restricted Lie algebra $L_i$ to a $p$-group $G_i$. Writing $U_i$ for the restricted enveloping algebra, he shows that the associated graded of $\FF_p[G_i]$ with respect to $I_i$ (the augmentation ideal) is $U_i$. In our case, we get $\mathbb{F}_{p^2} \otimes L_1 \cong \mathbb{F}_{p^2} \otimes L_2$ and so $\mathbb{F}_{p^2} \otimes U_1 \cong \mathbb{F}_{p^2} \otimes U_2$. Now, can we lift this back to $\FF_{p^2}[G_1] \cong \FF_{p^2}[G_2]$?

The most obvious reason this should be true is if $U \cong \FF_p[G]$ as algebras. That's true if $G$ is an abelian $p$-group. I think it is also true for the noncommutative $p$-torsion group of order $p^3$. But I don't think it should be true in our case. More generally, let's talk about $2$-step $p$-torsion nilpotents in general.

So, let $G$ be the group generated by $X_i$ and $Z_{ij}$ with the relations that everything is $p$-torsion, the $Z_{ij}$ are central, $Z_{ij} = Z_{ji}^{-1}$ and $X_i X_j = X_j X_i Z_{ij}$. We might, in addition, impose some further commutative relations between the $Z_{ij}$. (In our example, $X_i$ runs from $i=1$ to $4$, and the $Z_{ij}$ generate a group of rank $2$.) Let $\mathfrak{g}$ be the corresponding restricted Lie algebra on $x_i$ and $z_{ij}$, where all $[p]$-powers are $0$, the $z_{ij}$ are central, $z_{ij} = - z_{ji}$ and $[x_i, x_j] = z_{ij}$. We'd like to know whether there is an isomorphism $\FF_p[G] \to \FF_p\langle \mathfrak{g} \rangle$, where the right hand side is the restricted enveloping algebra. Let $A$ be the subring $k[z_{ij}]/z^{ij}^p=0$ of the right hand side. We want the $Z_{ij}$ to go to central elements on the right, so it seems plausible that the $Z_{ij}$ should map into $A^{\times}$.

For low degree terms in the $x_i$, we should clearly send $X_i$ to the initial terms of $\sum_s x_i^s/s!$. But this runs into trouble in degree $p$. Suppose we try fixing this by writing $X_i = \sum_{s<p} x_i^s/s! + Y_i$ where $Y_i$ appears in degrees $\geq p$. Then there are about $4 \dim k \langle \mathfrak{g} \rangle$ coefficients in $Y_i$. The condition that $X_i X_j X_i^{-1} X_j^{-1}$ lies in $A^{\times}$ is about $6 \dim k \langle \mathfrak{g} \rangle$ conditions. So we have more conditions then variables, I don't think a naive deformation approach will solve the problem.

share|improve this answer
I've not yet had time to think about your construction, but if it works then I think it would be quite a big deal for some people. The "modular isomorphism problem" asks whether the modular group algebras of non-isomorphic finite $p$-groups can be isomorphic. I think most people who've worked seriously on the problem have worked over $\mathbb{F}_p$, but I've certainly seen the problem stated over an arbitrary field of characteristic $p$, and I'm fairly sure it's open there. –  Jeremy Rickard Aug 24 at 15:47
Hmmm, so either I am wrong, or it is worth putting in the time to actually think through the deformation theory. Maybe you can help: My intuition is that group algebras of two step $p$-torsion nilpotent Lie groups are very similar to enveloping algebras of two-step nilpotent $p$-Lie algebras where the $p$-th power map is zero. This really is a counterexample in the $p$-Lie algebra world. Do you know theorems making this analogy precise? –  David Speyer Aug 24 at 15:55
Let's go for option (b)! It's not a problem I've ever worked on, so I'm not familiar with the details of what's known, but I'll see what I can find out. –  Jeremy Rickard Aug 24 at 16:00
Section 1.3 of igt.uni-stuttgart.de/LstDiffgeo/Hertweck/preprints/… shows how to canonically turn a group algebra $\mathbb{F}_p[G]$ into a $p$-Lie algebra $Jen(G)$. I am reasonably confident that I have constructed an example where $Jen(G_1) \not \cong Jen(G_2)$ but $\mathbb{F}_{p^2} \otimes Jen(G_1) \cong \mathbb{F}_{p^2} \otimes Jen(G_2)$; the remaining question is whether I can lift that last fact back to $\mathbb{F}_{p^2}[G]$. –  David Speyer Aug 24 at 16:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.