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

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 $\mathbb{F}_p[G]\simeq\mathbb{F}_p[H]$?

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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

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$\mathbb{F}_p$ 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.

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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 '15 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 '15 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 '15 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 '15 at 16:16
What happened to conjugacy class sums over non-trivial conjugacy classes in your $G_i$? They should contribute central elements in $\mathbb{F}_p[G_i]$ away from $\mathbb{F}_p[Z_i]$. – Konstantin Ardakov Oct 5 '15 at 7:35