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Let $\Gamma$ be a finite group of order $p^n$. Is there necessarily a unipotent algebraic group $G$ of dimension $n$, defined over $\mathbb{F}_p$, with $\Gamma \cong G(\mathbb{F}_p)$?

I have no real motivation for this question, it just came up in conversation and no one knew the answer. There does not appear to be any sort of uniqueness to $G$; both the groups $\mathbb{Z}/p^2 \mathbb{Z}$ and $(\mathbb{Z}/p \mathbb{Z})^2$ have infinitely many lifts to unipotent groups.

I am bumping this question to the top because I now suspect I have a counterexample, namely the dihedral group of order $16$, also known as $C_2 \ltimes C_8$, where $C_n$ is cyclic of order $n$. The "obvious" lift of $C_8$ is the additive group of the $3$-truncated Witt vectors, $\mathcal{W}_3$. So the obvious thing to try is a semidirect product $\mathbb{G}_a \ltimes \mathcal{W}_3$. But $\mathbb{G}_a$ has exponent $2$, so an action of $\mathbb{G}_a$ on $\mathcal{W}_3$ must live in the $2$-torsion elements of $\mathrm{Aut}(\mathcal{W}_3)$. I compute that the space of $2$-torsion elements in $\mathrm{Aut}(\mathcal{W}_3)$ has two connnected components, one of which maps $1$ to things that are $1 \bmod 4$ and one of which maps $1$ to things which are $-1 \bmod 4$. Since $\mathbb{G}_a$ is connected, it must land in one component, but the two elements of $C_2$ live in different components.

I've made some partial progress thinking about the structure of connected unipotent groups whose $\mathbb{F}_2$ points are $C_2 \ltimes C_8$, but I've gotten stuck, so I am putting up a bounty for progress either on this case or the questions as a whole.

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    $\begingroup$ If $U$ is smooth connected unipotent of dimension $n$ over $k=\mathbf{F}_p$ then $\#U(k)=p^n$ (use composition series with successive quotients $\mathbf{G}_a$). Thus, if $\Gamma=G(k)$ for a unipotent $k$-group $G$ of dimension $n$ then by passing to $G_{\rm{red}}$ so that $G$ is smooth we see that if $G$ is not connected then $G^0(k)$ of size $p^n$ must exhaust $G(k)$. In other words, we lose nothing by assuming $G$ is smooth and connected. So it is the same as asking for smooth connected unipotent $k$-groups $U$ if there is any constraint on $U(k)$ beyond its order being $p^n$. Good puzzler. $\endgroup$
    – nfdc23
    Commented Feb 21, 2017 at 3:55
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    $\begingroup$ This was asked a while ago, as well, but there was no answer. mathoverflow.net/questions/69397/… $\endgroup$
    – Gjergji Zaimi
    Commented Feb 21, 2017 at 5:09
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    $\begingroup$ Here is how I would try to prove this. There exists a faithful homomorphism, $\rho:\Gamma\to \textbf{SL}_n(\mathbb{F}_p)$. This follows, for instance, from the solution of the Abhyankar conjecture by Raynaud and Harbater. By counting $\mathbb{F}_p$-points of the flag variety of $\textbf{SL}_n$, there exists a Borel subgroup $B$ that contains the image of $\rho$. Thus, there exists a faithful homomorphism from $\Gamma$ to a unipotent group. Replace that unipotent group by the centralizer $U$ of $Z(\Gamma)$. Thus $Z(U)(\mathbb{F}_p)$ is an additive group that contains $Z(\Gamma)$ . . . $\endgroup$
    – Jason Starr
    Commented Feb 21, 2017 at 13:56
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    $\begingroup$ @JasonStarr: In view of your inductive suggestion it seems that what you call "$n$" (from SL$_n$) is not "$n$" as in the statement of the question, just some unknown integer, in which case such $\rho$ is provided by the regular representation of $\Gamma$ on itself over $\mathbf{F}_p$ (that gives $\rho$ into ${\rm{GL}}_N(\mathbf{F}_p)$, necessarily landing inside SL$_N({\mathbf{F}}_p)$ and even conjugating into $U_N(\mathbf{F}_p)$ since the latter is a $p$-Sylow subgroup). $\endgroup$
    – nfdc23
    Commented Feb 21, 2017 at 16:17
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    $\begingroup$ @DrorSpeiser I decided to put the bounty on my own version, rather than the duplicate because I'd rather edit my own question and because it has a bunch of useful comments. If I had (two years ago) seen Gjergji's link to the duplicate before a bunch of those comments showed up, I probably would have self-closed, but it doesn't seem like a big deal either way. Of course, if someone posts a good answer, I'll link it on that question as well to make sure searchers find it. $\endgroup$
    – David E Speyer
    Commented Mar 20, 2019 at 19:01

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