# Describe all extension groups of a given subgroup $H \trianglelefteq \mathrm{Heis}(\mathbb F_q)$ by $\mathrm{Heis}(\mathbb F_q)/H$

Let $\mathbb F_q$ be a finite field. Consider the group $$\mathrm{Heis}(\mathbb{F}_q):=\left\{ \begin{pmatrix} 1&a&b\\ 0&1&c\\ 0&0&1 \end{pmatrix} \colon a, b, c \in \mathbb F_q \right\}.$$

Put $$H := \left\{ \begin{pmatrix} 1&0&b\\ 0&1&c\\ 0&1&1 \end{pmatrix} \colon b,c \in \mathbb F_q \right\},$$ which is a normal subgroup of $\mathrm{Heis}(\mathbb F_q)$.

How to describe all extension groups of $H$ by $\mathrm{Heis}(\mathbb F_q)/H\simeq \mathbb F_q$?

• In other words, for a prime $p$ and $n>0$, you want to describe all extensions of $C_p^{2n}$ by $C_p^n$. I do not believe that this is feasible. – Derek Holt Sep 4 '16 at 7:18
• You seem to be studying group extensions, given your question math.stackexchange.com/questions/1804633/… (and your duplicate math.stackexchange.com/questions/1912385/…). It would probably be a good idea to follow @DietrichBurde's suggestion to learn about group cohomology. – LSpice Sep 4 '16 at 14:26
• Also, while I'm at it, please note the difference between $\mathbb{F_q}$—which produces the undesired result $\mathbb{F_q}$, which I edited out of your post (but which still appears in the old questions)—and $\mathbb{F}_q$ (or, as I prefer, $\mathbb F_q$)—which produces the desired result $\mathbb F_q$. – LSpice Sep 4 '16 at 14:29


Put $\FF = \FF_q$. For the moment, don't assume that $q$ is prime. Let $\wtilde H$ be an extension of the desired sort, and choose an isomorphism $i : \FF \to \wtilde H/H$. By the fact that $H = \FF^2$ is Abelian, we obtain for each element $a \in \mathbb F$ an automorphism $\Int(a) : \FF^2 \to \FF^2$ given by $\Int(a)h = \tilde h h\tilde h^{-1}$ for all $h \in \FF^2$, where $\tilde h$ is any lift of $i(a)$ to $\wtilde H$. It is clear that $\Int : \FF \to \Aut_\ZZ(\FF^2)$ is a homomorphism.

The thing that makes the case of $q$ prime easier than the general case is, of course, that $\Aut_\ZZ(\FF^2) = \Aut_\FF(\FF^2) = \GL_2(\FF)$; that is, that every group endomorphism of $\FF^2$ is a ($\FF$-)linear endomorphism. Now impose this assumption (that $q = p$ is a prime). Then the $p$-Sylow subgroups of $\GL_2(\FF)$ are the conjugates of its group of lower-triangular matrices, and (again because $q = p$ is prime) there is, up to isomorphism, only one embedding of this group in $\GL_2(\FF)$; so either $\Int$ is trivial, in which case (for consistency) I will put $\psi = 1$, or there is some $\psi \in \GL_2(\FF)$ so that $\Int(a)\psi(b, c) = \psi(b + a c, c)$ for all $a, b, c \in \FF$. (EDIT: Thanks to @DerekHolt for pointing out that I'd forgotten the trivial extension.)

Now let $\tilde h$ be any lift of $i(1)$ to $\wtilde H$, and put $\tilde 1 = \tilde h z^{-1}$, where $z = \tilde h^p \in H$. Note that $z$ is centralised by $H$ and $\Int(1)$, hence is central, so that $\tilde 1$ has order $p$. Then the map $\FF \ltimes_{\Int} \FF^2 \to \wtilde H$ given by $a \ltimes (b, c) \mapsto \psi(b, c)\tilde 1^a$ for $(a, b, c) \in \FF^3$ is an isomorphism. The source is $\FF^3$ or $\mathrm{Heis}(\FF)$, according as $\Int$ is trivial or not.

• But the elementary abelian group $C_p^3$ is also an extension of $C_p^2$ by $C_p$. It depends on whether the OP wanted all extensions, or just extensions with the same induced action as the Heisenberg group. – Derek Holt Sep 4 '16 at 15:52
• @DerekHolt, oops, of course you are right; thanks! I don't want to go counterexample-sniping, but I think my only error here was in assuming that the map into the $p$-Sylow subgroup was actually onto it, so that our two examples are in fact the only possible. I have updated my answer accordingly. – LSpice Sep 4 '16 at 16:07
• Can you give several examples of groups this type? – user272816 Sep 4 '16 at 21:01
• @Zudwa, several examples of two groups? – LSpice Sep 4 '16 at 21:58
• Examples of extension groups. – user272816 Sep 5 '16 at 7:38