$\textrm{GL}_2(\mathbb{Z}/p^2\mathbb{Z}) \to \textrm{GL}_2(\mathbb{Z}/p\mathbb{Z})$ has no section for $p > 3$

Question

Let $p > 3$ be a prime. I want to show that the exact sequence $$1 \to \ker \pi \to \textrm{GL}_2(\mathbb{Z}/p^2\mathbb{Z}) \xrightarrow{\pi} \textrm{GL}_2(\mathbb{Z}/p\mathbb{Z}) \to 1$$ is not split. It's easy to check that $A = \ker\pi$ is abelian, thus $G = \textrm{GL}_2(\mathbb{Z}/p\mathbb{Z})$ acts on $K$ by conjugation, making $A$ a $G$-module. If $$s \colon \textrm{GL}_2(\mathbb{Z}/p\mathbb{Z}) \to \textrm{GL}_2(\mathbb{Z}/p^2\mathbb{Z})$$ is a set-theoretic section of $\pi$, then $$\alpha_s(g,h) = s(g)s(h)s(gh)^{-1}$$ defines a $2$-cocyle $G \times G \to A$ whose class in $H^2(G,A)$ is independent of the choice of section. Since $\alpha_s$ is trivial if $s$ is a homomorphism, it suffices to show that $\alpha_s$ is not a coboundary, i.e. there does not exist a function $\phi \colon G \to A$ such that $$s(g)s(h)s(gh)^{-1} = s(g)\phi(h)s(g)^{-1}\phi(gh)^{-1}\phi(h)$$ for all $g,h \in G$.

I tried supposing such a $\phi$ exists and looking at the subgroup of $U$ unipotent matrices. Every element of $A$ is $p$-torsion, so $\alpha_s$ is a $p$-torsion elemeent of $H^2(G,A$). Since $U$ is the $p$-Sylow subgroup of $G$, the restriction map $H^2(G,A) \to H^2(U,A)$ is injective on the $p$-part of $H^2(G,A)$, so looking at the restriction of $\alpha_s$ to $U \times U$ should be sufficient.

Choose the section $s$ so that it maps $U$ to unipotent matrices in $\textrm{GL}_2(\mathbb{Z}/p^2\mathbb{Z})$. After playing around with the coboundary condition, I found that $\phi(g)$ and $s(g)$ should commute for all $g \in U$. Since $s(g)$ is unipotent, this means $$\phi(g) = \begin{pmatrix} 1 + a_gp & b_gp \\ 0 & 1 + a_gp \end{pmatrix}$$ for some $a_g,b_g \in \mathbb{Z}/p^2\mathbb{Z}$, well-defined modulo $p$. If $s_g$ denotes the upper-right entry of $s(g)$ for $g \in U$, then the coboundary condition tells us that $g \to a_g$ is a homomorphism and $$(b_g - b_{gh} + b_h)p = s_g - s_{gh} + s_h$$ for all $g,h \in U$. So we get a system of linear equations over $\mathbb{Z}/p\mathbb{Z}$ with $p$ variables. It's not clear to me how to show that this system is inconsistent, or how to use the hypothesis that $p > 3$. Moreover, lifting $0,1,-1$ in $\mathbb{Z}/3\mathbb{Z}$ to $0,1,-1$ in $\mathbb{Z}/9\mathbb{Z}$, I checked that this system is inconsistent over $\mathbb{Z}/3\mathbb{Z}$. But I thought that the exact sequence was supposed to be split in the case $p = 3$. So I'm unsure where I went wrong and also unsure how to proceed.

Answer 1

Let $S=\pmatrix{1&1\\0&1}\in\text{GL}_2(\Bbb Z/p\Bbb Z)$. This has order $p$, and one's intuition suggests that any lifting to $\text{GL}_2(\Bbb Z/p^2\Bbb Z)$ should have order $p^2$, meaning that there's no section for $\pi$. But is this true?

A lifting of $S$ has the form $S'=I+A$ where $$A=\pmatrix{ap&1+bp\\cp&dp}\in\text{GL}_2(\Bbb Z/p^2\Bbb Z).$$ Then $$A^2=\pmatrix{cp&(a+d)p\\0&cp},$$ $$A^3=\pmatrix{0&cp\\0&0}$$ and $A^4=0$. For $p\ge5$ then $$S'^p=I+pA+\binom p2A^2+\binom p3A^3=I+\pmatrix{0&p\\0&0}$$ which does mean that $S'$ does not have order $p$ in $\text{GL}_2(\Bbb Z/p^2\Bbb Z)$.

This argument breaks down for $p\in\{2,3\}$. For instance with $p=3$ one has $$S'^3=I+3A+3A^2+A^3=I+\pmatrix{0&(c+1)p\\0&0}$$ so one can take $c=-1$. Of course this is somewhat shy of proving that in this case $\pi$ has a section, but it shows that this argument does not refute it.

Answer 2

Here is a different proof that uses more theory and less calculation.

There is a class of $p$-groups called regular $p$-groups that have various nice properties. All groups of order $p^n$ for $n \le p$ are regular, and one of the properties of regular $p$-groups is that, for any fixed $k$, the elements of order dividing $p^k$ form a subgroup.

If the extension in the question were split, then a Sylow $p$-subgroup $P$ would also split. Now $|P|=p^5$, so $P$ is regular when $p \ge 5$.

Since $\ker \pi$ is elementary abelian, if the extension split, then $P$ would be generated by elements of order $p$, and hence $P$ would have exponent $p$. But that is clearly false, because the element $\left( \begin{array}{cc}1&1\\\ 0&1\end{array}\right)$ has order $p^2$. So the extension is non-split.