Palindromes in finite groups and the Explorer-Director game

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Problem: For a finite group $G$ of odd order, a twisted subgroup is a subset $P\subseteq G$ with $1\in P$ and $aba\in P$ for all $a,b\in P$. The paper defines

$$L(G)=\{|P|:P\subsetneq G,\ P\text{ a twisted subgroup}\}.$$

Literally, $L(G)=\{d:d\mid |G|\}$ is already false because $|G|$ is on the right but $G$ itself is excluded on the left. The natural repaired conjecture is that $L(G)$ is the set of proper divisors of $|G|$. This repaired conjecture is also false.

Result: Let $F=\mathbb F_{3^4}$, and choose $\alpha\in F$ of order $5$, e.g. a root of

$$x^4+x^3+x^2+x+1.$$

Let

$$G=F\rtimes C_5,$$

where $C_5=\langle t\rangle$ acts on the additive group of $F$ by $tvt^{-1}=\alpha v$. Thus $|G|=3^4\cdot 5=405$.

Write elements as $(v,i)$, $v\in F$, $i\in\mathbb Z/5\mathbb Z$, with

$$(v,i)(w,j)=(v+\alpha^i w,i+j).$$

First note that in an odd-order group every twisted subgroup is closed under inverses: if $x\in P$, then the exponents $n$ with $x^n\in P$ form a subset of $\mathbb Z/\operatorname{ord}(x)$ containing $0,1$ and closed under $a,b\mapsto 2a+b$; since multiplication by $2$ is bijective, this subset is all of $\mathbb Z/\operatorname{ord}(x)$.

Let $P\leq_{\mathrm{tw}} G$. Project $P$ to $C_5$. The image is either trivial or all of $C_5$.

If the image is trivial, then $P\subseteq F$. Since $F$ is elementary abelian of order $3^4$, $P$ is an $\mathbb F_3$-subspace, so $|P|$ is a power of $3$.

Now suppose the image is all of $C_5$. For $i\in\mathbb Z/5\mathbb Z$, set

$$P_i=\{v\in F:(v,i)\in P\}.$$

Each $P_i$ is nonempty. Closure under $(x,y)\mapsto xy^{-1}x$ gives, for $u,v\in P_i$,

$$2u-v\in P_i,$$

so $P_i$ is an affine $\mathbb F_3$-subspace, say $P_i=c_i+W_i$.

For $u\in P_i$, $v\in P_j$,

$$(u,i)(v,j)^{-1}(u,i)=((1+\alpha^{i-j})u-\alpha^{i-j}v,\,2i-j).$$

Taking differences gives

$$\alpha^{i-j}W_j\subseteq W_{2i-j}.$$

Hence, for all $i$,

$$W_i=\alpha^{i/2}W_0,$$

where $1/2$ is taken modulo $5$. Therefore all fibers have the same dimension $r=\dim_{\mathbb F_3}W_0$, and

$$|P|=5\cdot 3^r.$$

The other difference relation gives

$$(1+\alpha^{i-j})W_i\subseteq W_{2i-j}.$$

Taking $i=2,j=0$, we get

$$(\alpha+\alpha^{-1})W_0\subseteq W_0.$$

Let $\beta=\alpha+\alpha^{-1}$. Since $\alpha^5=1$ and $\alpha\neq1$,

$$\beta^2+\beta+2=0$$

over $\mathbb F_3$, and this polynomial is irreducible over $\mathbb F_3$. Thus $W_0$ is a vector space over $\mathbb F_9=\mathbb F_3(\beta)$, so $r$ is even.

Hence any twisted subgroup of $G$ has size either a power of $3$, or $5\cdot 3^r$ with $r\in\{0,2,4\}$. In particular, no twisted subgroup has order

$$15=5\cdot 3.$$

But $15\mid 405$. Therefore even the repaired conjecture is false.

Citation: Conjecture and definitions: Dagur Tómas Ásgeirsson and Pat Devlin, “Palindromes in finite groups and the Explorer-Director game,” arXiv:1904.00467. The counterexample above is self-contained.