Large counterexamples to "Non-isomorphic finite groups have verbal subgroups of different order"

Question

In this question, it was conjectured that for every pair of non-isomorphic finite groups $G$ and $H$, there exists some word $\omega$ such that $|V_{\omega}(G)|\ne|V_{\omega}(H)|$, i.e. their corresponding verbal subgroups have unequal order. The answer is no: the groups $D_4$ and $Q_8$ yield a counterexample, as explained in the link.

However, these are relatively small groups with relatively few subgroups. It seems reasonable that this conjecture fails for small groups as they don't have quite enough structure for verbal subgroups to tell them apart.

Is this conjecture true for sufficiently large groups? That is, does there exist some $n\in\mathbb{N}$ such that any counterexample pair $(G,H)$ must have order $|G|,|H|\le n$?

Answer 1

As it was pointed in the comments, if $G$ and $H$ are counterexamples and $K$ is an arbitrary finite group, then $G \times K$ and $H \times K$ also form a counterexample. And if $(G_0, H_0)$ and $(G_1, H_1)$ are both counterexample pairs, such that $G_0 \times G_1$ and $H_0 \times H_1$ are non-isomorphic, then $(G_0 \times G_1, H_0 \times H_1)$ is also a counterexample. So there are definitely infinitely many counterexamples (and your conjecture is equivalent to “there are finitely many counterexamples”, as for any $n \in \mathbb{N}$ there are only finitely many groups of order $n$).

Now let’s prove something stronger.

Let’s call counterexample pair $(G, H)$ non-primitive, if it is of the form ($G_0 \times G_1$, $H_0 \times H_1$), where $(G_0, H_0)$ is a counterexample, and primitive otherwise. It turned out, that, despite those restrictions, there are still infinitely many primitive counterexamples.

Suppose $Q_{8n} = \langle x, y | x^{4n} = y^4 = e, x^{2n} = y^2, y^{-1}xy = x^{-1} \rangle$ and $D_{4n} = \langle a \rangle_{4n} \rtimes \langle b \rangle_2$. These groups are counterexamples for any $n \in \mathbb{N}$. And they are also all primitive because there is no nontrivial direct decomposition of $Q_{8n}$.

One can see, that $Var(Q_{8n}) = Var(D_{4n})$ as $Q_{8n}$ is isomorphic to a subgroup of $\frac{(\langle a \rangle_{4n} \rtimes \langle b \rangle_2) \times (\langle c \rangle_{4n} \rtimes \langle d \rangle_2)}{\langle a^{2n}c^{2n} \rangle}$ (a homomorphic image of $D_{4n} \times D_{4n}$), generated by $a$ and $c^nb$, and $D_{4n}$ is isomorphic to a subgroup of $$\langle x, y, z, t| x^{4n} = y^4 = z^{4n} = t^4 = e, x^{2n} = y^2 = z^{2n} = t^2, y^{-1}xy = x^{-1}, t^{-1}zt = z^{-1}, [z, x] = [z, y] = [t, x] = [t, y] = e \rangle$$ (a homomorphic image of $Q_{8n} \times Q_{8n}$) generated by $x$ and $z^ny$.

One can also see, that both $D_{4n}$ and $Q_{8n}$ have the unique minimal nontrivial normal subgroup. In case of $Q_{8n}$ it is $\langle y^2 \rangle$ and in case of $D_{4n}$ it is $\langle a^2 \rangle$. And it is also quite obvious, that $\langle y^2 \rangle \cong \langle a^2 \rangle \cong C_2$ and that $\frac{Q_{8n}}{\langle y^2 \rangle} \cong \frac{D_{4n}}{\langle a^2 \rangle} \cong C_2 \times C_{2n}$.

Now suppose $A$ is some set of group words. If they are all identities in $D_{4n}$, then they are also identities in $Q_{8n}$, as $Var(D_{4n}) = Var(Q_{8n})$, which results in $|V_A(D_{4n})| = |V_A(Q_{8n})| = 1$. Now suppose, that some of them are not identities. Then $|V_A(D_{4_n})| > 1$ and $|V_A(Q_{8_n})| > 1$, which results in $\langle y^2 \rangle \leq V_A(Q_{8n})$ and $\langle a^2 \rangle \leq V_A(D_{4n})$. Now, as a homomorphic image of a verbal subgroup of a group is always the verbal subgroup of the homomorphic image of the group in respect to the same set of group words, we can conclude, that $$|V_A(D_{4n})| = |\langle a^2 \rangle||V_A(\frac{D_{4n}}{\langle a^2 \rangle})| = 2|V_A(C_2 \times C_{2n})| = |\langle y^2 \rangle||V_A(\frac{Q_{8n}}{\langle y^2 \rangle})| = |V_A(D_{8n})|$$