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Let’s define a group quasiword as an element of $F_\infty \times P(F_\infty)$. Suppose $Q \subset F_\infty \times P(F_\infty)$ is a set of quasiwords. Define a prevariety described by $Q$ as a class of all groups $G$, such that $\forall (w, A) \in Q, h \in Hom(F_\infty, G), (h(A) = \{e\} \to h(w) = e)$. One can easily see, that all group varieties actually are prevarieties.

Now, for a group $G$ let’s define $Pv(G)$ as the minimal prevariety, that contains $G$ (it always exists according to the Zorn lemma). Note, that it is always true, that $Pv(G) \subset Var(G)$, however the converse is generally false

My question is:

Do there exist two non-isomorphic finite groups $G$ and $H$ such that $|G| = |H|$ and $Pv(G) = Pv(H)$?

Note, that $Pv(G) = Pv(H)$ implies $Var(G) = Var(H)$. However, the converse is not always true. For example, it is known that $Q_{8n} := \langle x, y | x^{4n} = y^4 = e, x^{2n} = y^2, y^{-1}xy = x^{-1} \rangle$ and $D_{4n} := \langle a \rangle_{4n} \rtimes \langle b \rangle_2$ generate the same varieties. However, they do not fit our condition because they are distinguished by the quasiword $([x, y], \{y^2\})$.

Chain Markov
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1 Answers1

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Yes, you can take $G=\mathbb Z_2\times \mathbb Z_2\times \mathbb Z_4$ and let $H=\mathbb Z_4\times\mathbb Z_4$. It is easy to see that these groups have the same finite size, yet these groups are not isomorphic. The fact that each one embeds into the Cartesian square of the other is enough to imply that ${\it Pv}(G)={\it Pv}(H)$.

Keith Kearnes
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