Let $F_2$ be the non-abelian free group on two generators. Let $G\subset F_2\times F_2$ be a subgroup generated by three elements. Is there an algorithm deciding if a given element of $F_2\times F_2$ lies in $G$?
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1$\begingroup$ Why three in particular? $\endgroup$– Derek HoltCommented Apr 26, 2021 at 7:29
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4$\begingroup$ @DerekHolt because the number of generators for Mihailova subgroup is 2+the number of relations in a group with undecidable word problem and groups with one relation all have decidable word problems (and it is not known if groups with two relations have decidable word problems). $\endgroup$– user178109Commented Apr 26, 2021 at 7:49
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3$\begingroup$ Thanks! But you should really provide context like that when asking the question. $\endgroup$– Derek HoltCommented Apr 26, 2021 at 8:11
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3$\begingroup$ This could be true. By Goursat's lemma (en.wikipedia.org/wiki/Goursat%27s_lemma), "the subdirect product of two groups can be described as a fiber product and vice versa". It may be that one can use this to reduce the problem you describe to the word problem in one-relator groups. $\endgroup$– HJRWCommented Apr 26, 2021 at 9:06
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3$\begingroup$ @MattF.: your subgroup is free of rank $3$, freely generated by the above elements. This is because its projection to the first factor is free of rank $3$ (which can be easily checked, as it's a subgroup of the free group $F(a,b)$, generated by $a^3,b^3,b^2a^2$). This also shows that the subgroup has trivial intersection with the second factor. A symetric argument shows the same for the second factor. $\endgroup$– Ashot MinasyanCommented Apr 27, 2021 at 9:34
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