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I consider the Dyck congruence $\equiv$ on a parenthesis alphabet $\Sigma = \{a, \bar a, b, \bar b\}$, i.e. the least congruence on $\Sigma^*$ such that $a \bar a \equiv \varepsilon $ and $b \bar b \equiv \varepsilon$.

It appears that $\equiv$ can be cancelled, i.e. for every $v$, $w_1$, $w_2 \in \Sigma^*$, if $vw_1 \equiv vw_2$, then $w_1 \equiv w_2$ (and analogously for cancellation from the right).

Does someone know a literature reference for this property? I wasn't able to find anything along those lines.

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    $\begingroup$ I would check Lyndon and Schupp's Combinatorial Group Theory, but it's really not hard to prove by induction. $\endgroup$
    – Jeffε
    Commented Aug 16, 2017 at 14:13
  • $\begingroup$ You're right. I just wanted to omit the proof. Thank you for the suggestion! $\endgroup$
    – qbert
    Commented Aug 16, 2017 at 14:24
  • $\begingroup$ Ah, crap. I thought you meant the usual semigroup conguence. $\endgroup$
    – Jeffε
    Commented Aug 27, 2017 at 6:38

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Your assertion is wrong, the congruence $\equiv$ is not cancellable from the right: for instance $\bar a a \bar a \equiv \bar a$, but $\bar a a \not\equiv 1$.

By the way the quotient $\Sigma^*/{\equiv}$ is not a group, but a monoid, called the polycyclic monoid, first introduced in [1].

[1] M. Nivat and J-F. Perrot, Une généralisation du monoïde bicyclique, Comptes Rendus de l'Academie des Sciences de Paris 271 (1970), 824-827.

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  • $\begingroup$ I did not expect that. Thank you for the correction, and for the reference! $\endgroup$
    – qbert
    Commented Aug 25, 2017 at 13:59

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