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Jensen proved that under $\Diamond$ there is a homogeneous Suslin continuum, so the square of a ccc homogeneous space can fail to be ccc. What about ccc topological groups?

Is there a ccc topological group whose square is not ccc?

The obvious thing to try would be the free topological group over a Suslin continuum, but that doesn't work because it's a $\sigma$-compact group and, by a result of Tkachenko, $\sigma$-compact groups are ccc (and the square of a $\sigma$-compact group is clearly a $\sigma$-compact group).

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    $\begingroup$ In the book of Arkangelski and Tkachenko in Exercise 5.4.G it is written that Todorcevic constructed a ZFC-example of a topological group $G$ with $c(G)<c(G\times G)$. I have no access to Todorcevic' book "Partiotion Problems in Topology" to check if his group can be made countably cellular. $\endgroup$
    – Taras Banakh
    Commented Dec 27, 2017 at 20:24
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    $\begingroup$ Very nice, I'll have a look at that. Of course Todorcevic's ZFC-examples can't be made ccc, because under $MA_{\omega_1}$ the ccc is productive, but I don't know any example of a group where the cellularity increases in the square, so that is very interesting. $\endgroup$
    – Santi Spadaro
    Commented Dec 28, 2017 at 16:13

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Yes. There are such groups after adding a Cohen real (see Theorem $4$ in "Nonpreservation of properties of topological groups on taking their square", Malykhin, 1987) or under RVM (see Theorem $0^c$ in "Some applications of S and L combinatorics", Todorcevic, 1993).

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